LeanIMO.csv

Problem ID,Problem,Solution,Grading guidelines,Category,Level,Short Answer,Source,Lean Statement
PB-Basic-001,"Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all $x, y \in \mathbb{Z}$, we have \[ f(2x)+2f(y)=f(f(x+y)).\]","By taking $x = 0$, we get $f(0) + 2f(y) = f(f(y))$ for all $y\in \mathbb{Z}$. So, the problem becomes:
 $$ f(2x) + 2f(y) = f(0) + 2f(x + y) $$
 Taking $x = 1$ for this one, we quickly obtain that $2(f(y+1) - f(y)) = f(2) - f(0) \Leftrightarrow f(y+1)-f(y) = (f(2)-f(0))/2$, so $f$ is linear, that is there exists $k,c$ such that $f(x) = kx+c$. Plug this back into the original equation, we see $f(x) = 2x+c$ and $f(x)=0$ are all possible solutions, and these indeed satisfy the given condition","(Partial)
 1. Guessed the solution correctly
 2. Proved $f(f(y)) = 2f(y) + f(0)$ and $f(2x) + 2f(0) = f(f(x))$
(Almost)
 1. Solved correctly, but did not check that the finding candidates are indeed the solutions to the given equation.",Algebra,IMO-easy,$f(x) = 0$ and $f(x) = 2x + c$ for all integer $x$ and some constant $c$.,"(Modified) IMO 2019, P1","import Mathlib

/-
Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all $x, y \in \mathbb{Z}$, we have \[ f(2x)+2f(y)=f(f(x+y)).\]

Solution: $f(x) = 0$ and $f(x) = 2x + c$ for all integer $x$ and some constant $c$.
-/
theorem PBBasic001 :
    {f : ℤ → ℤ | ∀ x y, f (2 * x) + 2 * f y = f (f (x + y))}
      = {0} ∪ {(fun x ↦ 2 * x + c)| (c : ℤ)} := by sorry"
PB-Basic-002,"Show that $x^2 + y^2 + z^2 + t^2 \ge xyzt$ for any positive real numbers $x, y, z, t$ that satisfy $2(x + y + z + t) \ge xyzt$.","Assume the contrary that $xyzt> x^2+y^2+z^2+t^2$. Using AM-GM, we have $xyzt> x^2+y^2+z^2+t^2 \geq 4\sqrt{xyzt}$, which means that $xyzt > 16$. From here, this implies $2(x+y+z+t) >16$ from the condition $2(x+y+z+t) \geq xyzt$.
 However, from Cauchy-Scwarz, we have $x^2+y^2+z^2+t^2 \ge\frac{1}{4}(x+y+z+t)^2>2(x+y+z+t) \geq xyzt$. This is a contradiction.

 Therefore: $x^2+y^2+z^2+t^2\ge xyzt$.","(Partial)
 1. Attempted the ""proof by contradiction"" and made a positive progress based on it.
 2. Applied the AM-GM inequality correctly.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Algebra,pre-IMO, ,"(Modified)  All-Russian MO 2013, Grade 11, P6","import Mathlib

/-
Show that $x^2 + y^2 + z^2 + t^2 \ge xyzt$ for any positive real numbers $x, y, z, t$ that satisfy $2(x + y + z + t) \ge xyzt$.
-/
theorem PBBasic002
    (x y z t : ℝ)
    (hx : 0 < x) (hy : 0 < y) (hz : 0 < z) (ht : 0 < t)
    (H : x * y * z * t ≤ 2 * (x + y + z + t)) :
    x * y * z * t ≤ x ^ 2 + y ^ 2 + z ^ 2 + t ^ 2 := by sorry"
PB-Basic-003,"Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $a,b \in \mathbb{R}$,
 \[ (b - a)f(f(a)) = a f(a + f(b)). \]""","Let $P(a,b)$ denote the given assertion.
 $P(a,a) \implies f(a+f(a))=0$
 $P(0,b) \implies f(f(0))=0$
 $P(f(0),b) \implies f(0)f(f(0)+f(b))=(b-f(0))f(0)$

 So $f(0)=0$ or $ f(f(0)+f(b))=(b-f(0))$

 If $ f(f(0)+f(b))=(b-f(0))$, setting $b \rightarrow f(b)+b \implies f(b)=-b+f(0)$ and checking we see that $f(a)=-a+k$ truly satisfies the condition
 Suppose now $f(0)=0$

 $P(a,0) \implies f(f(a))=-f(a)$
 So $P(a,b)$ becomes $af(a+f(b))=(a-b)f(a)$
 So $P(a,f(b)+b)$ gives us $f(a)=0$ or $f(a)=-a$

 Finally $f(a)=0, f(a)=-a+k$ where $k$ is a constant","(Partial)
 1. Proved that either $f(0)=0$ or $f(x)=-x+k$ for some constant $k$
(Almost)
 1. Solved correctly by handling the case $f(0)=0$, but did not check that the candidates are indeed the solutions to given equation.",Algebra,IMO-easy,"$f(x)=0, f(x)=-x+k$ where $k$ is a constant","(Modified) Balkan MO 2023, P1","import Mathlib

/-
Find all functions $f\colon \mathbb{R} \rightarrow \mathbb{R}$ such that for all $a,b \in \mathbb{R}$,
\[ (b - a)f(f(a)) = a f(a + f(b)). \]

Solution: $f(x)=0, f(x)=-x+k$ where $k$ is a constant
-/
theorem PBBasic003 :
    {f : ℝ → ℝ | ∀ a b, (b - a) * f (f a) = a * f (a + f b)}
      = {0} ∪ {(fun x ↦ - x + k) | (k : ℝ)} := by sorry"
PB-Basic-004,"Find all strictly increasing functions $g:\mathbb{R} \to \mathbb{R}$ such that:
 (a) $g$ is surjective
 (b) $g(g(x))=g(x)+20x.$","Note that the function is bijective, so setting $x=0$ gives $g(0)=0$ and since it is strictly increasing, we further derive $g(x)>0$ for $x>0$ and $g(x)<0$ for $x<0$.

 Consider the sequence $u_0 = x, u_{n+1} = g(u_n)$, we have that $u_{n+1} = u_n + 20u_{n-1}$, which gives $u_n = a (-4)^n + b 5^n$ for some real number $a, b$. Moreover, since $g$ is a bijective, there exists a real number $r$ such that $u_{-1} = r$, and so the sequence is well-defined for negative integer $n$ by defining the sequence backward. And since the sequence is increasing, $u_0 > 0$ implies $u_n > 0$ for all integer $n$.

 We will show that $a = 0$, which implies $g(x) = u_1 = 5u_0 = 5x$.

 Indeed, assume that $a > 0$, if $u_0 = x > 0$ then we choose a sufficiently small odd negative integer $n$ such that $(-4/5)^n < -b/a$, this gives $u_n < 0$. However, since $g$ is increasing and $u_0 > 0$, we must have that $u_n > 0$, which is a contradiction. Similarly, if $u_0 = a+b < 0$ then we must have $b < 0$ given $a > 0$, and so for a sufficiently small negative even integer $n$, we have $(-4/5)^n > -b/a$, which gives $u_n > 0$, which is a contradiction. Thus, we must have $a \leq 0$.

 Similarly, we can show that $a < 0$ will lead to a contradiction, and so, we must have $a \geq 0$.

 Together, this implies that $a = 0$. Hence proved.","(Partial)
 1. Proved $g(0)=0$
 2. Constructed the sequence $u_n =g^{(n)}(x)$ and the recurrence relation.
(Almost)
 1. Solved correctly, but did not check that the candidates are indeed the solutions to given equation.",Algebra,IMO-easy,$g(x) = 5x$ for all x,"(Modified) VMO 2012, Day2 P3","import Mathlib

/-
Find all strictly increasing functions $g:\mathbb{R} \to \mathbb{R}$ such that:
(a) $g$ is surjective
(b) $g(g(x))=g(x)+20x.$

Solution: $g(x) = 5x$ for all x
-/
theorem PBBasic004 :
    {g : ℝ → ℝ | StrictMono g ∧ g.Surjective ∧ ∀ x, g (g x) = g x + 20 * x}
      = {(fun x ↦ 5 * x)} := by sorry"
PB-Basic-005,"Let $P$ be a polynomial with real coefficients whose leading coefficient is $1$. Suppose that for all nonzero real numbers $x$, we have $P(x) + P(1/x) = \frac{P(x + 1/x) + P(x - 1/x)}{2}$. Determine all possibilities for $P$.","Substitute $\frac{1}{x}$ in the place of $x$. Comparing with the original equation, one can see that $P(x-\frac{1}{x})=P(-x+\frac{1}{x})$ . This implies that $P$ is even.

 Suppose $\deg P\geq 6$. Let $P(x)= x^n + a x^{n-2} + b x^{n-4} +Q(x)$, where $\deg Q < n-4$. Comparing the coefficient of $x^{n-4}$ in the given condition, we have that $b = \frac{2\binom{n}{2}+2b}{2}$, which is a contradiction. Therefore, $\deg P \leq 4$.

 If $\deg P = 4$, let $P(x) = x^4 + ax^2 +b$. Comparing the constant term, we can conclude that $b=6$. The possible solutions are $P(x) = x^4 +a x^2 +6$ for all real numbers $a$.

 If $\deg P = 2$, let $P(x) = x^2 +a$. Comparing the constant term, we can conclude that $a=0$. The only possible solution is $P(x)=x^2$.

 Therefore, the possible solutions are $P(x)= x^4 +ax^2 +6$, $P(x)=x^2$ which satisfies the condition.","(Partial)
 1. Proved that $P$ is even.
 2. Correctly identified all solutions.
(Almost)
 1. Proved $\deg(P)\le 4$, but did not conclude the proof.",Algebra,IMO-easy,"$P(x)= x^4 +ax^2 +6$, $P(x)=x^2$",(Modified) ELMO 2023 Shortlist A1,"import Mathlib

/-
Let $P$ be a polynomial with real coefficients whose leading coefficient is $1$. Suppose that for all nonzero real numbers $x$, we have $P(x) + P(1/x) = \frac{P(x + 1/x) + P(x - 1/x)}{2}$. Determine all possibilities for $P$.

Solution: $P(x)= x^4 +ax^2 +6$, $P(x)=x^2$
-/
open Polynomial

theorem PBBasic005 :
    {P : ℝ[X] | P.Monic ∧ ∀ x ≠ 0, P.eval x + P.eval (1 / x)
      = 1 / 2 * (P.eval (x + 1 / x) + P.eval (x - 1 / x))}
    = {X ^ 4 + a • X ^ 2 + 6 | (a : ℝ) } ∪ {X ^ 2} := by sorry"
PB-Basic-006,"Consider any infinite sequence of integers $c_0, c_1, c_2, \ldots $ such that $c_0 \neq 0$. Prove that for some integer $k \geq 0$, the polynomial $P(x) = \sum_{i = 0}^k c_i x^i$ has fewer than $k$ distinct real roots.","If there exists an integer $k>0$ such that $c_k = 0$, $P(x)=\sum_{i=0}^k c_i x^i$ has less than $k$ roots. So it suffices to see the cases where $c_k \ne 0$ for all integers $k>0$. Let's prove the following lemma:
 If a polynomial $\sum_{i=0}^n a_i x^i$ has $n$ different real roots, $(\frac{a_{1}}{a_0})^2 - 2 \frac{a_{2}}{a_0} \geq n\sqrt[n]{\frac{a_n ^2}{a_0^2}}$

 (Proof) The left hand side is the sum of squares of inverses of all roots. The right hand side is $n$ times the $n$th root of the product of square of inverses of roots. By AM-GM, we can see that the inequality holds.

 Going back to the main proof, let's assume that all $P(x)=\sum_{i=0}^k c_i x^i$ has exactly $k$ distinct real roots.
 By the lemma, $\frac{c_{1}^2 - 2c_0 c_{2}}{\sqrt[n]{c_0^{2n-2}}} \geq n\sqrt[n]{c_n^2}$, which shows that $1\leq c_n^2 \leq (\frac{\frac{c_{1}^2 - 2c_0 c_{2}}{\sqrt[n]{c_0^{2n-2}}}}{n})^n$. We can easily see that the righthandside tends to $0$ as $n$ goes to infinity. It's a contradiction.","(Partial)
 1. Derived a useful inequality of between $c_n$ based on Vieta's formula (along with the proof by contradiction).
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Algebra,IMO-medium, ,(Modified) ELMO 2023 Shortlist A3,"import Mathlib

/-
Consider any infinite sequence of integers $c_0, c_1, c_2, \ldots $ such that $c_0 \neq 0$. Prove that for some integer $k \geq 0$, the polynomial $P(x) = \sum_{i = 0}^k c_i x^i$ has fewer than $k$ distinct real roots.
-/
open Polynomial

theorem PBBasic006 (c : ℕ → ℤ) (hc : c 0 ≠ 0) :
    ∃ k, ((∑ i ∈ Finset.Icc 0 k, monomial i (c i)).rootSet ℝ).ncard < k := by sorry"
PB-Basic-007,"Determine all positive integers $n$ and integer sequences $a_0, a_1,\ldots, a_n$ such that $a_n = 3$ and
 \[f(a_{i-1}) = a_i\]
 for all $i = 1,2,\ldots,n$, where $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$.","$n=2$ with $\left(a_{0}, a_{1}, a_{2}\right)=(-1,1,3)$.

 We provide a solution when we relax the condition $a_n=3$. When we add back the condition, the only solution is $n=2$ with $\left(a_{0}, a_{1}, a_{2}\right)=(-1,1,3)$.

 When relaxing the condition $a_n=3$, the short answers are $n=1$ with $\left(a_{0}, a_{1}\right)=(2,-2) ; n=2$ with $\left(a_{0}, a_{1}, a_{2}\right)=(-1,1,3)$; or $n$ even with $a_{0}=\cdots=a_{n}=-1$.

 It is clear that $a_{0} \neq 0$ as otherwise $a_{n}=0$. For any $k=1, \ldots, n+1$, let $I_{k}$ be the convex hull of $0, a_{0}, \ldots, a_{k-1}$. We will define $a_{-1}=0$ for convenience.

 We first show that $a_{k}$ is not in the interior of $I_{k}$. Otherwise, let $i, j \in\{-1,0, \ldots, k-$ $1\}$ be such that $a_{i}, a_{j}$ are the endpoints of $I_{k}$. Then as $a_{i}-a_{j} \mid f\left(a_{i}\right)-f\left(a_{j}\right)=$ $a_{i+1}-a_{j+1}$, we know that $a_{i+1}$ and $a_{j+1}$ are also the endpoints of $I_{k}$, and by continuing this argument we eventually get that $a_{k}$ is an endpoint of $I_{k}$, which is a contradiction. A consequence is that the endpoints of $I_{k}$ are $a_{k-2}$ and $a_{k-1}$ unless $a_{k-2}=a_{k-1}$. Now if $a_{i}=0$ for some $i>0$, then it is clear that the nonzero terms of $a_{0}, \ldots, a_{n}$ all have the same sign. Then $f\left(a_{i-1}\right)=0$ gives a contradiction if we take $i$ to be the smallest index with $a_{i}=0$.

 We first assume that $a_{n} \neq a_{0}$. If $\left|a_{i}\right|<2$ for any $i<n$, then we have several possibilities: either $n \leqslant 2$; or $a_{0}, \ldots, a_{n}=-1,1, \ldots,-1,1$, or $a_{0}, \ldots, a_{n}=1,-1, \ldots, 1,-1$, or $a_{0}, \ldots, a_{n}=-1,1,1, \ldots, 1$, or $a_{0}, \ldots, a_{n}=1,-1, \ldots,-1$, for $n>2$. It is easy to verify that the last four cases cannot occur by evaluating $f(1)$. Therefore we have $n=2, a_{0}=-a_{1} \in\{1,-1\}$, or $n=1, a_{0}= \pm 1$. In the first case, we have $a_{2}-a_{1}=f\left(a_{1}\right)-f\left(a_{0}\right)=2 a_{1}^{2}=2$ and so $a_{2}=a_{1}+2$. Therefore $a_{1}=f\left(a_{0}\right)=$ $\left(a_{0}+a_{1}+2\right)+a_{0} a_{1}=2-1=1$, and we get the solution $a_{0}, a_{1}, a_{2}=-1,1,3$. In the second case, we have $a_{1}=f\left(a_{0}\right)= \pm a_{1} \pm 1$, and we get no solution by parity.

 Now if $a_{n} \neq a_{0}$ and $a_{n}$ has the largest absolute value, pick $i<n$ such that $\left|a_{i}\right| \geqslant 2$. Then

 \[
 \left|f\left(a_{i}\right)\right| \geqslant\left|a_{n}\right|\left(\left|a_{i}\right|^{n}-\left|a_{i}\right|^{n-1}-\cdots-1\right) \geqslant\left|a_{n}\right|,
 \]

 and $\left|f\left(a_{i}\right)\right|=\left|a_{i+1}\right|\left|a_{n}\right|$. Therefore all equalities should hold, and we get $a_{0}=\cdots=$ $a_{n-1}=-a_{n}$ and $\left|a_{i}\right|=2$. Note that if $n \geqslant 2$, then $a_{0}=a_{1}$, which shows that

 $a_{0}=\cdots=a_{n}$, which is a contradiction. Therefore $n=1$ and we get $\left(a_{0}, a_{1}\right)=(2,-2)$ or $(-2,2)$. The latter is not a solution, so we get $a_{0}=2$ and $a_{1}=-2$.

 So now in the case that $a_{n} \neq a_{0}$, we may assume that $a_{n}$ does not have the largest absolute value. Let $a_{k}$ be the least index such that $a_{k}=a_{k+1}=\cdots=a_{n}$. Then we must have $\left|a_{k-1}\right|>\left|a_{k}\right|$ as $a_{k-1}$ and $a_{k}$ are the endpoints of $I_{k-1}$. We know that $a_{0} \mid a_{k-1}$ and $a_{0} \mid a_{k}$, and so $\left|a_{k-1}\right| \geqslant\left|a_{k}\right|+\left|a_{0}\right|$. As $a_{k-1} \mid f\left(a_{k-1}\right)-f(0)=a_{k}-a_{0}$, we have $\left|a_{k-1}\right| \leqslant\left|a_{k}-a_{0}\right|$ since $a_{k} \neq a_{0}$. This shows that $\left|a_{k}-a_{0}\right|=\left|a_{k}\right|+\left|a_{0}\right|=\left|a_{k-1}\right|$. As a consequence, we have $\left|a_{i}\right| \leqslant\left|a_{k}\right|+\left|a_{0}\right|$ for any $i$. Now we have

 \[
 \left|a_{k}\right|=\left|f\left(a_{k-1}\right)\right| \geqslant\left|a_{k}\right|\left(\left|a_{k-1}\right|^{n}-\left|a_{k-1}\right|^{n-1}-\cdots-1\right)-\left|a_{0}\right|\left(\left|a_{k-1}\right|^{n-1}+\cdots+1\right) .
 \]

 Hence

 \[
 \frac{\left|a_{k}\right|}{\left|a_{0}\right|} \leqslant \frac{\left|a_{k-1}\right|^{n-1}+\cdots+1}{\left|a_{k-1}\right|^{n}-\left|a_{k-1}\right|^{n-1}-\cdots-1} .
 \]

 Note that $\left|a_{k}\right| /\left|a_{0}\right|$ is a positive integer. If it is greater than 1 , then as $\left|a_{k-1}\right|=$ $\left|a_{k}\right|+\left|a_{0}\right| \geqslant 3$, we have

 \[
 \frac{\left|a_{k-1}\right|^{n-1}+\cdots+1}{\left|a_{k-1}\right|^{n}-\left|a_{k-1}\right|^{n-1}-\cdots-1} \leqslant \frac{3^{n-1}+\cdots+1}{3^{n}-3^{n-1}-\cdots-1}<1
 \]

 which is a contradiction. Therefore $\left|a_{k}\right|=\left|a_{0}\right|$, and so $a_{k}=-a_{0}$. This also shows that $a_{k-1}=2 a_{0}$, and by the inequality we see that $\left|a_{0}\right|=1$. With these constraints, we know that $a_{0}, \ldots, a_{n} \in\left\{a_{0},-a_{0}, 2 a_{0}\right\}$. By enumerating, we only have the possibilities $a_{0}, \ldots, a_{n}= \pm 1, \pm 2, \mp 1, \pm 2, \mp 1, \ldots, \pm 2, \mp 1$, or $a_{0}, \ldots, a_{n}= \pm 1, \mp 1, \pm 2, \mp 1, \ldots, \pm 2, \mp 1$, or $a_{0}, \ldots, a_{n}= \pm 1, \pm 2, \mp 1, \ldots, \mp 1$. For the first case, we have $n=2 t$ adn $2 a_{0}=a_{1}=$ $f\left(a_{0}\right)=-(t-1) a_{0}+2 t$, showing that $t=1, a_{0}=1$, and by plugging in $a_{1}$ we get a contradiction. For the second case, we have $n=2 t+1$ and $-a_{0}=a_{1}=f\left(a_{0}\right)=$ $(2 t+1) a_{0}-(t+1)$, which has no solutions. For the third case, if $a_{0}=1$ then we get $n=2$ and $a_{0}, a_{1}, a_{2}=1,2,-1$ by the equation $a_{1}=f\left(a_{0}\right)$, which is not a solution. Thus $a_{0}=-1$, and by plugging in $a_{0}$ we also get a contradiction.

 The remaining case is $a_{n}=a_{0}$. If $a_{n}=a_{n-1}$, then we must have $a_{0}=\cdots=a_{n}$. By plugging in $a_{0}$ we have $a_{0}^{n+1}+\cdots+a_{0}^{2}=0$, and so $a_{0}=-1$ and $n$ is even. Now assume that $a_{n} \neq a_{n-1}$. Then $a_{n}, a_{n-1}$ are the endpoints of $I_{n+1}$. Note that if $a_{n-2}=a_{n-1}$

 then $a_{n}=a_{n-1}$, which is a contradiction. Therefore $a_{n-2}, a_{n-1}$ are also endpoints of $I_{n+1}$. By induction we may show that $a_{k}, a_{k-1}$ are the endpoints of $I_{k+1}$. As $a_{0} \neq 0$, we must have $a_{n}=a_{n-2}=\cdots=a_{0}$, and so $n$ is even. This shows that $a_{n-1}=\cdots=a_{1}$. Now we have

 \[
 a_{0}^{n+1}+a_{0}^{n-1} a_{1}+\cdots+a_{0}^{3}+a_{0} a_{1}+a_{0}=a_{1}
 \]

 and

 \[
 a_{0} a_{1}^{n}+a_{1}^{n}+\cdots+a_{0} a_{1}^{2}+a_{1}^{2}+a_{0}=a_{0}
 \]

 The latter can be rewritten as $\left(a_{0}+1\right)\left(a_{1}^{n-2}+a_{1}^{n-4}+\cdots+1\right)=0$. Therefore $a_{0}=-1$ or $a_{1}=-1$. If $a_{0}=-1$, then $-\left(a_{1}+1\right) n / 2-1=a_{1}$, or equivalently, $\left(a_{1}+1\right)(n+2)=0$. This shows that $a_{1}=-1$, which was already obtained above. If $a_{1}=-1$, then the first equation gives $a_{0}^{n+1}=-1$ and so $a_{0}=-1$ too.

 In conclusion, all the solutions are: $n=1$ with $\left(a_{0}, a_{1}\right)=(2,-2) ; n=2$ with $\left(a_{0}, a_{1}, a_{2}\right)=(-1,1,3)$; or $n$ even with $a_{0}=\cdots=a_{n}=-1$.","(Incorrect)
 1. We do not give partial credits for handling the cases when $n$ is small.
(Partial)
 1. Observed that $a_k$ should be greater or less than all $a_0, \ldots, a_{k-1}$.
 2. Dealt with the case $a_n = a_0$.
 3. Any other useful observation regarding behavior of the sequnece $(a_i)$ which is useful in the possible solution.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Algebra,IMO-medium,"$n=2$ with $\left(a_{0}, a_{1}, a_{2}\right)=(-1,1,3)$","(Modified)  Taiwan TST 2024, 3J Day2 P6","import Mathlib

/-
Determine all positive integers $n$ and integer sequences $a_0, a_1,\ldots, a_n$ such that $a_n = 3$ and
\[f(a_{i-1}) = a_i\]
for all $i = 1,2,\ldots,n$, where $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$.

Solution: $n=2$ with $\left(a_{0}, a_{1}, a_{2}\right)=(-1,1,3)$
-/
open Polynomial

theorem PBBasic007 :
    { f : ℤ[X] | f.leadingCoeff = 3 ∧ 1 ≤ f.degree ∧
      ∀ (i : ℕ), i < f.degree → f.coeff (i + 1) = f.eval (f.coeff i) } =
    {3 * X ^ 2 + 1 * X - 1} := by sorry"
PB-Basic-008,"Let $a,b,c$ be positive reals such that $a+b+c = 1$, prove that $\sqrt{a}+\sqrt{b}+\sqrt{c} \geq 3\sqrt{3}(ab+bc+ca)$.","Renormalizing, we may suppose $a + b + c = 3$ and show $\sqrt{a}+\sqrt{b}+\sqrt{c} \geq ab+bc+ca$. Note that $\sqrt{a}+\sqrt{b}+\sqrt{c} \geq ab+bc+ca$ is equivalent to $2\sqrt{a}+2\sqrt{b}+2\sqrt{c} \geq 2ab+2bc+2ca \Leftrightarrow a^2+2\sqrt{a}+b^2+2\sqrt{b}+c^2+2\sqrt{c} \geq a^2+2ab+b^2+2bc+c^2+2ca = (a+b+c)^2$. The inequality is now equivalent to the inequality $a^2+2\sqrt{a}+b^2+2\sqrt{b}+c^2+2\sqrt{c} \geq 9$. Observe that from AM-GM, we have $a^2+\sqrt{a}+\sqrt{a} \geq 3a$. Similarly, we have $b^2+\sqrt{b}+\sqrt{b} \geq 3b$ and $c^2+\sqrt{c}+\sqrt{c} \geq 3c$. Adding these together, we have $a^2+2\sqrt{a}+b^2+2\sqrt{b}+c^2+2\sqrt{c} \geq 3(a+b+c) = 9$. Hence proved.","(Partial)
 1. Reduced the inequality to a more managable form so that one can complete the proof by applying one of the well-knnown inequalities (e.g., Cauchy-Schwarz, AM-GM, Muirhead, or Schur)
(Almost)
 1. Made a minor mistake when applying an inequality, or skipped rigorous justification for such a step.",Algebra,pre-IMO, ,(Modified) All-Russia MO 2002.,"import Mathlib

/-
Let $a,b,c$ be positive reals such that $a+b+c = 1$, prove that $\sqrt{a}+\sqrt{b}+\sqrt{c} \geq 3\sqrt{3}(ab+bc+ca)$.
-/
theorem PBBasic008
    (a b c : ℝ)
    (ha : 0 < a) (hb : 0 < b) (hc : 0 < c)
    (H : a + b + c = 1) :
    3 * √3 * (a * b + b * c + c * a) ≤ √a + √b + √c := by sorry"
PB-Basic-009,"Let $a_1,a_2,...,a_{18}$ be 18 real numbers, not necessarily distinct, with average $m$. Let $A$ denote the number of triples $1 \le i < j < k \le 18$ for which $a_i + a_j + a_k \ge 3m$. What is the minimum possible value of $A$?","I claim the answer is $136$. Take the construction $17, -1,...,-1,-1,-1,-1,-1,-1,-1,-1$, which clearly works.
 Now I will show that this is the minimum. For any partition of $\{a_1, a_2, ..., a_{18}\}$ into six triplets, at least one triplet must satisfy that $a_i + a_j +a_k \ge 0$.
 Therefore there are $\frac{18!}{(3!)^6 6!}$ partitions, and a fixed triplet is contatined in $\frac{15!}{(3!)^5 5!}$ partitions, so there are at least $\frac{18!}{(3!)^6 6!}/\frac{15!}{(3!)^5 5!} = 136$ partitions satisfying the condition.","(Partial)
 1. Guessed the answer $136$, and constructed the equality case.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,IMO-easy,$136$,(Modified) ELMO 2013,"import Mathlib

-- Let $a_1,a_2,...,a_{18}$ be 18 real numbers, not necessarily distinct, with average $m$. Let $A$ denote the number of triples $1 \le i < j < k \le 18$ for which $a_i + a_j + a_k \ge 3m$. What is the minimum possible value of $A$?
-- Answer: $136$

-- Let $a_1,a_2,...,a_{18}$ be 18 real numbers, not necessarily distinct
-- the indexing doesn't matter, so we use Lean's default indexing from zero
variable (a : Fin 18 → ℝ)

-- $m$ is the average of $a$
noncomputable
def m : ℝ := (∑ i : Fin 18, a i) / 18

-- Let $A$ denote the number of triples $1 \le i < j < k \le 18$ for which $a_i + a_j + a_k \ge 3m$.
noncomputable
def A : ℕ :=
    { (i, j, k) : Fin 18 × Fin 18 × Fin 18 |
    i < j ∧ j < k ∧ a i + a j + a k ≥ 3 * (m a) }.ncard

-- prove that the minimum is 136
theorem PBBasic009 : iInf A = 136 := by sorry"
PB-Basic-010,"The numbers $\{ 1, 2, 3, \ldots , 2022 \}$ are partitioned into two sets $A$ and $B$ of size $1011$ each. Let $S_{AB}$ denote the set of all pairs $(a, b) \in A \times B$ where $a < b$, and let $S_{BA}$ denote the set of all pairs $(a, b) \in A \times B$ where $b < a$.

 Prove that $\sum_{(a, b) \in S_{AB}} (b - a) \neq \sum_{(a, b) \in S_{BA}} (a - b)$.","Let $X = \sum_{(a, b) \in S_{AB}} (b - a)$, $Y=\sum_{(a, b) \in S_{BA}} (a -b)$ and $Z=X-Y$. We first analyze the case where all elements of $A$ are smaller than all elements of $B$. In that case, $S_{BA} = 0$, and
 \[ X = \sum_{i = 1012}^{2022} \sum_{j = 1}^{1011} (i - j) = 1011 \cdot \left(\sum_{i = 1012}^{2022} i - \sum_{j = 1}^{1011} j \right) = 1011^3, \]
 and so $Z = 1011^3$.

 Now, arrange the numbers $1$ to $2022$ on a line in increasing order from left to right, and consider how the quantity $Z$ changes if we swap the set of two adjacent numbers, where the left number is in $A$ and the right number is in $B$. Let $A_l$ and $B_l$ be the number of numbers from $A$ and $B$ to the left of this adjacent pair, and similarly let $A_r$ and $B_r$ be the number of numbers from $A$ and $B$ to the right of this adjacent pair,

 First think of how $X$ changes. By moving the number from $A$ one unit to the right, we decrease each $A$-$B$ segment of $B_r$ by one unit. By moving the number from $B$ one unit to the left, we shorten each $A$-$B$ segments of $A_l$ by one unit. Also, we eliminated the length-1 $A$-$B$ segment formed by the swapped points themselves. Thus, $X$ decreases by $B_r + A_l + 1$.

 Similarly, $Y$ increases by $A_r + B_l + 1$, after the swap. Thus, $Z$ decreases by $B_r + R_l + B_r + B_l + 2 = 2022$.

 Now, starting from the initial configuration where all the 1011 left-most points are $A$, we can reach any other configuration by a series of swaps of adjacent pairs, where the left number is $A$ and the right number is $B$. In the initial configuration, we have $Z = 1011^3$, and it decreases by $2022$ with each swap. Thus, $Z$ always remains an odd number, so it can never be zero. We conclude that $X \neq Y$.","(Partial)
 1. Found and proved that the parity of the difference is invariant
 2. Found an invariant equivalent to it and proved that it is indeed an invariant.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,IMO-easy, ,"(Modified) Tournament of Towns, Fall 2022 Senior O-level, P3","import Mathlib

/-
The numbers $\{ 1, 2, 3, \ldots , 2022 \}$ are partitioned into two sets $A$ and $B$ of size $1011$ each. Let $S_{AB}$ denote the set of all pairs $(a, b) \in A \times B$ where $a < b$, and let $S_{BA}$ denote the set of all pairs $(a, b) \in A \times B$ where $b < a$.

Prove that $\sum_{(a, b) \in S_{AB}} (b - a) \neq \sum_{(a, b) \in S_{BA}} (a - b)$.""
-/

theorem PBBasic010 (A B : Finset ℕ) (hdisj : Disjoint A B) (hunion : A ∪ B = Finset.Icc 1 2022)
    (ha : A.card = 1011) (hb : B.card = 1011) :
    -- we don't need to do the restriction to S_AB / S_BA, it corresponds to
    -- natural number subtraction in Lean, because if a > b, then b - a = 0
    ∑ a ∈ A, ∑ b ∈ B, (b - a : ℕ) ≠
    ∑ a ∈ A, ∑ b ∈ B, (a - b : ℕ) := by sorry"
PB-Basic-011,"Let $A\subset \{1,2,\ldots,2000\}$, $|A|=1000$, such that $a$ does not divide $b$ for all distinct elements $a,b\in A$. For a set $X$ as above let us denote with $m_{X}$ the smallest element in $X$. Find $\min m_{A}$ (for all $A$ with the above properties).","Partition the set $\{ 1,2,3,\cdots ,2000\}$ into 1000 parts $P_{1}\cup P_{2}\cup P_{3}\cup \cdots \cup P_{1000}$ such that $P_{a}$ contains all numbers of the form $2^{b}(2a-1)$ where $b$ is a nonnegative integer.

 $A$ cannot have two elements from the same part (otherwise one would divide the other by a power of 2). So $A$ must have exactly one element from each part.

 Let $t_{a}$ be the element of $A$ contained in $P_{a}$. Then consider $t_{1},t_{2},t_{5},\cdots $, each being a product of a power of 3 and a power of 2. The highest power of 2 dividing $t_{1}$ must be strictly greater than the highest power of 2 dividing $t_{2}$ (otherwise $t_{1}$ divides $t_{2}$). Similarly, the highest powers of 2 dividing $t_{1},t_{2},t_{5},\cdots $ must be a strictly decreasing sequence. In particular, there are 7 numbers in the sequence, so $t_{1}$ must be at least $2^6=64$. In a similar manner, if $666<3^{b}(2a-1)<2000$, then $t_{a}\geq 2^{b}(2a-1)$. (Here we look at the numbers where the odd factors are $(2a-1)$ times a power of 3.)

 Suppose one of the $t_{a}$ is less than 64. Then we have
 \[666<3^{b}(2a-1)\leq (3/2)^{b}t_{a}<64(3/2)^{b}\]
 so that $b\geq 6$. However, this means either $a=1$ (we know $t_{1}\geq 64$) or $t_{a}\geq 3\cdot 2^{6}$, which is a contradiction. Therefore, every element must be at least 64.


 We could attain a minimum by setting $t_{a}=2^{f(a)}(2a-1)$ where $f(a)$ satisfies $666<3^{f(a)}(2a-1)<2000$ if we can show it works. To do this, we must show that $t_{x}$ cannot divide $t_{y}$ for any $x\neq y$. Suppose there were such $x$ and $y$. Then
 \[2^{u}(2x-1)|2^{v}(2y-1)\]
 where $u=f(x)$ and $v=f(y)$. In particular, $u\leq v$ and $(2x-1)|(2y-1)$.

 In order to have $(2x-1)|(2y-1)$, we must have $2y-1\geq 3(2x-1)$, but this implies
 \[3^{u+1}(2x-1) \ge 2001 >3^{v}(2y-1)\geq 3^{v+1}(2x-1)\]
 and $u>v$, a contradiction! Therefore $\boxed{64}$ is the minimum.","(Partial)
 1. Constructed the example for the answer $64$.
 2. Decomposed the set $\{1,2,...,2000\}$ as in the solution and analyzed intersections of $A$ and these decomposed subsets.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,IMO-easy,$64$,(Modified) Vietnam TST 2007 P5,"import Mathlib

-- Let $A\subset \{1,2,\ldots,2000\}$, $|A|=1000$, such that $a$ does not divide $b$ for all distinct elements $a,b\in A$. For a set $X$ as above let us denote with $m_{X}$ the smallest element in $X$. Find $\min m_{A}$ (for all $A$ with the above properties).
-- Answer: 64

variable (A : Finset ℕ)

def Acond := A ⊆ Finset.Icc 1 2000 ∧ A.card = 1000 ∧ ∀ a ∈ A, ∀ b ∈ A, a ≠ b → ¬ a ∣ b

theorem PBBasic011 : sInf { Finset.min A | (A : Finset ℕ) (_ : Acond A) } = 64 := by sorry"
PB-Basic-012,"Consider a positive integer $n$. We define $f(n)$ as the number of pairs of paths on an $n \times n$ grid that:

 (1) Both paths start at $(0, 0)$ (bottom left corner) and end at $(n, n)$ (top right corner).

 (2) Both paths allow only right or up movements (one unit each).

 (3) The $y$ coordinate of the first path never exceeds the y coordinate of the second path at any timestep.

 For example, when $n = 2$, consider the following pair of paths:

 The first path: $(0,0) \rightarrow (1,0) \rightarrow (1,1) \rightarrow (2,1) \rightarrow (2,2)$
 The second path: $(0,0) \rightarrow (1,0) \rightarrow (2,0) \rightarrow (2,1) \rightarrow(2,2)$
 The example is invalid because after 2 steps, the y coordinate of the first path (1) is larger than the y coordinate of the second path (0).

 However, the following example is valid,

 The first path: $(0,0) \rightarrow (1,0) \rightarrow (2,0) \rightarrow (2,1) \rightarrow (2,2)$
 The second path: $(0,0) \rightarrow (1,0) \rightarrow (1,1) \rightarrow (2,1) \rightarrow (2,2)$

 since the y coordinate of the first path is never larger than the second path. Find $f(10)$.","If we move the second path up-left by one unit, the pair of paths corresponds to a pair of paths from (0,0) to (10,10) and from (-1,1) to (9,11) which are non-intersecting. The total number of path pairs is $\binom{20}{10}^2$. The number of intersecting pairs corresponds to the number of pair of paths from (0,0) to (9, 11) and paths from (-1, 1) to (10,10), because interchanging the two paths at the first intersection point gives a one-to-one correspondence, and thus equals $\binom{20}{9}^2$. Therefore there are $\binom{20}{10}^2 - \binom{20}{9}^2$ paths.","(Partial)
 1. Came up with an idea of correcting shifting (shifting only one path by $+1$ in the $y$-coordinate) or found an equivalent idea with it.
(Almost)
 1. Made a minor mistake when counting.",Combinatorics,IMO-medium,$\binom{20}{10}^2 - \binom{20}{9}^2$ ,Novel Problem,"import Mathlib

/-
Consider a positive integer $n$. We define $f(n)$ as the number of pairs of paths on an $n \times n$ grid that:
 (1) Both paths start at $(0, 0)$ (bottom left corner) and end at $(n, n)$ (top right corner).
 (2) Both paths allow only right or up movements (one unit each).
 (3) The $y$ coordinate of the first path never exceeds the y coordinate of the second path at any timestep.

 For example, when $n = 2$, consider the following pair of paths:

 The first path: $(0,0) \rightarrow (1,0) \rightarrow (1,1) \rightarrow (2,1) \rightarrow (2,2)$
 The second path: $(0,0) \rightarrow (1,0) \rightarrow (2,0) \rightarrow (2,1) \rightarrow(2,2)$
 The example is invalid because after 2 steps, the y coordinate of the first path (1) is larger than the y coordinate of the second path (0).

 However, the following example is valid,

 The first path: $(0,0) \rightarrow (1,0) \rightarrow (2,0) \rightarrow (2,1) \rightarrow (2,2)$
 The second path: $(0,0) \rightarrow (1,0) \rightarrow (1,1) \rightarrow (2,1) \rightarrow (2,2)$

 since the y coordinate of the first path is never larger than the second path. Find $f(10)$.""

Answer: $\binom{20}{10}^2 - \binom{20}{9}^2$
-/

-- a predicate asserting that a path only goes up or right by a single step
structure ProblemPath (n : ℕ) where
  -- points of the path, (0,0), (0,1), (1,1), ...
  points : List (ℕ × ℕ)
  nonempty : points ≠ [] -- auxuliary condition to easier state the rest
  -- starts at (0,0)
  start : points.head nonempty = (0,0)
  -- ends at (n,n)
  finish : points.getLast nonempty = (n,n)
  -- every step goes up or right
  up_right : points.Chain' (fun (x₁,y₁) (x₂,y₂) ↦ (x₂,y₂) = (x₁+1,y₁) ∨ (x₂,y₂) = (x₁,y₁+1))

-- a pair of paths satisfying the condition from the problem statement:
-- The $y$ coordinate of the first path never exceeds the y coordinate of the second path at any timestep.
structure ProblemPath.pair n where
  (path1 path2 : ProblemPath n)
  (cond : ∀ (i : ℕ) (_ : i < path1.points.length) (_ : i < path2.points.length),
    path1.points[i].2 ≤ path2.points[i].2)

noncomputable
def f (n : ℕ) : ℕ := Nat.card (ProblemPath.pair n)

theorem PBBasic012 : f 10 = (Nat.choose 20 10)^2 - (Nat.choose 20 9)^2 := by sorry"
PB-Basic-013,"Each of 8 boxes contains 6 balls. Each ball has been colored with one of 22 colors. If no two balls in the same box are the same color, prove that there are two colors that occur together in more than one box.","Let’s say a color ""connects"" two boxes if it appears in both. We want to show there are two colors that connect the same pair of boxes.

 Let $c_i$ denote the number of balls with color $i$. Then, the total number of times a color connects two boxes is $N = \sum_{i=1}^{22} \binom{c_i}{2}$. We also know the total number of balls is $48$, so $\sum_{i=1}^{22} c_i = 48$. We now try to find the minimum possible value of $N$.

 To do this, we first analyze the quantity $X = \sum_{i=1}^{22} (c_i - 2)^2$. If at $|c_i - 2| >= 2$ for some $i$, then clearly $X \ge 4$. Otherwise, all the $c_i$ are 1, 2, or 3. Then, in order for them to sum to 48, at most 18 of them can be 2, so once again $X \ge 4$. We then have

 $4 \le X = 2N - 3 \sum_{i=1}^{22} c_i + 4 * 22 = 2N - 3 * 48 + 4 * 22 = 2N - 56$,

 so that $N \ge 30$. But note that there are only $\binom{8}{2} = 28$ pairs of boxes. Thus, by the pigeonhole principle, some two colors connect the same pair of boxes, as desired.","(Partial)
 1. Applied some double counting argument but did not progress further.
(Almost)
 1. Solved the problem with some mistakes (such as calculation mistakes).",Combinatorics,pre-IMO, ,(Modified) USAMO 2001 P1,"import Mathlib

-- Each of 8 boxes contains 6 balls. Each ball has been colored with one of 22 colors. If no two balls in the same box are the same color, prove that there are two colors that occur together in more than one box.

theorem PBBasic013 (Box Color : Type)
    -- there are 8 boxes and 22 colors
    (num_boxes : Box ≃ Fin 8) (num_colors : Color ≃ Fin 22)
    -- each box contains 6 distinct balls
    (balls : Box → (Finset Color)) (num_balls : ∀ box : Box, (balls box).card = 6) :
    -- then there exist two colors and two boxes, both containing both of the two colors
    ∃ (color1 color2 : Color) (box1 box2 : Box), color1 ≠ color2 ∧ box1 ≠ box2 ∧
    color1 ∈ balls box1 ∧ color2 ∈ balls box1 ∧ color1 ∈ balls box2 ∧ color2 ∈ balls box2 := by sorry"
PB-Basic-014,"In Shoreline Amphitheatre, 2024 Googlers each hold up to five flags of various colors. Any group of three Googlers will always have at least two people holding flags of the same color. Prove that one specific flag color is held by at least 200 Googlers.","Take any Googler. He holds at most 5 flags. If everyone holds at least one of those flags but not more than 199 people speak have each flag, then there are at most 1 + 5·198 = 991 Googlers. But there are 2024 Googlers, so we must be able to find someone who holds none of the 5 flags. He too has at most 5 flags, giving a total of 10 between the two Googlers selected. Now we are told that every other Googler holds at least one of these 10 flags. So if no more than 199 Googlers holds each flag, then there are at most 2 + 10·198 = 1982 Googlers. But there are 2024 Googlers, so one of the 10 flags must be held by 200 or more people.","(Partial)
 1. Established that there is a pair of people not sharing a flag of the same color.
 (Almost)
 1.  Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,pre-IMO, ,(Modified) Balkan MO 1985 P4,"import Mathlib

-- In Shoreline Amphitheatre, 2024 Googlers each hold up to five flags of various colors. Any group of three Googlers will always have at least two people holding flags of the same color. Prove that one specific flag color is held by at least 200 Googlers.

variable (Googler Color : Type) (flags : Googler → Finset Color)

-- the main condition: Any group of three Googlers will always have at least two people holding flags of the same color
def tripleCond : Prop :=
  ∀ triple : Finset Googler, triple.card = 3 →
    ∃ g1 ∈ triple, ∃ g2 ∈ triple, g1 ≠ g2 ∧ ¬ Disjoint (flags g1) (flags g2)

theorem PBBasic014
    -- there are 2024 Googlers
    (num_googlers : Googler ≃ Fin 2024)
    -- each holding flags of up to 5 colors
    (num_flags : ∀ googler, (flags googler).card ≤ 5)
    (cond : tripleCond Googler Color flags) :
    -- one specific flag color is held by at least 200 Googlers.
    ∃ color : Color, Nat.card {g : Googler | color ∈ flags g } ≥ 200 := by sorry"
PB-Basic-015,"Adrian is lining up $n \geq 2$ toy cars in a row. Each car has a width and length, and no two cars have the same width or length. Initially, the cars are arranged in increasing order of length from left to right. Adrian repeatedly does the following: he chooses some two adjacent cars where the left car is shorter in length and greater in width than the right car, and he swaps them. He keeps doing this until no further moves are possible. Prove that no matter how Adrian chooses his swaps, the procedure will eventually terminate, and in the end, the cars will be sorted in increasing order of width from left to right.","Define an inversion to be a pair of cars where the one on the left is wider than the one on the right. Clearly, each move strictly decreases the number of inversions, so the procedure must terminate.

 Now, suppose for sake of contradiction that we can make no more moves, but the cars are not sorted in increasing order of width. Consider the smallest index $i$ such that car $i$ is wider than car $i + 1$. Since we cannot make any further swaps, it must also be the case that car $i$ is longer than book $i + 1$. This means that originally, car $i$ was to the right of car $i + 1$, and at some point they switched relative positions. However, this is not possible because the rules prevent us from ever swapping two books when one is both taller and wider than the other.","(Partial)
 1. Found an increasing quantity (e.g. the number of inversion according to the width or length)
 2. Tried to use minimality argument by taking the first book not sorted by width.
(Almost)
 1.  Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,pre-IMO, ,(Modified) 2020 USOJMO P1,"import Mathlib

-- Adrian is lining up $n \geq 2$ toy cars in a row. Each car has a width and length, and no two cars have the same width or length. Initially, the cars are arranged in increasing order of length from left to right. Adrian repeatedly does the following: he chooses some two adjacent cars where the left car is shorter in length and greater in width than the right car, and he swaps them. He keeps doing this until no further moves are possible. Prove that no matter how Adrian chooses his swaps, the procedure will eventually terminate, and in the end, the cars will be sorted in increasing order of width from left to right.

structure Car where (width length : Nat)

def Cars (n : ℕ) := Fin n → Car

variable {n : ℕ}

-- 1. Setup: Initial Conditions
structure Cars.IniCond (cars : Cars n) : Prop where
  distinct_width : (Car.width ∘ cars).Injective
  distinct_length : (Car.length ∘ cars).Injective
  increasing_length : Monotone (Car.length ∘ cars)

-- 2. Setup: The Swap Move
structure Cars.Swap (cars : Cars n) where
  (i1 i2 : Fin n)
  adjacent : i1.val + 1 = i2.val
  shorter_length : (cars i1).length < (cars i2).length
  greater_width : (cars i1).width > (cars i2).width

def Cars.Swap.apply {cars : Cars n} (step : Swap cars) : Cars n :=
  cars ∘ Equiv.swap step.i1 step.i2

-- The relation: `next` is reachable from `prev` via one swap
def Cars.swapRel (prev next : Cars n) : Prop :=
  ∃ step : prev.Swap, step.apply = next

-- 3. The Combined Theorem
theorem PBBasic015
  (start : Cars n)
  (ini : start.IniCond) :
  -- Part 1: Termination (No infinite chain of swaps starts from `start`)
  (Acc (fun next prev => Cars.swapRel prev next) start) ∧
  -- Part 2: Correctness (Any reachable state with no moves left is sorted)
  (∀ final, Relation.ReflTransGen Cars.swapRel start final →
    IsEmpty final.Swap → Monotone (Car.width ∘ final)) := by sorry"
PB-Basic-016,"101 stones are arranged in a circle, numbered 1 through 101 in order (so that stone 1 is next to stone 101). Each stone is painted either red, white, or blue. Initially, stone 101 is the only blue stone. Among the remaining stones, even-numbered stones are red, and odd-numbered stones are white.

We perform a sequence of modifications to the coloring, where in each step, we may choose a stone and repaint it a different color, as long as we ensure that no two adjacent stones are ever the same color. Prove that it is not possible to eventually reach a state where again stone 101 is the only blue stone, but among the remaining stones, all even-numbered stones are white and all odd-numbered stones are red.","Let $A$ be the number of indices $i$ where $i$th stone is blue and $i+1$th stone is white (we define the order in mod 101). Similarly, let $C$ be the number of white stones followed by blue stones. In the initial state, we have $A = 1$ and $C = 0$, and in the final state, we have $A = 0$ and $C = 1$.

 We claim that $A - C$ is invariant. Indeed, consider all possible colorings of three consecutive stones, and let us enumerate what color the middle side can be changed to. Writing $R, W, B$ to represent red, white, and blue, we have the following possibilities:
 \[ RBR \to RWR, \qquad RWR \to RBR, \]
 \[ BWB \to BRB, \qquad BRB \to BWB, \]
 \[ WBW \to WRW, \qquad WRW \to WBW. \]
 In each case, we find that the quantity $A - C$ doesn't change. However, the initial state has $A - C = 1$, and the final state has $A - C = -1$, so it is not possible to reach the final state.","(Partial)
 1. Analyzed all the possible cases for when a stone changes color in terms of its neighbors' colors.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,pre-IMO, ,(Modified) USAMO 1994 Problem 2,"import Mathlib

/-
101 stones are arranged in a circle, numbered 1 through 101 in order (so that stone 1 is next to stone 101). Each stone is painted either red, white, or blue. Initially, stone 101 is the only blue stone. Among the remaining stones, even-numbered stones are red, and odd-numbered stones are white.

We perform a sequence of modifications to the coloring, where in each step, we may choose a stone and repaint it a different color, as long as we ensure that no two adjacent stones are ever the same color. Prove that it is not possible to eventually reach a state where again stone 101 is the only blue stone, but among the remaining stones, all even-numbered stones are white and all odd-numbered stones are red.""
-/
open Relation

inductive Color
| red
| white
| blue
deriving DecidableEq

def Coloring := Fin 101 → Color

-- a coloring is valid if no adjacent stones have the same color
def Coloring.Valid (coloring : Coloring) :=
  ∀ i : Fin 101, coloring i ≠ coloring (i+1)

-- the initial coloring, we swap even / odd from the problem statement because we use 0-based indexing
def Coloring.initial : Coloring := fun i ↦
  if i = 100 then .blue else if i % 2 = 1 then .red else .white
-- the final coloring is the same but with white & red swapped
def Coloring.final : Coloring := fun i ↦
  if i = 100 then .blue else if i % 2 = 1 then .white else .red

-- `c2` can be obtained with a single step from `c1`
def Coloring.Step (c1 c2 : Coloring) : Prop :=
  ∃ i : Fin 101, c2.Valid ∧ ∀ j, j ≠ i ↔ c1 j = c2 j

-- The main problem: the final coloring is not reachable, that is, the edge (initial, final)
-- is not in the transitive closure.
theorem PBBasic016 : ¬ ReflTransGen Coloring.Step .initial .final := by sorry"
PB-Basic-017,Find all non-negative integer $n$ such that $A_n = 1 + 3^{20(n^2+n+1)} + 9^{14(n^2+n+1)}$ is a prime number.,"Substitute $a = 3^{4(n^2+n+1)} \geq 81 \in Z^+$, we have $A_n = a^7 + a^5 + 1$. Observe that $a^7 + a^5 + 1 = (a^2+a+1)(a^5-a^4+a^3-a+1)$. Since $a \geq 81$, we have $a^2+a+1 > 1$ and $a^5-a^4+a^3-a+1 > 1$. Hence, $A_n$ is not a prime number for all $n \geq 0$.","(Partial)
 1. Applied correct substitution to $x^7+x^5+1$ or $x^{14}+x^{10}+1$ or $x^{28}+x^{20}+1$
(Almost)
 1. Correctly applied substitution and factorization but with made minor mistakes.",Number theory,pre-IMO,There is no such $n$.,folklore,"import Mathlib

/-
Find all non-negative integer $n$ such that $A_n = 1 + 3^{20(n^2+n+1)} + 9^{14(n^2+n+1)}$ is a prime number.

Solution: There is no such $n$.
-/
theorem PBBasic017
    (A : ℕ → ℕ)
    (hA : ∀ n, A n = 1 + 3 ^ (20 * (n ^ 2 + n + 1))
      + 9 ^ (14 * (n ^ 2 + n + 1))) :
    {n | (A n).Prime} = {} := by sorry"
PB-Basic-018,Let $x$ and $y$ be positive integers satisfying $2x^2 + x = 3y^2 + y$. Prove that $2x+2y+1$ is a perfect square.,"Lemma: If $ab$ is a perfect square for coprime integers $a,b$ then they are perfect squares.

 Observe that $2x^2+x=3y^2+y$ is equivalent to $y^2 = 2x^2-2y^2+x-y = (x-y)(2x+2y+1)$. Let $d$ be the greatest common divisior of $x-y,2x+2y+1$, we have $d | x-y$ and $d | 2x+2y+1$. On the other hand, observe that $y^2 = (x-y)(2x+2y+1)$. Hence, we must have that $d^2 | y^2$, which implies $d | y$. Since $d | y$ and $d | x-y$, we have $d | x$. Furthermore, since $d | x, d | y, d | 2x+2y+1$, we have that $d | 1$ and thus, $d = 1$, meaning that $x-y,2x+2y+1$ are coprime integers. From the Lemma, we have that $x-y,2x+2y+1$ are perfect squares. Hence proved.","(Partial)
 1. Correctly factorized $(x-y)(2x+2y+1) = y^2$ but did not get other non-trivial results.
(Almost)
 1. Proved the problem but made minor mistakes or did not prove $d = 1$",Number theory,pre-IMO, ,(Modified) Polish MO 1965 P4,"import Mathlib

/-
Let $x$ and $y$ be positive integers satisfying $2x^2 + x = 3y^2 + y$. Prove that $2x+2y+1$ is a perfect square.
-/
theorem PBBasic018
    (x y : ℕ) (hx : x ≠ 0) (hy : y ≠ 0)
    (H : 2 * x ^ 2 + x = 3 * y ^ 2 + y) :
    IsSquare (2 * x + 2 * y + 1) := by sorry"
PB-Basic-019,"For any positive integer $n$, let $\delta(n)$ be the largest odd divisor of $n$. Prove for any $N>0$ that we have
 \[
 \left| \sum_{n=1}^N \frac{\delta(n)}{n}- \frac{2}{3}N \right| <1.
 \]","Observe the following: If the greatest power of $2$ that divides $n$ is $2^k$, the greatest odd divisor dividing $n$ is $\frac{n}{2^k}$.

 Let $\sum_{n=1}^N \frac{\delta(n)}{n}$ be $S(N)$.

 Let the greatest power of $2$ smaller or equal to $N$ be $2^k$. Then, $2^k \leq N <2^{k+1}$ holds. Using the observation, we get the following equation:
 $\sum_{n=1}^{2^k} \frac{\delta(n)}{n} = \sum_{i=0}^{k-1} \frac{2^{k-1-i}}{2^i} + \frac{1}{2^k} = 2^{k-1}(\frac{1-\frac{1}{4^{k}}}{\frac{3}{4}}) + \frac{1}{2^k}$

 Now observe that for $2^k < n <2^{k+1}$, $\frac{\delta(n)}{n} = \frac{\delta(n-2^k)}{n-2^k}$. Therefore, we have that $S(N)= S(N-2^k)+ (\frac{2^{k+1}}{3})(1-\frac{1}{4^{k}}) + \frac{1}{2^k}$

 Using this, we have that $S(N) - \frac{2N}{3} = S(N-2^k) - \frac{2(N-2^k)}{3} + \frac{1}{3\cdot 2^k}$

 Therefore, $|S(N) - \frac{2N}{3}|\leq \sum_{k=0}^\infty \frac{1}{3\cdot 2^k} \leq 1$","(Partial)
 1. Expressed $\sum \delta(n)/n$ as negative powers of 2's and attempted to rearrange the summation.
(Almost)
 1. Successfully rearranged the summation, and counted $2^{-k}$ correctly.",Number theory,IMO-easy, ,(translated) 2003 Korea Winter Program Practice Test,"import Mathlib

/-
For any positive integer $n$, let $\delta(n)$ be the largest odd divisor of $n$. Prove for any $N>0$ that we have
\[
\left| \sum_{n=1}^N \frac{\delta(n)}{n}- \frac{2}{3}N \right| <1.
\]
-/
theorem PBBasic019
    (δ : ℕ → ℕ) (hδ : ∀ n, δ n = (n.divisors.filter Odd).sup id)
    (N : ℕ) (hN : 0 < N) :
    |∑ n ∈ Finset.Icc 1 N, (δ n : ℚ) / n - 2 / 3 * N| < 1 := by sorry"
PB-Basic-020,"Find all pairs of primes $(a, b)$ such that $a^2 - ab - b^3 = 1$.","Step 1. Lemma. If for a prime $p$, $p\equiv 2 \pmod{3}$, then $p$ cannot divide $q^2-q+1$ for any prime $q$.
 Proof. Assume $p \mid q^2-q+1$ (so $p$ is odd); then $p \mid q^3+1$, so $q^3 \equiv -1 \pmod{p}$, thus $q^6 \equiv 1 \pmod{p}$. Let $\nu$ be the multiplicative order of $q$ modulo $p$; then $\nu \mid 6$, and $\nu \mid p-1$ (by Fermat's Little theorem). That forces $\nu = 2$ (since $\nu = 1$ means $q \equiv 1 \pmod{p}$, so $-1 \equiv q^3 \equiv 1 \pmod{p}$, forcing $p=2$, absurd), and so $q \equiv -1 \pmod{p}$. But then $0\equiv q^2-q+1 \equiv 1 + 1 + 1 = 3 \pmod{p}$, forcing $p=3$, absurd.

 Step 2. Easily $a>3$, then $a^2=ab+b^3+1>2b+b^3+1>2b+b^2+1=(b+1)^2$, hence $a>b+1$; but then from $a\mid b^3+1=(b+1)(b^2-b+1)$ follows that $a\mid b^2-b+1$, hence $a\equiv 1 \pmod{3}$.

 Step 3. If $b\equiv 1\pmod{3}$, then $1=a^2-ab-b^3 \equiv 1-1-1 = -1 \pmod{3}$, a contradiction. If $b\equiv 2\pmod{3}$, then $1=a^2 - ab - b^3 \equiv 1 - 2 + 1 = 0 \pmod{3}$, a contradiction again. Hence $b=3$ (the only moment where the primality of $b$ is actually used), and from this we easily get $a=7$. So $(a,b)=(7,3)$ is the only solution.","(Partial)
 1. Proved that $a$ should divide $b^2 - b+1$. Also partial points for proving that $a-2$ cannot be a multiple of $3$.
(Almost)
 1. Verification contains minor mistakes only.",Number theory,IMO-easy,"$(p,q) = (7,3)$","(Modified) Tuymaada 2013, Seniors, P6","import Mathlib

/-
Find all pairs of primes $(a, b)$ such that $a^2 - ab - b^3 = 1$.

Solution: $(p,q) = (7,3)$
-/
theorem PBBasic020 :
    {(a, b) : ℕ × ℕ | a.Prime ∧ b.Prime ∧ (a ^ 2 - a * b - b ^ 3 : ℕ) = 1}
      = {(7, 3)} := by sorry"
PB-Basic-021,Define the sequence $x_1 = 6$ and $x_n = 2^{x_{n-1}} + 2$ for all integers $n \ge 2$. Prove that $x_{n-1}$ divides $x_n$ for all integers $n \ge 2$.,"We will prove this using induction. Note that $x_{2}=2^{6}+2=66$
 which is divisible by $x_{1}$ and $x_{3}=2^{66}+2=2(2^{65}+1)$,
 which is divisible by $2(2^{5}+1)=2^{6}+2=x_{2}$.


 For $a,b\in\mathbb{N}$, it is well known that $2^{a}+1|2^{ab}+1$
 and $2^{a}-1|2^{ab}-1$ if $2\not|b$.


 We will prove the result by induction. Note that $\nu_{2}(x_{i})=1$,
 so $\frac{x_{n-1}}{x_{n-2}}$ is odd. Also, note that $x_{1}|x_{2}$,
 as $2|6$, and $x_{2}|x_{3}$, as $6|66$.

 $x_{n}|x_{n+1}\Longleftrightarrow2^{x_{n-1}-1}+1|2^{x_{n}-1}+1\Longleftarrow x_{n-1}-1|x_{n}-1\Longleftrightarrow2^{x_{n-2}}+1|2^{x_{n-1}}+1\Longleftarrow x_{n-2}|x_{n-1}\text{ and that \ensuremath{\frac{x_{n-1}}{x_{n-2}}} is odd.}$
 Thus $x_{n-2}|x_{n-1}\implies x_{n}|x_{n+1}$, we can prove by induction
 that $x_{i}|x_{i+1}$ for all $i\in\mathbb{N}$.","(Partial)
 1. Proved $x_{n-1}-1|x_{n}-1$.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Number theory,IMO-easy, ,(Modified) ELMO 2015 P1,"import Mathlib

/-
Define the sequence $x_1 = 6$ and $x_n = 2^{x_{n-1}} + 2$ for all integers $n \ge 2$. Prove that $x_{n-1}$ divides $x_n$ for all integers $n \ge 2$.
-/
theorem PBBasic021
    (x : ℕ → ℕ) (hx₁ : x 1 = 6)
    (hx_rec : ∀ n ≥ 1, x (n + 1) = 2 ^ (x n) + 2)
    (n : ℕ) (hn : 1 ≤ n) :
    x n ∣ x (n + 1) := by sorry"
PB-Basic-022,"Let $S=\{0,1,2^{2024},3^{2024},\ldots\}$ be the set of all perfect
 2024-th powers. Find all polynomials $P$ with integer coefficients
 such that $P(x)=s$ has an integer solution $x$ for every $s\in S$.","Since $0$ is in $S$, there exists an integer $c$ such that $P(c)=0$.
 We can write $P(x)=(x-c)^{n}Q(x)$ for some positive integer $n$
 and integer coefficient polynomial $Q$ such that $Q(c)\ne0$.

 \noindent Pick a prime number $q$ bigger than $|Q(c)|,|Q(c+1)|,|Q(c-1)|$.
 For every positive integer $m$, there exists an integer $a$ such
 that $P(a)=q^{2024m}$. Then, $(a-c)$ is a power of $q$.

 \noindent If it is 1, it is a contradiction to the fact that $|Q(c+1)|,|Q(c-1)|<q$.
 Therefore, $q$ divides $a-c$ and so $Q(a)-Q(c)$ is a multiple of
 $q$, which shows that $Q(a)$ is not a multiple of $q$. Therefore,
 $|Q(a)|=1$ and $|(a-c)^{n}|=q^{2024m}$. Therefore, $|Q(x)|=1$ for
 infinitely many integers $x$ and so $Q(x)=1$ or $Q(x)=-1$. Then
 $|P(x)|$ is always a $n$th power.

 \noindent In this case we can easily see that $n$ needs to divide
 $2024$. Moreover if $n$ is even and $Q(x)=-1$, $P$ is always negative
 or 0 . Hence, the only possible solutions are of the form: $P(x)=(\pm(x+c))^{n}$
 for some divisor $n$ of 2024. We can easily see that this satisfies
 the problem conditions.","(Partial)
 1. Observed that there must be an integer root $c$ and tried to analyze $P$ in terms of it (e.g. reducing without loss of generality to the case $P(0)=0$).
 (Almost)
 1. Verification contains minor mistakes only.",Number theory,IMO-easy,"$P(x) = (x - a)^d or P(x) = (-x - a)^d$, where $d | 2024$.",(Modified) ELMO 2023 Shortlist N1,"import Mathlib

/-
Let $S = \{ 0, 1, 2^{2024}, 3^{2024}, \ldots \}$ be the set of all perfect 2024-th powers. Find all polynomials $P$ with integer coefficients such that $P(x) = s$ has an integer solution $x$ for every $s \in S$.

Solution: $P(x) = (x - a)^d or P(x) = (-x - a)^d$, where $d | 2024$.
-/
open Polynomial

theorem PBBasic022 :
    {P : ℤ[X] | ∀ n ≥ 0, ∃ x, P.eval x = n ^ 2024} =
      {(X - C a) ^ d | (a : ℤ) (d : ℕ) (_ : d ∣ 2024)}
        ∪ {(- X - C a) ^ d | (a : ℤ) (d : ℕ) (_ : d ∣ 2024)} := by sorry"
PB-Basic-023,"Suppose $a,b,c$ are positive integers such that $2^{a}+1=7^{b}+2^{c}$.
 Find all possible values of $(a,b,c)$.","From the given equation, observe that $a>c$. The equation can be
 rewritten as:
 \[
 2^{c}\left(2^{a-c}-1\right)=7^{b}-1.
 \]

 We consider the following cases:


 \textbf{Case 1: $b$ is odd}

 In this case, $7^{b}-1\equiv2\pmod 4$, so $2^{c}=2$, which implies
 $c=1$. Substituting back into the equation, we have:
 \[
 2^{a}=7^{b}+1.
 \]

 On the other hand, note that $7^{b}+1=8(7^{b-1}-7^{b-2}+\cdots+7+1)$.
 If $b\ge3$ the second factor is an odd number greater than 1 which
 cannot divide $2^{a}$, contradiction. Therefore we have $b=1$ and
 $2^{a}=8$, so $a=3$. Hence, this case gives the solution $(a,b,c)=(3,1,1)$.


 \textbf{Case 2: $b$ is even but not divisible by $4$}

 Let $b=4k+2$, where $k\in\mathbb{N}$. Then:
 \[
 7^{b}-1=7^{4k+2}-1=\left(7^{2k+1}-1\right)\left(7^{2k+1}+1\right).
 \]

 Reasoning similarly to the previous case, we find that $7^{2k+1}-1$
 is divisible by $2$ but not by $4$, $7^{2k+1}+1$ is divisible by
 $8$ but not by $16$. Therefore, $7^{b}-1$ is divisible by $16$
 but not by $32$, which implies $c=4$. Substituting back into the
 equation, we have:
 \[
 2^{a}=7^{4k+2}+15.
 \]

 Since $7^{4k+2}+15\equiv1\pmod 3$, it follows that $2^{a}\equiv1\pmod 3$.
 Thus, $a$ is even. Let $a=2\ell$, where $\ell\in\mathbb{Z}^{+}$,
 $\ell\geq2$ (since $x>z=4$). We can write:
 \[
 15=2^{2\ell}-7^{4k+2}=\left(2^{\ell}-7^{2k+1}\right)\left(2^{\ell}+7^{2k+1}\right).
 \]

 From here we easily obtain $\ell=3$ and $k=0$ which gives $a=6$
 and $b=2$. Thus, this case gives the solution $(a,b,c)=(6,2,4)$.


 \textbf{Case 3: $b$ is divisible by $4$}

 In this case, $7^{b}-1$ is divisible by $4^{2}=16$. Since $2^{a-c}-1$
 is divisible by $25$, $a-c$ is divisible by $\text{ord}_{25}(2)=20$.
 Then, $2^{a-c}-1$ is divisible by 31. Additionally:
 \[
 2^{a-c}-1\text{ is divisible by }31\implies7^{b}-1\text{ is divisible by }31.
 \]

 Note that $\text{ord}_{31}(7)=15$, so $b$ is divisible by $15$.
 However, in this case $7^{b}-1$ is also divisible by $9$, implying
 $a-c$ must be divisible by $6$ which implies that $2^{a-c}-1$ is
 divisible by $7$. Then, we find:
 \[
 7^{b}-1\text{ is divisible by }7,\text{ leading to a contradiction.}
 \]

 Hence, there are no solutions in this case.


 \textbf{Conclusion}

 The two valid solutions are:
 \[
 (a,b,c)=(3,1,1)\quad\text{and}\quad(a,b,c)=(6,2,4).
 \]","(Partial)
 1. Found all answers.
 2. Handled the case of $b$ is odd and $2\pmod 4$.
(Almost)
1. Verification contains minor mistakes only or missed a trivial case.",Number theory,IMO-medium,"$(a,b,c) =( 3,1,1) ,( 6,2,4)$.",(Modified) Vietnam TST 2019 P4,"import Mathlib

/-
Suppose $a, b, c$ are positive integers such that $2^a + 1 = 7^b + 2^c$. Find all possible values of $(a, b, c)$.

Solution: $(x,y,z) = (3,1,1), (6,2,4)$.
-/
theorem PBBasic023 :
    {(a, b, c) : ℕ × ℕ × ℕ | 0 < a ∧ 0 < b ∧ 0 < c ∧ 2 ^ a + 1 = 7 ^ b + 2 ^ c}
      = {(3,1,1), (6,2,4)} := by sorry"
PB-Basic-024,"Find all non-negative integers $a,b,c$ such that $20^{a}+b^{4}=2024^{c}$","We state the following well-known lemma: If $p=4k+3$ is a prime
 number for some positive integer $k$, and there exists integers $x,y$
 such that $x^{2}+y^{2}$ is divisible by $p$, then we have $x,y$
 are both divisible by $p$.

 We omit the proof of this lemma since it is a well-known result.

 Now, return to the main problem. It is obvious that $(a,b,c)=(0,0,0)$
 is a solution. Now, consider $c>0$, we have the following cases:


 \textbf{Case 1: $a$ is even.}

 In this case, we have $a=2k$ for some non-negative integer $k$.
 This gives $(20^{k})^{2}+(b^{2})^{2}$ is divisible by $2024$, which
 is divisible by $11$. Since $11$ is a prime in the form of $4n+3$
 for some positive integer $n$, we must have that $20^{k},b^{2}$
 are all divisible by $11$, which is a contradiction since $\gcd(20,11)=1$.
 Hence, $a$ cannot be even.


 \textbf{Case 2: $a$ is odd.}

 In this case, we have $a,c\geq1$, and so $b$ must be even. Let $b=2^{t}q$
 for some non-negative integer $t,q$ where $q$ is odd, we have the
 equation becomes $2^{2a}5^{a}+2^{4t}q^{4}=2^{3c}253^{c}$. Consider
 quadratic residue modulo $11$, we have the LHS gives us $-5$ is
 a quadratic residue modulo $11$. However, since $11$ has the form
 $4n+3$ we have $-1$ is a quadratic nonresdue, while $5$ is a quadratic
 residue modulo $11$ from the Law of Quadratic Reciprocity. Hence,
 we have that $-5$ is a quadratic nonresidue modulo $11$, which a
 contradiction.

 Hence, $(a,b,c)=(0,0,0)$ is the only solution.","(Partial)
 1. Correctly analyzed either the $a$ is even case or $a$ is odd case but not both.
(Almost)
 1. Verification contains minor mistakes only or missed a trivial case.",Number theory,IMO-medium,"$(a,b,c) = (0,0,0)$","(Modified) Serbia MO 2008, P1","import Mathlib

/-
Find all non-negative integers $a, b, c$ such that $20^a + b^4=2024^c$

Solution: $(x,y,z) = (0,0,0)$
-/
theorem PBBasic024 :
    {(a, b, c) : ℕ × ℕ × ℕ | 20 ^ a + b ^ 4 = 2024 ^ c}
      = {(0, 0, 0)} := by sorry"
PB-Basic-025,"Given a triangle $XYZ$ with circumcenter $O$, the incircle of triangle
 $XYZ$ has center $I$. Let $M,N$ on the sides $XY,XZ$
 respectively such that $YM=ZN=YZ$. If $\gamma$ is the angle created
 by two lines $MN,OI$, what is $\frac{\gamma}{2}$ in terms of degree?","Let the incircle of $XYZ$ meet the sides of $XYZ$ at points $D,E,F$. Let $G$ be the centroid of triangle $DEF$, it is well-known that
 $I,G,O$ are collinear. Let $r$ be the inradius and $\vec{v}$ be
 the unit vector perpendicular to $MN$ pointing outward, we have $YZ\vec{ID}+ZN\vec{IE}+YM\vec{IF}+rMN\vec{v}=\vec{0}$.
 Since $YZ=YM=ZN=a$, we have $YZ\vec{ID}+ZN\vec{IE}+YM\vec{IF}+rMN\vec{v}=\vec{0}$
 is equivalent to $3a\vec{IG}=-rMN\vec{v}$, which implies that $IG$
 is parallel with $\vec{v}$. Since $\vec{v}$ is perpendicular to
 $MN$, this implies that $MN$ is perpendicular to $IG$. Combining
 this with the fact that $I,G,O$ is collinear, we have $\gamma=90^{\circ}$,
 which means that $\frac{\gamma}{2}=45^{\circ}$.","(Partial)
 1. Considered the point $G$, the centroid of the triangle formed by the points of tangency of the incircle and the sides of $XYZ$, and explained that $I,G,O$ is collinear.
(Almost)
 1. Verification contains minor mistakes only.",Geometry,IMO-easy,45,folklore,"import Mathlib

/-
Given a triangle $XYZ$ with circumcenter $O$, the incircle of triangle
$XYZ$ has center $I$. Let $M,N$ on the sides $XY,XZ$
respectively such that $YM=ZN=YZ$. If $\gamma$ is the angle created
by two lines $MN,OI$, what is $\frac{\gamma}{2}$ in terms of degree?
-/
open Real Affine Simplex EuclideanGeometry
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)
local notation degrees ""ᵒ"" => (degrees * π / 180)

theorem PBBasic025
    -- Given a triangle $XYZ$
    (X Y Z : ℝ²) (tri : AffineIndependent ℝ ![X, Y, Z]) :
    -- with circumcenter $O$,
    let O := circumcenter ⟨![X, Y, Z], tri⟩
    -- the incircle of triangle $XYZ$ has center $I$.
    let I := incenter ⟨![X, Y, Z], tri⟩
    -- Let $M,N$ on the sides $XY,XZ$ respectively such that $YM=ZN=YZ$.
    ∀ M N : ℝ², Sbtw ℝ X M Y → Sbtw ℝ X N Z →
    dist Y M = dist Y Z → dist Z N = dist Y Z →
    -- take an auxiliary point `p` at the intersection of `MN` and `OI`
    -- and two helper points `p₁ p₂` on the lines to define the angle `gamma`
    ∀ (p p₁ p₂ : ℝ²), p ∈ affineSpan ℝ {M, N} → p ∈ affineSpan ℝ {O, I} →
    p₁ ∈ affineSpan ℝ {M, N} → p₂ ∈ affineSpan ℝ {O, I} →
    let γ := ∠ p₁ p p₂
    -- it is true that once we know `γ / 2` is supposed to equal `45ᵒ`, we don't
    -- need the condition γ ≤ 90ᵒ but in general, we probably mean to take
    -- the non-obtuse angle when measuring an angle between lines
    γ ≤ 90ᵒ → γ / 2 = 45ᵒ
      := by sorry"
PB-Basic-026,"Let $\triangle ABC$ be an inscribed triangle in $(O)$ and circumscribed
 around $(I)$. The incircle $(I)$ touches $BC,CA,AB$ at $D,E,F$,
 respectively. Construct the circle $(W_{a})$ passing through $B,C$
 and tangent to $(I)$ at $X$, and let $D'$ be the reflection of
 $D$ across $AI$. Define $Y,Z,E',F'$ similarly. Prove that the lines
 $D'X,E'Y,F'Z$ are concurrent on the line $OI$.","Lemma: Let $ABC$ be an acute triangle with circumcircle $\Omega$.
 Let $B_{0}$ be the midpoint of $AC$ and let $C_{0}$ be the midpoint
 of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$
 be the centroid of the triangle $ABC$. Let $\omega$ be a circle
 through $B_{0}$ and $C_{0}$ that is tangent to the circle $\Omega$
 at a point $X\not=A$. Prove that the points $D,G$ and $X$ are collinear.

 This is the well-known IMO Shortlist 2011 G4 problem, we omit the
 proof here.

 Consider an inversion center $I$ with power $r^{2}$ where $r$ is
 the inradius. This maps $B,C$ to the midpoint $M_{B},M_{C}$ of $DF,DE$
 respectively. Moreover, we have $DD'$ is parallel to $EF$. Let $G$
 be the centroid of $\triangle DEF$ and $H_{D}$ be the foot of $D$
 onto $EF$. We have $D',G,H_{D}$ are collinear from Thales Theorem,
 and $X,G,H_{D}$ are collinear. And so, we have $XD'$ passes through
 $G$. Similarly, we have $YE',ZF'$ passes through $G$ as well. Finally,
 it is well-known that $I,G,O$ are collinear. And so, we have $D'X,E'Y,F'Z$
 are concurrent on the line $OI$.","(Partial)
 1. Applied correct inversion and explain that $D',G,H_{D}$ is collinear, where $H_{D}$ is the foot of $D$ onto $EF$.
 2. Obtained equivalent information without inversion.
(Almost)
 1. Verification contains minor mistakes only.",Geometry,IMO-medium, ,Novel Problem,"import Mathlib

/-
Let $\triangle ABC$ be an inscribed triangle in $(O)$ and circumscribed
around $(I)$. The incircle $(I)$ touches $BC,CA,AB$ at $D,E,F$,
respectively. Construct the circle $(W_{a})$ passing through $B,C$
and tangent to $(I)$ at $X$, and let $D'$ be the reflection of
$D$ across $AI$. Define $Y,Z,E',F'$ similarly. Prove that the lines
$D'X,E'Y,F'Z$ are concurrent on the line $OI$.
-/
open Real Affine Simplex EuclideanGeometry
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

-- We start by defining the auxiliary construction.
-- Given points A, B, C, we define a predicate that a given line
-- was constructed as the line `D'X`.
def isLineDX (A B C : ℝ²) (dx : AffineSubspace ℝ ℝ²) : Prop :=
  ∃ (tri : AffineIndependent ℝ ![A, B, C]) (W : Sphere ℝ²) (X : ℝ²),
  B ∈ W ∧ C ∈ W ∧
  let I := incenter ⟨![A, B, C], tri⟩
  let D := touchpoint ⟨![A, B, C], tri⟩ ∅ 0
  (insphere ⟨![A, B, C], tri⟩).IsIntTangentAt W X ∧
  let D' := reflection (affineSpan ℝ {A, I}) D
  dx = affineSpan ℝ {D', X}

-- The main problem does this construction 3-times, and claims the result
-- is on the line O I
theorem PBBasic026
    (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
    (l1 l2 l3 : AffineSubspace ℝ ℝ²) :
    isLineDX A B C l1 →
    isLineDX B C A l2 →
    isLineDX C A B l3 →
    let O := circumcenter ⟨![A, B, C], tri⟩
    let I := incenter ⟨![A, B, C], tri⟩
    ∃ p : ℝ², p ∈ l1 ∧ p ∈ l2 ∧ p ∈ l3 ∧ p ∈ affineSpan ℝ {O, I} := by sorry"
PB-Basic-027,"Given an acute triangle $ABC$, let $D$ and $E$ be the feet of
 the altitudes from $B$ to $AC$ and $C$ to $AB$, respectively.
 Let $E_{1}$ and $E_{2}$ be the reflections of $E$ with respect
 to $AC$ and $BC$, respectively. If $X$ (not equal to $C$) is an
 intersection point of the circumcircle of $\triangle CE_{1}E_{2}$
 and $AC$, and $O$ is the circumcenter of $\triangle CE_{1}E_{2}$,
 prove that $XO$ is perpendicular to $DE$.","\textbf{Claim.} $X,E,E_{2}$ are colinear.


 \noindent\emph{Proof. }Let $T$ be a intersection of $EE_{2}$ and
 $AC$. Observe that $CE_{1}=CE=CE_{2}$ and $(B,C,E,E_{1},E_{2})$
 are cyclic. This gives $\angle E_{2}E_{1}C=\frac{1}{2}(180^{\circ}-2\angle C)=\angle E_{2}TC$,
 so $(E_{2},T,C,E_{1})$ are cyclic. Thus, $T=X$, and we have $X,E,E_{2}$
 colinear.


 Now, observe $\angle OXC=90-\angle XE_{2}C=90-\angle EE_{2}C=90-\angle ABC=90-\angle ADE=90-\angle XDE$.
 Thus, $\angle OXC+\angle XDE=90^{\circ}$, and we have $XO\perp DE$
 as desired.","(Partial)
 1. Proved that $X,E,E_{2}$ are colinear.
(Almost)
 1. Verification contains minor mistakes only.",Geometry,IMO-easy, ,(Modified) FKMO 2016 P1,"import Mathlib

/-
Given an acute triangle $ABC$, let $D$ and $E$ be the feet of
the altitudes from $B$ to $AC$ and $C$ to $AB$, respectively.
Let $E_{1}$ and $E_{2}$ be the reflections of $E$ with respect
to $AC$ and $BC$, respectively. If $X$ (not equal to $C$) is an
intersection point of the circumcircle of $\triangle CE_{1}E_{2}$
and $AC$, and $O$ is the circumcenter of $\triangle CE_{1}E_{2}$,
prove that $XO$ is perpendicular to $DE$.
-/
open Real Affine Simplex EuclideanGeometry
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

theorem PBBasic027
    (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
    (acute : AcuteAngled ⟨![A, B, C], tri⟩) :
    let D := altitudeFoot ⟨![A, B, C], tri⟩ 1
    let E := altitudeFoot ⟨![A, B, C], tri⟩ 2
    let E₁ := reflection (affineSpan ℝ {A, C}) E
    let E₂ := reflection (affineSpan ℝ {B, C}) E
    ∀ X : ℝ², X ≠ C → Cospherical {X, C, E₁, E₂} → Collinear ℝ {X, A, C} →
    ∀ (tri2 : AffineIndependent ℝ ![C, E₁, E₂]),
    let O := circumcenter ⟨![C, E₁, E₂], tri2⟩
    (affineSpan ℝ {X, O}).direction ⟂ (affineSpan ℝ {D, E}).direction := by sorry"
PB-Basic-028,"In $\triangle ABC$ the altitudes $BE$ and $CF$ intersect at $H$. A circle $(W)$ is
 externally tangent to the Euler circle $(E)$ of $\triangle ABC$ and also tangent
 to the sides $AB$ and $AC$ at $X$ and $Y$, respectively, with
 $(W)$ being closer to $A$ than the Euler circle. Let $I'$ be the
 incenter of $\triangle AEF$. Prove that $AXI'Y$ is a rhombus.","Consider the inversion centered at $A$ with power $\sqrt{\frac{1}{2} AE \times AC}$. Under
 this inversion, the Euler circle remains unchanged, while $B$ and
 $C$ are mapped to $B'$ and $C'$, which are the midpoints of $AF$
 and $AE$, respectively.


 Observe that if $AXI'Y$ is a rhombus, then $(W)$ is the image of
 the mixtilinear incircle of $\triangle AEF$ under a homothety centered
 at $A$ with ratio $\frac{1}{2}$. The $A$-excircle of $\triangle ABC$,
 which is tangent to $AB,AC$, and $BC$, is mapped to the circle $(V)$,
 which is tangent to $AB,AC$, and $(AB'C')$, i.e., the mixtilinear
 incircle of $\triangle AEF$.


 It remains to prove that $(V)$ is tangent to the Euler circle, which
 would imply that $(V)$ coincides with $(W)$. Indeed, by Feuerbach's
 theorem, the $A$-excircle is externally tangent to $(E)$. Thus,
 under the inversion, $(V)$ is also tangent to $(E)$, which implies
 that $(V)$ coincides with $(W)$, as desired.","(Partial)
 1. Applied correct inversion and modified the goal to showing $(V)$ is tangent to the Euler circle.
 2. Obtained equivalent information without inversion.
(Almost)
 1. Verification contains minor mistakes only.",Geometry,IMO-medium, ,Novel Problem,"import Mathlib

/-
In $\triangle ABC$ the altitudes $BE$ and $CF$ intersect at $H$. A circle $(W)$ is
externally tangent to the Euler circle $(E)$ of $\triangle ABC$ and also tangent
to the sides $AB$ and $AC$ at $X$ and $Y$, respectively, with
$(W)$ being closer to $A$ than the Euler circle. Let $I'$ be the
incenter of $\triangle AEF$. Prove that $AXI'Y$ is a rhombus.
-/
open Real Affine Simplex EuclideanGeometry
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

theorem PBBasic028
    (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
    (ω : Sphere ℝ²) -- ω is the 9-point (Euler) circle
    (hω : midpoint ℝ A B ∈ ω ∧ midpoint ℝ B C ∈ ω ∧ midpoint ℝ C A ∈ ω)
    (W : Sphere ℝ²) (X Y : ℝ²) -- W is a circle that
    (hX : W.IsTangentAt X (affineSpan ℝ {A, B})) -- is tangent to AB at X
    (hY : W.IsTangentAt Y (affineSpan ℝ {A, C})) -- is tangent to AC at Y
    (tangent : W.IsExtTangent ω) -- is externally tangent to the 9-point circle
    (closer : dist A W.center < dist A ω.center) : -- and is closer to A than ω
    let E := altitudeFoot ⟨![A, B, C], tri⟩ 1 -- E is the foot from B
    let F := altitudeFoot ⟨![A, B, C], tri⟩ 2 -- F is the foot from C
    let H := Triangle.orthocenter ⟨![A, B, C], tri⟩ -- H is the orthocenter
    ∀ tri2 : AffineIndependent ℝ ![A, E, F], -- aux assumption to take the incenter
    let I' := incenter ⟨![A, E, F], tri2⟩ -- I' is the incenter of AEF
    -- then all `dist A X`, `dist X I'`, `dist I' Y`, `dist Y A` are equal
    List.Pairwise (· = ·) [dist A X, dist X I', dist I' Y, dist Y A] := by sorry"
PB-Basic-029,"Let $p$ and $n$ be integers with $0\le p\le n-2$. Consider a set
 $S$ of $n$ lines in the plane such that no two of them are parallel
 and no three have a common point. Denote by $I$ the set of intersections
 of lines in $S$. Let $O$ be a point in the plane not lying on any
 line of $S$. A point $X\in I$ is colored red if the open line segment
 $OX$ intersects at most $p$ lines in $S$. What is the minimum number
 of red points that is contained in $I$?","Let $\ell$ be the nearest to $O$ line from $L$ and $\ell\cap I=\left\{ X_{1},X_{2},...,X_{n-1}\right\} $
 with $|OX_{i}|\leq|OX_{i+1}|$ for every $i.$

 Claim. Open segment $OX_{i}$ intersects at most $i-1$ line from
 $S$.

 Proof. Suppose the contrary, so $OX_{i}$ intersects $i$ lines and
 in particular it intersects line $\ell'$ such that $Y=\ell\cap\ell'\notin\bigcup_{t=1}^{i-1}\{X_{t}\}.$
 If $Z=OX_{i}\cap\ell'$ we get the contradiction with
 \[
 |OY|\geq|OX_{i}|\implies\angle OYX_{i}<90^{\circ}\implies d(O,\ell')=|OY|\cdot\sin\angle OYZ<|OY|\cdot\sin\angle OYX_{i}=d(O,\ell)\text{ }\Box
 \]
 Now with fixed $n$ we apply induction by $p.$ By the claim $X_{1}$
 is always red, so the base case $p=0$ follows. Now consider problem
 statement for $S\backslash\ell$ and $p=k,$ so there exist at least
 $C_{p+2}^{2}$ red points. After backing $\ell$ to $L$ and increasing
 $p$ to $p+1$ all red points are preserved and by claim all points
 $X_{1},X_{2},...,X_{p+2}$ are red, so totally there are at least
 $C_{p+3}^{2}$ red points.

 For the example construction, we take $X=(-1,-1)$ and $n$ lines
 to be the lines connecting $(0,n-1-i)$ and $(i,0)$ for $0\le i\le n-1$.","(Partial)
 1. Guessed the correct answer along with the optimal case construction.
(Almost)
 1. Verification contains minor mistakes only.
 2. Proof is correct, but does not provided the optimal case construction.",Geometry,IMO-medium,$\frac{(p + 1)(p + 2)}{2}$,(modified) IMO Shortlist 2008 G5,"import Mathlib

/-
Let $p$ and $n$ be integers with $0\le p\le n-2$. Consider a set
$S$ of $n$ lines in the plane such that no two of them are parallel
and no three have a common point. Denote by $I$ the set of intersections
of lines in $S$. Let $O$ be a point in the plane not lying on any
line of $S$. A point $X\in I$ is colored red if the open line segment
$OX$ intersects at most $p$ lines in $S$. What is the minimum number
of red points that is contained in $I$?
Answer: $\frac{(p + 1)(p + 2)}{2}$
-/
open Affine
abbrev Point := EuclideanSpace ℝ (Fin 2)

def Line := {l : AffineSubspace ℝ Point // Module.rank ℝ l.direction = 1}
local instance : Membership Point Line where mem l X := X ∈ l.1

/-
A set of lines is in a general position if no two lines are parallel, and no three lines are concurrent.
-/
abbrev GeneralPosition (lines : Set Line) : Prop :=
  (∀ l1 ∈ lines, ∀ l2 ∈ lines, l1 ≠ l2 → ¬ l1.1 ∥ l2.1) ∧
  (∀ l1 ∈ lines, ∀ l2 ∈ lines, ∀ l3 ∈ lines, l1 ≠ l2 → l1 ≠ l3 → l2 ≠ l3 →
    (l1.1 : Set Point) ∩ l2.1 ∩ l3.1 = ∅)

-- The set of all intersections of a set of lines
abbrev intersections (lines : Set Line) : Set Point :=
  { X : Point | ∃ l1 ∈ lines, ∃ l2 ∈ lines, l1 ≠ l2 ∧ X ∈ l1 ∧ X ∈ l2 }

-- A point avoids a set of lines if it is not contained in any
abbrev Point.avoids (lines : Set Line) (X : Point) := ∀ l ∈ lines, X ∉ l

-- A predicate claiming that there at most `k` lines strictly between `O` `X`
abbrev atMostBetween (lines : Set Line) (p : ℕ) (O X : Point) : Prop :=
  { line ∈ lines | ∃ inter : Point, Sbtw ℝ O inter X ∧ inter ∈ line }.encard ≤ p

-- All points in the plane that have at most `p` lines
abbrev redPoints (lines : Set Line) (p : ℕ) (O : Point) : Set Point :=
  atMostBetween lines p O

-- The minimum possible number of red intersections given `p` and `n`
noncomputable
def minRed (p n : ℕ) : ℕ := sInf {
  (intersections lines ∩ redPoints lines p O).encard |
  -- we consider all general positions of `n` lines
  (lines : Set Line) (_ : lines.encard = n) (_ : GeneralPosition lines)
  -- and all positions of the point `O`
  (O : Point) (_ : O.avoids lines)
}

theorem PBBasic029 (p n : ℕ) (le : p+2 ≤ n) :
  minRed p n = (p+1) * (p+2) / 2 := by sorry"
PB-Basic-030,"Given a trapezoid $ABCD$ with $AB,CD$ as the two legs. Circle $(W_{1})$ passes through $A,B$, and $(W_{2})$ passes through $C,D$ so that they are tangent to each other. The inscribed angle on circle $W_1$ corresponding to the arc AB on the side opposite to C and D is alpha, and the inscribed angle on circle $W_2$ corresponding to the arc CD on the side opposite to  A and B is beta. Construct $(W_{3})$ passing through $A,B$, $(W_{4})$ passing through $C,D$ such that the inscribed angle on circle W3 corresponding to the arc AB on the side opposite to C and D is $\beta$, and the inscribed angle on circle $W_4$ corresponding to the arc CD on the side opposite to  A and B is b $\alpha$. Prove that $(W_{3}),(W_{4})$ are tangent to each other.","If $AB,CD$ are parallel to each other then $(W_{1})$ maps to $(W_{4}),(W_{2})$
 maps to $(W_{3})$ via a reflection w.r.t. the center $O$ of the
 parallelogram $ABCD$.

 If $AB,CD$ are not parallel to each other then let $X$ be the intersection
 of $AB,CD$. Consider the inversion with power $XA.XC$ and then reflect
 it w.r.t. to the internal bisector $\angle AXD$, we have $A\rightarrow C$,
 $B\rightarrow D$ (1). Note that inversion preserves the angle between
 the circles and the lines, and so $(W_{1})$ is mapped to $(W_{4})$
 since we have (1) and the angle between $(W_{1})$ and $AB$ is $\alpha$,
 which is also the angle between $(W_{4})$ and $CD$. Similarly, we
 have $(W_{2})$ is mapped to $(W_{3})$. Since $(W_{1})$ is tangent
 to $(W_{2})$, we have $(W_{3})$ is tangent to $(W_{4})$. Hence
 proved.","(Partial)
 1. Performed an inversion w.r.t. $AB\cap CD$ and re-analyzed the given problem.
(Almost)
 1. Verification contains minor mistakes only.
 2. Did not consider the case when $AB$ and $CD$ are parallel.",Geometry,IMO-easy, ,Novel Problem,"import Mathlib

/-
Given a trapezoid $ABCD$ with $AB,CD$ as the two legs. Circle $(W_{1})$ passes through $A,B$, and $(W_{2})$ passes through $C,D$ so that they are tangent to each other. The inscribed angle on circle $W_1$ corresponding to the arc AB on the side opposite to C and D is alpha, and the inscribed angle on circle $W_2$ corresponding to the arc CD on the side opposite to  A and B is beta. Construct $(W_{3})$ passing through $A,B$, $(W_{4})$ passing through $C,D$ such that the inscribed angle on circle W3 corresponding to the arc AB on the side opposite to C and D is $\beta$, and the inscribed angle on circle $W_4$ corresponding to the arc CD on the side opposite to  A and B is b $\alpha$. Prove that $(W_{3}),(W_{4})$ are tangent to each other.
-/
open Real Affine Simplex EuclideanGeometry
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

-- A predicate that ABCD is a trapezoid with legs AB, CD
structure IsTrapezoid (A B C D : ℝ²) : Prop where
  nodup : [A,B,C,D].Nodup
  ncoll : ¬ Collinear ℝ {A,B,C,D}
  para : (affineSpan ℝ {B, C}) ∥ (affineSpan ℝ {D, A})
  order : (affineSpan ℝ {A, B}).WSameSide C D

-- A predicate that two sphere are tangent somehow (externally or internally)
def SphereTangent (s1 s2 : Sphere ℝ²) :=
  s1.IsExtTangent s2 ∨ s1.IsIntTangent s2 ∨ s2.IsIntTangent s1

theorem PBBasic030 (A B C D : ℝ²) (trapezoid : IsTrapezoid A B C D)
    (W₁ W₂ W₃ W₄ : Sphere ℝ²)
    (h_AW₁ : A ∈ W₁) (h_BW₁ : B ∈ W₁) (h_CW₂ : C ∈ W₂) (h_DW₂ : D ∈ W₂)
    (h_AW₃ : A ∈ W₃) (h_BW₃ : B ∈ W₃) (h_CW₄ : C ∈ W₄) (h_DW₄ : D ∈ W₄)
    (tangent_W : SphereTangent W₁ W₂) (α β : Angle)
    (X₁ X₂ X₃ X₄ : ℝ²) -- X₁ ... X₄ will be the points to measure the inscribed angles
    (h_XW₁ : X₁ ∈ W₁) (h_XW₂ : X₂ ∈ W₂) (h_XW₃ : X₃ ∈ W₃) (h_XW₄ : X₄ ∈ W₄)
    -- since a point measures inscribed angle of an arc opposite to the given point,
    -- X₁, X₃ should be on the same side as AB, and X₂, X₄ on the same side as CD
    (h_side_X₁ : (affineSpan ℝ {A, B}).SSameSide X₁ C)
    (h_side_X₂ : (affineSpan ℝ {C, D}).SSameSide X₂ A)
    (h_side_X₃ : (affineSpan ℝ {A, B}).SSameSide X₃ C)
    (h_side_X₄ : (affineSpan ℝ {C, D}).SSameSide X₄ A)
    (angle_X₁ : ∠ A X₁ B = α) (angle_X₂ : ∠ C X₂ D = β)
    (angle_X₃ : ∠ A X₃ B = β) (angle_X₄ : ∠ C X₄ D = α) :
    SphereTangent W₃ W₄ := by sorry"
PB-Advanced-001,"For a positive integer $n$, let $A_{n}$ be the number of perfect powers less than or equal to $n$. Here, a perfect power is a number that can be expressed in the form $a^{b}$, where $a$ is a positive integer and $b$ is an integer greater than or equal to 2. Prove that there are infinitely many $n$ such that $A_{n}$ divides $n+2024$.","Let's look at the following lemma using the intermediate value property.

 <Lemma 1> If a sequence of non-negative integers $\left(x_{n}\right)_{n=1}^{\infty}$ satisfies $x_{n+1}-x_{n} \in\{0,1\}$ and for any $M>0$, there exists a positive integer $n$ such that $\frac{n}{x_{n}}>M$, then there are infinitely many $n$ such that $\frac{n}{x_{n}}$ is a positive integer.

 <Proof of Lemma 1> First, let $k_{0}$ be the smallest $k$ such that $x_{k} \geq 1$. Now, we will prove that for any integer $m \geq k_{0}$, there exists an $n$ such that $\frac{n}{x_{n}}=m$.

 (1) There exists a positive integer $n$ such that $\frac{n}{x_{n}}>m$, and let $n_{0}$ be the smallest such $n$. Since $\frac{k_{0}}{x_{k_{0}}} \leq k_{0}$, we have $n_{0} \geq k_{0}+1$.

 (2) By the minimality of $n_{0}$ and $n_{0} \geq k_{0}+1$, we have $\frac{n_{0}-1}{x_{n_{0}-1}} \leq m$. If $\frac{n_{0}-1}{x_{n_{0}-1}}<m$, then

 \[
 n_{0}<m x_{n_{0}-1}+1 \quad \Rightarrow \quad n_{0} \leq m x_{n_{0}-1}.
 \]

 However, from $\frac{n_{0}}{x_{n_{0}}}>m$, we have $m x_{n_{0}}<n_{0}$, so combining the two results gives

 \[
 m x_{n_{0}-1} \geq n_{0}>m x_{n_{0}},
 \]

 which leads to $x_{n_{0}-1}>x_{n_{0}}$. This contradicts the condition of the problem that $x_{n+1}-x_{n} \in\{0,1\}$. Therefore, we must have $\frac{n_{0}-1}{x_{n_{0}-1}}=m$, and the proof is complete.

 Therefore, for any integer $m \geq k_{0}$, there exists an $n$ such that $\frac{n}{x_{n}}=m$, and it is obvious that these values of $n$ are different for different $m$, so the proof is complete. \qed

 The following lemma is a Bernoulli-type inequality.

 <Lemma 2> For any integer $k \geq 2$, we have $2^{k} \geq \frac{k^{2}}{2}$.

 <Proof of Lemma 2> Since $k \geq 2$, by the binomial theorem, we have

 \[
 2^{k} \geq 1+k+\binom{k}{2}=1+\frac{k}{2}+\frac{k^{2}}{2}>\frac{k^{2}-k}{2},
 \]

 so the proof is complete. $\qed$

 Now, let's prove the main problem.

 <Step 1> Finding an upper bound for $A_{n}$

 <Step 1.1> For a positive integer $a$ and $k \geq 2$, let a number of the form $a^{k}$ be called a $k$-th power. Then, for any $k$-th power $a^{k}$ to be less than or equal to $n$, we must have

 \[
 a^{k} \leq n \quad \Rightarrow \quad a \leq n^{\frac{1}{k}},
 \]

 so the number of $k$-th powers less than or equal to $n$ is less than or equal to $n^{\frac{1}{k}}$.

 <Step 1.2> If any $k$-th power other than 1 is less than or equal to $n$, then we must have $2^{k} \leq n$, so $k \leq \log _{2} n$.

 <Step 1.3> By (1) and (2) above, $A_{n}$ satisfies the following:

 \[
 A_{n} \leq \sum_{k=2}^{\left[\log _{2} n\right]} n^{\frac{1}{k}} \leq\left(\left[\log _{2} n\right]-1\right) n^{\frac{1}{2}}<\log _{2} n \cdot n^{\frac{1}{2}}.
 \]

 <Step 2> Solving the problem using the lemma

 Now, let the sequence $\left(x_{n}\right)_{n=1}^{\infty}$ be defined as $x_{n}=0$ if $n \leq 2024$ and $x_{n}=A_{n-2024}$ for $n \geq 2025$.

 <Step 2.1> Since $A_{1}=1$ and $A_{n+1}-A_{n}$ is 1 if $n+1$ is a perfect power and 0 otherwise, $x_{n}$ satisfies $x_{n+1}-x_{n} \in\{0,1\}$ for any $n \geq 1$.

 <Step 2.2> By <Step 1>, we have $A_{2^{2 k}} \leq 2 k \cdot 2^{k}$, so by <Lemma 2>, we have

 \[
 \frac{2^{2 k}}{A_{2^{2 k}}} \geq \frac{2^{k}}{2 k} \geq \frac{k}{4}.
 \]

 Since $A_{n}$ is an increasing sequence, we have $A_{n} \geq x_{n}$, so $\frac{n}{x_{n}} \geq \frac{n}{A_{n}}$. Combining these results gives

 \[
 \frac{2^{2 k}}{x_{2^{2 k}}} \geq \frac{2^{2 k}}{A_{2^{2 k}}} \geq \frac{k}{4},
 \]

 so for any $M>0$, there exists an $n$ such that $\frac{n}{x_{n}}>M$.

 <Step 2.3> By (1), (2), and <Lemma 1> above, there are infinitely many $n$ such that $x_{n} \mid n$. Therefore, there are also infinitely many $n$ such that

 \[
 \frac{n+2024}{A_{n}}=\frac{n+2024}{x_{n+2024}}
 \]

 is a positive integer.","(Partial)
 1. Observed that $A_n$ is a slowly increasing sequence, and from there guessed that the sequence satisfied some property such as Lemma 1 of the given solution.
(Almost)
 1. Applied intermediate value property to get useful information, but couldn't complete the proof.",Algebra,IMO-easy, ,Novel Problem,"import Mathlib

/-
For a positive integer $n$, let $A_{n}$ be the number of perfect powers less than or equal to $n$. Here, a perfect power is a number that can be expressed in the form $a^{b}$, where $a$ is a positive integer and $b$ is an integer greater than or equal to 2. Prove that there are infinitely many $n$ such that $A_{n}$ divides $n+2024$.
-/
open scoped Classical

theorem PBAdvanced001 (A : ℕ → ℕ)
    (hA : ∀ n, A n = (Finset.Icc 1 n |>.filter fun m => ∃ b a, 0 < a ∧ 2 ≤ b ∧ m = a ^ b).card) :
    {n : ℕ | 0 < n ∧ A n ∣ n + 2024}.Infinite := by sorry"
PB-Advanced-002,"There are 120 students in DeepMind highschool, and each student `loves' some of the students. Here `love' is not always a mutual relationship; a student A may love B while B may not love A, and it is also possible that a student loves oneself. For the set $F$ of all $2^{120}$ subsets of the students, define the function $f: F \rightarrow F$ as the function that maps each $X \in F$ to the set of students loved by one or more students in $X$. For $A, B \in F$, $A$ and $B$ are said to have a 'lovely relationship' iff there exists a natural number $k$ such that $f^{k}(A)=B$. Now we want to select distinct elements $A_{1}, A_{2}, \cdots A_{t}$ from $F$ such that for any $1 \leq i<j \leq t$, $A_{i}$ and $A_{j}$ have a lovely relationship. Let $M(f)$ be the maximum possible value of $t$.  Show that $M(f) \leq 2^{70}$. ","To being with, we prove there must be some $A \in F$ such that $\left\{A_{1}, \cdots A_{t}\right\} \subset \cup_{i=0}^{\infty}\left\{f^{i}(A)\right\}$. Why is this? Consider a graph $H$ with $A_{1}, A_{2}, \cdots A_{t}$ as its vertices. If $f^{k}\left(A_{i}\right)=A_{j}$ for some $k$, draw a directed edge $A_{i} \rightarrow A_{j}$ to create a directed graph. This graph is a tournament. Now, if there is a vertex $A$ such that $d^{-}(A)=0$, then it holds directly. If not, then in this graph, the in-degree of any vertex is at least 1. This means the graph contains a cycle. Having a cycle means that for each element $A$ in the cycle, there is some $T$ such that $f^{T}(A)=A$. Therefore, for two vertices $A, B$ in the cycle, if $f^{i}(A)=B$, then $f^{T-i}(B)=A$. This means any two vertices in the cycle can reach each other via applications of $f$. Now, if we remove all but one vertex from the cycle and then use induction on the number of vertices, we can immediately find that there is a vertex $A$ such that $d^{+}(A)=t-1$. Thus, we can obtain our desired conclusion.

Now, considering a directed graph $G$ whose vertices are the students and where an edge $i \rightarrow j$ is drawn if student $i$ loves student $j$, we need to find $\max_{A_0 \in F} \left| \{f^i(A_0) \mid i \in \mathbb{Z}_{\ge 0} \} \right|$. Now, in $G$, let's perform the following procedure: \begin{itemize} \item Pick any cycle in $G$ and call it $C_{1}$. \item Pick any cycle in $G-C_{1}$ (the graph after removing all vertices of $C_1$) and call it $C_{2}$. \item Repeat this procedure until no more cycles can be found. \end{itemize} As a result of this procedure, we will have disjoint cycles $C_{1}, C_{2}, \cdots, C_{m}$, and a graph $H$ (which is $G - \bigcup C_i$) that has no cycles. Let $a_i$ be the number of vertices in $C_i$, and let $|V(H)|=x$. Then $a_{1}+a_{2}+\cdots+a_{m}=n-x \leq n$ (where $n=120$). Let $L =c\operatorname{lcm}\left[a_{1}, a_{2}, \cdots, a_{m}\right]$, where $c$ is the smallest positive integer such that $L\ge n$.
\\Consider some nonempty $A \subset V(G)$. Now, if we define $T(k) = \bigcup_{j=1}^{m} (f^{k}(A) \cap C_{j})$, then since each $|f^{k}(A) \cap C_{j}|$ is non-decreasing, $|T(k)|$ is a non-decreasing function. Now let's look at $|T(L)|, \cdots, |T((2n+1)L)|$. Since $|T(k)| \leq |V(G)|=n$ and $|T(k)|$ is non-decreasing, there exists some $i$ such that $|T(iL)|=|T((i+1)L)|=|T((i+2)L)|$. At this point, by the definition of $L$, it must be that $T(iL) \subset T((i+1)L) \subset T((i+2)L)$, and therefore $T(iL)=T((i+1)L)=T((i+2)L)$. Therefore, for each $1 \leq r < L$, $T(iL+r)=T((i+1)L+r)$ also holds.
\\Now, if $f^{(i+2)L}(A) \neq f^{(i+1)L}(A)$, then there must be a $v_{0} \in f^{(i+2)L}(A) \triangle f^{(i+1)L}(A)$ (symmetric difference; $A \triangle B=(A-B) \cup (B-A)$), and in this case, $v_{0} \in H$ must hold. Therefore, there also exists $v_1 \in f^{(i+2)L-1}(A) \triangle f^{(i+1)L-1}(A)$ (from which $v_0$ is reached), and $v_2 \in \dots$, so we can keep finding $v_j$ such that there is a path $v_k \rightarrow \dots \rightarrow v_1 \rightarrow v_0$. If $v_1 , v_2, \cdots, v_{|V(H)|+1}$ are all in $H$, then some point should appear at least twice, which is a contradiction to the fact that $H$ has no cycles. Therefore, there exists some $k \le n$ such that $v_k \not \in H$, or equivalently  $v_k \in \bigcup C_j$. However, since $v_k \in f^{(i+2)L -k} (A) \triangle f^{(i+1)L -k} (A)$ where $k \le n \le L$, this is a contradiction to the observation above that $T(iL+r)=T((i+1)L+r)$ for each $1\le r \le L$.

Thus, $f^{(i+2)L}(A)=f^{(i+1)L}(A)$ and thus it follows that $\left\{f^{j}(A) \mid j \in \mathbb{Z}_{\ge 0}\right\}=\left\{f^{j}(A) \mid j \leq (i+1)L \right\}$. Since $i <2n$, we can conclude that $M(f) \leq 2nL$.
\\Now, $\operatorname{lcm}[a_1, a_2, \cdots, a_m] \leq 120 \cdot 60 \cdot 40 \cdot 30 \cdot 24 \cdot 20 \cdot 17 \cdot 15 \cdot 13 \cdot \operatorname{lcm}[1,2,3, \cdots, 12]<2^{60}$ so $L \le \max\{2^{60}, 2n\} = 2^{60}$ Since $2n=240<2^8$, $2nL < 2^8 \cdot 2^{60} = 2^{68}$.","(Partial)
 1. Proved there must be some $A \in F$ such that $\left\{A_{1}, \cdots A_{t}\right\} \subset \cup_{i=0}^{\infty}\left\{f^{i}(A)\right\}$.
 2. Considered the non-intersecting cycles
(Almost)
 1.  Introduced a quantity like $\lcm[a_1, a_2, \cdots, a_n]$, but failed to find the desired numerical bound.
 2.  Verification contains minor mistakes only.",Combinatorics,IMO-medium, ,Novel Problem,"import Mathlib

/--
  M(f) is the maximum possible number of distinct subsets A_1, ..., A_t
  such that any pair has a lovely relationship, where f is derived from the love relation L on S.
  This definition is made robust by taking S and L as explicit parameters,
  and defining all intermediate concepts (f, LovelyRelationship, Configuration) locally.
-/
noncomputable
def M_func (S : Type) [Fintype S] (L : S → S → Prop) : ℕ :=
  -- 1. Define f: The function that maps a set X to the set of students loved by someone in X.
  let f : Set S → Set S := fun X => {y : S | ∃ x ∈ X, L x y}

  -- 2. Define Lovely Relationship (reachability)
  let LovelyRelationship (A B : Set S) : Prop := ∃ k : ℕ, f^[k] A = B

  -- 3. Define Mutually Lovely Relationship (symmetric reachability)
  let MutuallyLovelyRelationship (A B : Set S) : Prop :=
    LovelyRelationship A B ∨ LovelyRelationship B A

  -- 4. Type for indexed collection of subsets
  let Configuration (t : ℕ) := Fin t → Set S

  -- 5. Define the predicate for a Lovely Configuration
  let IsLovelyConfiguration (t : ℕ) (A : Configuration t) : Prop :=
    -- Distinctness condition
    (∀ i j : Fin t, i ≠ j → A i ≠ A j) ∧
    -- Mutually related condition
    (∀ i j : Fin t, i ≠ j → MutuallyLovelyRelationship (A i) (A j))

  -- 6. M is the supremum of possible sizes t
  sSup {t : ℕ | ∃ (A : Configuration t), IsLovelyConfiguration t A}

-- The main claim: M(f) <= 2^70 when the number of students is 120.
theorem PBAdvanced002 {S : Type} [Fintype S] {L : S → S → Prop} (h_card : Fintype.card S = 120) :
  M_func S L ≤ 2^70 := by sorry"
PB-Advanced-003,"Let $ ABC $ be an acute triangle which is not an isosceles.Let $ I $ be the incenter and let $ \omega $ be the circumcircle of $ABC$. Let the intersections of lines $ AI $, $ BI $, and $ CI $ with $ BC $, $ CA $, and $ AB $ be $ D $, $ E $, and $ F $ respectively. Also, let $ \omega_A $ be the circle that lies inside $\angle BAC$, tangent to lines $ AB $ and $ AC $, and internally tangent to the circumcircle $ \omega $ at $ T_A $. Similarly, define $ T_B $ and $ T_C $ for points $ B $ and $ C $ respectively. Prove that there exist two points $ X $ and $ Y $ such that the circumcircles of triangles $ ADT_A $, $ BET_B $, and $ CFT_C $ all pass through $ X $ and $ Y $.","Let's define some points as follows:
 \begin{enumerate}
  \item Let $A_1$, $B_1$, and $C_1$ be the points of tangency of the incircle with sides $BC$, $CA$, and $AB$, respectively.
  \item Let $N$ be the midpoint of arc $BC$.
  \item Let $S_A$ be the intersection of line $NT_A$ and side $BC$ (if they are parallel, consider $S_A$ as the point at infinity).
  \item Let $\Gamma_A$, $\Gamma_B$, and $\Gamma_C$ be the circumcircles of triangles $ADT$, $BET$, and $CFT$, respectively.
 \end{enumerate}

 <Step 1> Finding a point on the radical axis of circles $\Gamma_A$, $\Gamma_B$, and $\Gamma_C$

 By the well-known property of Mixtilinear circles, $AT_A$ is isogonal to the line joining the point of tangency of the excircle opposite to vertex $A$ and vertex $A$.
 Similarly, consider the lines joining the points of tangency of the excircles opposite to vertices $B$ and $C$ with their respective vertices. It is well-known that these three lines intersect at a single point, the Nagel point.
 Therefore, lines $AT_A$, $BT_B$, and $CT_C$ also intersect at a single point, which is the isogonal conjugate of the Nagel point. Let this point be $T$.
 Clearly, $T$ lies on segments $AT_A$, $BT_B$, and $CT_C$, and is inside circles $\Gamma_A$, $\Gamma_B$, and $\Gamma_C$.

 Furthermore, by the power of a point theorem with respect to the circumcircle of triangle $ABC$, we have
 \[AT \cdot TT_A = BT \cdot TT_B = CT \cdot TT_C.\]
 Therefore, point $T$ has the same power with respect to circles $\Gamma_A$, $\Gamma_B$, and $\Gamma_C.$

 Hence, to prove the statement, it is sufficient to find another point that has the same power with respect to circles $\Gamma_A$, $\Gamma_B$, and $\Gamma_C$.

 <Step 2> $(A, D, T_A, S_A)$ is cyclic

 $\angle ADS_A = \frac{\angle A}{2} + \angle C = \angle ACN = \angle A T_A SA$, so $(A, D, T_A, S_A)$ is cyclic.

 <Step 3> $\angle AIS_A = 90^\circ$

 This is also a well-known property of Mixtilinear circles. By angle chasing, we can show that the circumcircles of triangles $BIC$ and $NIT_A$ are tangent to each other at point $I$.
 Let's consider the radical axes of these three circles:
 \begin{itemize}
  \item The radical axis of the circumcircles of triangles $BIC$ and $NITA$ is the line passing through $I$ and perpendicular to $AI$.
  \item The radical axis of the circumcircles of triangles $BIC$ and $ABC$ is $BC$.
  \item The radical axis of the circumcircles of triangles $ABC$ and $NITA$ is $NT_A$.
 \end{itemize}
 Therefore, the intersection point $S_A$ of $BC$ and $NT_A$ satisfies $\angle AIS_A = 90^\circ$.

 <Step 4> Inversion with respect to the incircle of triangle $ABC$

 Now, let's invert the figure with respect to the incircle $I$. In particular, let's pay attention to which circle $I$ maps to under this inversion.
 Let $A'$, $B'$, and $C'$ be the images of points $A$, $B$, and $C$ under the inversion. Also, let $D'$ and $S_A'$ be the images of points $D$ and $S_A$ under the inversion.

 Let's investigate the properties of the inverted points.
 Since $DA_1$ is tangent to the circle, $D'$ is the foot of the perpendicular from $A_1$ to $ID$. $A'$ is the midpoint of $B_1C_1$. Finally, since $SA_1$ is tangent to the circle, $S_A'$ is the foot of the perpendicular from $A_1$ to $IS_A$.
 By the conclusion of Step 3, $A_1S_A'$ is parallel to $AI$.
 By the observation from step 2, the inversion of the circumcircle of $ADT_A$ is the circumcircle of triangle $A'S_A'D'$. Similarly, the circumcircles of $B'S_B'E'$ and $C'S_C'F'$ are our desired circles. Let's prove that the three circumcircles pass through two common points.

 <Step 5> The circumcircles of triangles $A'S_A'D', B'S_B'E'$, and $C'S_C'F'$ pass through two common points

 Now, let's consider the figure with respect to triangle $A_1B_1C_1$.

 In triangle $A_1B_1C_1$, point $I$ is the circumcenter. Let $H_1$ be the foot of the perpendicular from $A_1$ to $B_1C_1$.
 Then, points $A'$, $S_A'$, and $D'$ are constructed as follows:
 \begin{itemize}
  \item Point $A'$ is the midpoint of $B_1C_1$.
  \item Point $D'$ is the intersection of the line parallel to $B_1C_1$ passing through $A_1$ and the perpendicular bisector of $B_1C_1$.
  \item Point $S_A'$ is the foot of the perpendicular from $I$ to $AH_1$.
 \end{itemize}
 Through this construction, we can find that the circumcenter $O_A$ of triangle $A'S_A'D'$ lies on the perpendicular bisector of $A'D'$, which is the line $B'C'$.
 Now, let's consider the point $O_A'$ which is the reflection of $O_A$ with respect to the midpoint of $B'C'$.

 <Step 6> $O_A'$ is the circumcenter of triangle $A_1IH_1$

 Since $A'$, $B'$, $C'$, and $H_1$ lie on the nine-point circle of triangle $ABC$, these four points are concyclic. Therefore, quadrilateral $A'H_1 B'C'$ is an isosceles trapezoid, and points $A'$ and $H_1$ are symmetric with respect to the perpendicular bisector of line $B_1 C_1$.
 Furthermore, $A_1$ and $D'$ are symmetric with respect to the perpendicular bisector of line $B'C'$.
 Therefore, triangle $AIH_1$ is the reflection of triangle $A'S_A'D'$ across the perpendicular bisector of line $B'C'$, and the circumcenters $O_A$ and $O_A'$ are also symmetric with respect to each other.

 <Step 7> $O_A, O_B$, and $O_C$ are collinear

 Now, let's define $O_B, O_C, O_B'$, and $O_C'$ similarly, and show that $O_A, O_B$, and $O_C$ are collinear.
 First, let's show that $O_A'$, $O_B'$, and $O_C'$ are collinear. It is sufficient to prove that the circumcircles of triangles $AIH_1$, $BIH_B$, and $C_1IH_C$ have a common radical axis.
 On the other hand, the circumcenter $I$ of $ABC$ and the orthocenter $A_1H_A \cap B_1H_B \cap C_1H_C$ have the same power with respect to these three circumcircles. Therefore, these three circles have a common radical axis. Hence, the centers of these three circles, $O_A'$, $O_B'$, and $O_C'$, are collinear.
 Now, by Menelaus' theorem, we have
 \[\frac{C'O_A' \cdot B'O_C' \cdot A'O_B'}{O_A'B' \cdot O_C'A' \cdot O_B'C'} = 1\]
 \[\frac{C'O_A \cdot B'O_C \cdot A'O_B}{O_AB' \cdot O_CA' \cdot O_BC'} = 1\]
 Therefore, $O_A, O_B$, and $O_C$ are also collinear.

 <Step 8> Circles $\Gamma_A, \Gamma_B$, and $\Gamma_C$ have two common points

 Now, the circles $\Gamma_A, \Gamma_B$, and $\Gamma_C$ are inverted into circles. The centers of these three circles are collinear, as proven earlier. Therefore, these three circles either have a common radical axis or their radical axes are parallel. This property remains true before the inversion.
 However, in Step 1, we found point $T$ that has the same power with respect to these three circles. Therefore, circles $\Gamma_A, \Gamma_B$, and $\Gamma_C$ have a common radical axis.
 Furthermore, since point $T$ lies inside all three circles, this common radical axis intersects each circle at two distinct points. Therefore, the three circles have two common points.","(Partial)
 1. Showed that $AT_a$, $BT_b, CT_c$ are concurrent (there are many possible solutions).
(Almost)
 1. Verification contains minor mistakes only.",Geometry,IMO-hard, ,Novel Problem,"import Mathlib

/-
Let $ ABC $ be an acute triangle which is not an isosceles.Let $ I $ be the incenter and let $ \omega $ be the circumcircle of $ABC$. Let the intersections of lines $ AI $, $ BI $, and $ CI $ with $ BC $, $ CA $, and $ AB $ be $ D $, $ E $, and $ F $ respectively. Also, let $ \omega_A $ be the circle that lies inside $\angle BAC$, tangent to lines $ AB $ and $ AC $, and internally tangent to the circumcircle $ \omega $ at $ T_A $. Similarly, define $ T_B $ and $ T_C $ for points $ B $ and $ C $ respectively. Prove that there exist two points $ X $ and $ Y $ such that the circumcircles of triangles $ ADT_A $, $ BET_B $, and $ CFT_C $ all pass through $ X $ and $ Y $.
-/

local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)
open EuclideanGeometry Real InnerProductSpace Set Affine Simplex

/--
A helper predicate to define the point of tangency `T` of the `A`-mixtilinear incircle
with the circumcircle `ω`.
The `A`-mixtilinear incircle is the circle lying inside angle `A`, tangent to sides `AB` and `AC`,
and internally tangent to the circumcircle `ω`.
-/
def IsMixtilinearTouchPoint (A B C T : ℝ²) (ω : Sphere ℝ²) : Prop :=
  ∃ (c : Sphere ℝ²),
    -- The circle c is tangent to lines AB and AC
    c.IsTangent (affineSpan ℝ {A, B}) ∧
    c.IsTangent (affineSpan ℝ {A, C}) ∧
    c.IsIntTangentAt ω T

/--
The main statement
-/
theorem PBAdvanced003
  (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
  -- Triangle ABC is acute
  (h_acute : AcuteAngled ⟨![A, B, C], tri⟩)
  -- Triangle ABC is not isosceles (scalene)
  (h_not_iso : dist A B ≠ dist B C ∧ dist B C ≠ dist C A ∧ dist C A ≠ dist A B)
  (I : ℝ²) (hI : I = incenter ⟨![A, B, C], tri⟩)
  (ω : Sphere ℝ²) (hω : ω = circumsphere ⟨![A, B, C], tri⟩)
  -- D, E, F are intersections of angle bisectors with opposite sides
  (D : ℝ²) (hD : D ∈ affineSegment ℝ B C ∩ affineSpan ℝ {A, I})
  (E : ℝ²) (hE : E ∈ affineSegment ℝ C A ∩ affineSpan ℝ {B, I})
  (F : ℝ²) (hF : F ∈ affineSegment ℝ A B ∩ affineSpan ℝ {C, I})
  -- T_A, T_B, T_C are the mixtilinear touch points
  (T_A : ℝ²) (hT_A : IsMixtilinearTouchPoint A B C T_A ω)
  (T_B : ℝ²) (hT_B : IsMixtilinearTouchPoint B C A T_B ω)
  (T_C : ℝ²) (hT_C : IsMixtilinearTouchPoint C A B T_C ω) :
  -- Conclusion: There exist two distinct points X and Y common to the circumcircles
  ∃ X Y : ℝ², X ≠ Y ∧
    Cospherical {A, D, T_A, X} ∧ Cospherical {A, D, T_A, Y} ∧
    Cospherical {B, E, T_B, X} ∧ Cospherical {B, E, T_B, Y} ∧
    Cospherical {C, F, T_C, X} ∧ Cospherical {C, F, T_C, Y} := by sorry"
PB-Advanced-004,"For a positive integer $n$, a convex $18n+2$-gon $P$ is divided into $18n$ triangles by drawing $18n-1$ diagonals. Prove that we can choose two of these diagonals such that the three parts of $P$ divided by these two diagonals each contain at least $3n$ and at most $9n$ triangles.","First, let's look at the following well-known theorem.

 <Lemma>
 For a graph $G$ with $n$ vertices, the following four statements are equivalent. Therefore, if one of them holds, $G$ is a tree.

 (A) $G$ is connected and has no cycles.

 (B) $G$ is connected and has $n-1$ edges.

 (C) $G$ has $n-1$ edges and no cycles.

 (D) There is exactly one path between any two vertices in $G$.

 The following lemma is a general fact about the division of trees.

 For a positive integer $k \geq 2$, if the degree of each vertex in a tree is at most $k$, then we can remove an edge from the tree so that both resulting connected components have size at least $\frac{n-1}{k}$.

 <Proof of Lemma>

 (1) Let $H_{e}, K_{e}$ be the two connected components obtained by removing an edge $e$ from the tree, and let $h_{e}, k_{e}$ be the number of vertices in the two components. (Of course, $h_{e}+k_{e}=n$.) Let $l=\{x, y\}$ be the edge for which $\min \left(h_{e}, k_{e}\right)$ is maximized. (If there are multiple such edges, choose one arbitrarily)

 (2) Assume for contradiction that the smaller component obtained by removing $l$ has size less than $\frac{n-1}{k}$. Without loss of generality, let this be the component containing $x$. Let this component be $A$.

 (3) Now, for any edge $e$ other than $l$ connected to vertex $y$, if we remove that edge instead of $l$, let $H_{e}$ be the component containing $l$ and $K_{e}$ be the other component. Then $H_{e}$ contains $A$, so it is larger than $A$. If the number of vertices in $H_{e}$ is less than or equal to the number of vertices in $K_{e}$, then it contradicts the maximality of $A$. Therefore, the number of vertices in $K_{e}$ must be less than the number of vertices in $H_{e}$. Therefore, by the maximality of $A$, the number of vertices in $K_{e}$ is less than or equal to the number of vertices in $A$, and therefore less than $\frac{n-1}{k}$.

 (4) Therefore, when we remove each edge adjacent to vertex $y$, each resulting connected component has less than $\frac{n-1}{k}$ vertices. Since the degree of vertex $y$ is at most $k$, there are at most $k$ such components, and therefore the number of vertices excluding $y$ is less than $\frac{n-1}{k} \times k=n-1$, which is a contradiction.

 Therefore, the proof is complete. \qed

 Now, let's prove the problem. First, let's look at the general properties of triangulation of a convex $n$-gon before looking at the convex $18n+2$-gon.

 <Step 1> Basic properties of triangulation of a convex polygon

 <Step 1.1> The sum of the interior angles of a convex $n$-gon is $(n-2) \pi$. Therefore, in order for triangles with an interior angle sum of $\pi$ to divide this sum, we need a total of $n-2$ triangles. Since each time we draw a diagonal, the division of the convex $n$-gon increases by one, we know that if we have divided a convex $n$-gon into $n-2$ triangles, we have drawn a total of $n-3$ diagonals. In summary, to triangulate a convex $n$-gon, we need to draw $n-3$ diagonals to divide it into $n-2$ triangles.

 <Step 1.2> If $n \geq 4$, then there are $n-2$ triangles, and each triangle cannot have all 3 sides as sides of the convex $n$-gon. Therefore, there must be at least 2 triangles that share 2 sides with the convex $n$-gon.

 <Step 2> Mapping triangulation of a convex polygon to a tree

 The problem of dividing a convex $n$-gon $P$ into triangles is directly related to trees. For convenience, assume $n \geq 4$. Let's consider the triangles as vertices and connect two vertices if the corresponding triangles share a side to draw a graph $G$.

 <Step 2.1> By <Step 1>, this graph has $n-2$ vertices and $n-3$ edges. (This is because each diagonal drawn during the division corresponds to one edge in $G$.)

 <Step 2.2> We can confirm that this graph $G$ is connected by mathematical induction. The case $n=4$ is trivial. Now, assume that $G$ is connected for triangulations of convex $n-1$-gons for $n \geq 5$, and consider a triangulation $T$ of a convex $n$-gon $P$. By (2) of <Step 1>, this triangulation includes a triangle $X$ that has two consecutive sides of $P$.

 <Step 2.3> The vertex corresponding to this triangle $X$ has degree 1 in graph $G$. Let $Y$ be the vertex connected to $X$. Now, the remaining part of the $n$-gon $P$ excluding triangle $X$ is an $n-1$-gon $P^{\prime}$, and if we remove triangle $X$ from triangulation $T$, the remaining part is $T^{\prime}$, which is a triangulation of $P^{\prime}$. Let $G^{\prime}$ be the graph for this triangulation. Clearly, $G^{\prime}$ is the graph obtained by removing vertex $X$ and edge $XY$ from $G$. By the inductive hypothesis, $G^{\prime}$ is connected, so the graph $G$ obtained by adding edge $XY$ to it is also connected.

 In conclusion, graph $G$ is a connected graph with $n-2$ vertices and $n-3$ edges. Therefore, by -(B), $G$ is a tree.

 <Step 3> Analysis of the problem situation

 <Step 3.1> According to the above analysis, we can see that the triangulation given in the problem corresponds to a tree $T$ with $18n$ vertices. It is very important that the degree of each vertex in this tree is at most 3.

 <Step 3.2> Choosing two diagonals to divide the convex $18n+2$-gon can be thought of as removing two edges from this tree to divide it into three connected components. The question is whether we can do this so that the size of each connected component is at least $3n-1$ and at most $8n+1$.

 <Step 4> Solving the problem

 <Step 4.1> First, the tree $T$ corresponding to the triangulation has $18n$ vertices, and the degree of each vertex is at most 3. Therefore, by Lemma, we can remove one edge to make the sizes of the two resulting components at least $6n-1$.

 <Step 4.2> Let $A$ be the component with fewer vertices and $B$ be the component with more vertices (if the sizes are the same, assign $A$ and $B$ arbitrarily). The number of vertices in $A$ is at least $6n-1$ and at most $9n$, and the number of vertices in $B$ is at least $9n$ and at most $12n+1$.

 <Step 4.3> Since $B$ is also a tree with degree at most 3, we can apply again to $B$. By removing one edge from $B$, we can divide it into $B_{1}, B_{2}$ so that the size of each component is at least $\frac{9n-1}{3}$ and at most $\frac{2(12 n+1)+1}{3}=8 n+1$. Therefore, the sizes of $B_{1}, B_{2}$ are at least $3n$ and at most $8n+1$.

 <Step 4.4> Therefore, we can divide it into $A, B_{1}, B_{2}$.","(Partial)
 1. Found a first diagonal that cuts the polygon into at least $3n$ triangles, or attempted an argument where an unbalanced cut can be modified to create a more balanced cut.
 2. Successfully connected the triangulation with a tree.
(Almost)
 1. Established first cut diagonal as and in the solution attempts to successively apply the argument again on one of the sides, but with errors (e.g. not fulfilling the same hypotheses as the original cut).",Combinatorics,IMO-easy, ,Novel Problem,"import Mathlib

/-
For a positive integer $n$, a convex $18n+2$-gon $P$ is divided into $18n$ triangles by drawing $18n-1$ diagonals. Prove that we can choose two of these diagonals such that the three parts of $P$ divided by these two diagonals each contain at least $3n$ and at most $9n$ triangles.
-/

noncomputable -- helper instance to allow computing degrees
local instance {V : Type} [Finite V] (g : SimpleGraph V) (v : V)
    : Fintype (g.neighborSet v) := by
  apply Fintype.ofFinite

theorem PBAdvanced004 (n : ℕ) (n_pos : n > 0)
    /-
    We will represent a triangulation with a tree of triangles. This captures enough
    data for the problem statement, even if it doesn't tell the order of the triangles.
    -/
    (Tri : Type) -- all the triangles
    [finTri : Fintype Tri]
    (g : SimpleGraph Tri) -- the adjacency of triangles
    (is_tree : g.IsTree)
    -- every triangle can be adjacent with at most three other triangles
    (degree_bound : ∀ tri : Tri, g.degree tri ≤ 3)
    -- There are 18n triangles in the triangulation
    (num_triangles : Fintype.card Tri = 18*n) :
    -- We can choose two of the diagonals
    ∃ cut ⊆ g.edgeSet, cut.ncard = 2 ∧
    -- such that every component
    ∀ comp : (g.deleteEdges cut).ConnectedComponent,
    -- contain at least $3n$ and at most $9n$ triangles
    3*n ≤ Nat.card comp ∧ Nat.card comp ≤ 9*n
    := by sorry"
PB-Advanced-005,"Let $\angle XYZ$ be an acute angle with $\angle XYZ \ne 60^\circ$, and let $A$ be a point inside $\angle XYZ$. Prove that there exists $D\ne A$ inside $\angle XYZ$ and $\theta\in (0,2\pi )$ satisfying the following condition:

 For points $B$ and $C$ on the rays $\overrightarrow{YX}$ and $\overrightarrow{YZ}$ respectively, then
 \[
 \angle BAC = \angle XYZ \quad \implies \quad \angle BDC = \theta.
 \]","step 1: Construction of $D$ and $\theta$

 step 1-1: Without loss of generality we may assume that $XA\perp YZ$
 and $ZA\perp YX$.

 step 1-2: Let $D$ be the circumcenter of $\triangle XYZ$.

 step 1-3: Let $\theta=180^{\circ}-2\angle XYZ$.

 \medskip{}

 step 2: We prove that if $\angle BAC=\angle XYZ$, then $\angle BDC=\theta$.

 \medskip{}

 step 3: Lemma. A point $K$ inside a convex quadrilateral $PQRS$
 has an isogonal conjugate with respect to the $PQRS$ if and only
 if $\angle PKQ+\angle RKS=180^{\circ}$.

 \medskip{}

 step 4: Proof of Lemma.

 step 4-1: Let $H_{1},H_{2},H_{3},H_{4}$ be the feet of perpendicular
 from $K$ onto $PQ,QR,RS,SP$, respectively.

 step 4-2: First assume that $K^{\prime}$ is the isogonal conjugate
 of $K$

 step 4-2-1: Let $H_{1}^{\prime},H_{2}^{\prime},H_{3}^{\prime},H_{4}^{\prime}$
 be the feet of perpendicular from $K^{\prime}$ onto $PQ,QR,RS,SP$,
 respectively.

 step 4-2-2: We prove that $H_{1},H_{2},H_{1}^{\prime},H_{2}^{\prime}$
 are cyclic.

 step 4-2-2-1: We have $\angle KQH_{1}=\angle K^{\prime}QH_{2}^{\prime}$
 by assumption.

 step 4-2-2-2: We have $\angle KH_{1}Q=\angle K^{\prime}H_{2}^{\prime}Q=90^{\circ}$.

 step 4-2-2-3: By steps 4-2-2-1 and 4-2-2-2, we have $\triangle KQH_{1}\sim\triangle K^{\prime}QH_{2}^{\prime}$.

 step 4-2-2-4: Thus $QH_{1}:QH_{2}^{\prime}=KQ:K^{\prime}Q$.

 step 4-2-2-5: Similarly $QH_{1}^{\prime}:QH_{2}=KQ:K^{\prime}Q$.

 step 4-2-2-6: By steps 4-2-2-4 and 4-2-2-5, we have $QH_{1}\times QH_{1}^{\prime}=QH_{2}\times QH_{2}^{\prime}$.

 step 4-2-2-7: Hence $H_{1},H_{2},H_{1}^{\prime},H_{2}^{\prime}$ are
 cyclic.

 step 4-2-3: Denote $M$ as the midpoint of $KK^{\prime}$.

 step 4-2-4: We prove that $H_{1},H_{1}^{\prime},H_{2},H_{2}^{\prime}$
 are equidistant to $M$.

 step 4-2-4-1: Let $N$ be the midpoint of $H_{1}H_{1}^{\prime}$.

 step 4-2-4-2: Since $H_{1}K\parallel H_{1}^{\prime}K^{\prime}$ and
 both are perpendicular to $PQ$, we have $NM\perp PQ$.

 step 4-2-4-3: Hence $M$ lies on the perpendicular bisector of $H_{1}H_{1}^{\prime}$.

 step 4-2-4-4: Similarly $M$ lies on the perpendicular bisector of
 $H_{2}H_{2}^{\prime}$.

 step 4-2-4-5: By steps 4-2-2, 4-2-4-3, 4-2-4-4 and the fact that $PQ$
 and $QR$ are not parallel, we have that the center should be $M$.

 step 4-2-5: Similarly $H_{2},H_{2}^{\prime},H_{3},H_{3}^{\prime}$
 are equidistant to $M$, and $H_{3},H_{3}^{\prime},H_{4},H_{4}^{\prime}$
 are equidistant to $M$.

 step 4-2-6: By steps 4-2-4 and 4-2-5, $H_{1},H_{1}^{\prime},H_{2},H_{2}^{\prime},H_{3},H_{3}^{\prime},H_{4},H_{4}^{\prime}$
 all lie on a circle centered at $M$.

 step 4-2-7: Since $\angle KH_{1}Q=\angle KH_{2}Q=90^{\circ}$, $K,H_{1},Q,H_{2}$
 are cyclic.

 step 4-2-8: Similarly $K,H_{2},R,H_{3}$ are cyclic.

 step 4-2-9: Similarly $K,H_{3},S,H_{4}$ are cyclic.

 step 4-2-10: Similarly $K,H_{4},P,H_{1}$ are cyclic.

 step 4-2-11: By steps 4-2-6, 4-2-7, 4-2-8, 4-2-9, and 4-2-10, we have
 \begin{align*}
 \angle PKQ+\angle RKS&=(\angle PKH_{1}+\angle QKH_{1})+(\angle RKH_{3}+\angle SKH_{3})
 &=\angle PH_{4}H_{1}+\angle QH_{2}H_{1}+\angle RH_{2}H_{3}+\angle SH_{4}H_{3}
 &=360^{\circ}-(\angle H_{1}H_{2}H_{3}+\angle H_{3}H_{4}H_{1})=180^{\circ}.
 \end{align*}
 step 4-3: Assume $\angle PKQ+\angle RKS=180^{\circ}$.

 step 4-3-1: By the same logic as in step 4-2, $H_{1},H_{2},H_{3},H_{4}$ lie on a circle $\omega$.

 step 4-3-2: Let $O$ be the center of $\omega$.

 step 4-3-3: Denote $J_{1},J_{2},J_{3},J_{4}$as the second intersection
 of $\omega$ and $PQ,QR,RS,SP$, respectively.

 step 4-3-4: Let $K^{\prime}$ be the reflection of $K$ with respect
 to $O$.

 step 4-3-5: Let $N^{\prime}$ be the midpoint of $H_{1}J_{1}$.

 step 4-3-6: Since $KH_{1}\parallel ON^{\prime}$ and both are perpendicular
 to $PQ$, we have $K^{\prime}J_{1}\perp PQ$.

 step 4-3-7: Similarly $K^{\prime}J_{2}\perp QR$.

 step 4-3-8: Hence $\angle PQK=\angle H_{1}QK=90^{\circ}-\angle H_{1}KQ=90^{\circ}-\angle H_{1}H_{2}Q=90^{\circ}-\angle J_{2}J_{1}Q=\angle RQK^{\prime}$.

 step 4-3-9: Similarly $K,K^{\prime}$ are isogonal with respect to
 all four angles of $PQRS$.

 step 4-3-10: Thus $K^{\prime}$ is the isogonal conjugate of $K$
 with respect to $PQRS$.\medskip{}

 step 5: Now suppose $\angle BAC=\angle XYZ$.

 step 5-1: Since $\angle XYZ$ is acute, $\triangle XYZ$ is acute,
 and $\angle XAY,\angle ZAY>90^{\circ}>\angle BAC$.

 step 5-2: Hence $\square BCZX$ is convex, and $K$ is inside of $\square BCZX$.\medskip{}

 step 6: We prove that $D$ is the isogonal conjugate of $A$ with
 respect to $\square BCZX$.

 step 6-1: We have $\angle ZAX+\angle BAC=(180^{\circ}-\angle XYZ)+\angle XYZ=180^{\circ}$.

 step 6-2: Hence by Lemma and steps 5-2, 6-1, $A$ has an isogonal
 conjugate with respect to $\square BCZX$.

 step 6-3: $A,D$ are isogonal conjugates with respect to $\triangle XYZ$
 as orthocenter and circumcenter.

 step 6-4: By step 6-3, $XA,XD$ are isogonal lines with respect to
 $\angle BXZ$, and $ZA,ZD$ are isogonal lines with respect to $\angle CZX$.

 step 6-5: By steps 6-2, 6-4, and the fact that there is at most one
 isogonal conjugate, we conclude that $D$ is indeed the isogonal conjugate.

 \medskip{}

 step 7: We prove that $\angle BDC=180^{\circ}-2\angle XYZ$.

 step 7-1: By step 6, $A$ is the isogonal conjugate of $D$ with respect
 to $\square BCZX$.

 step 7-2: By Lemma, we have $\angle BDC+\angle XDZ=180^{\circ}$.

 step 7-3: Since $D$ is the circumcenter of $\triangle XYZ$, we have
 $\angle XDZ=2\angle XYZ$.

 step 7-4: By step 7-2 and 7-3, we have $\angle BDC=180^{\circ}-2\angle XYZ$.
 step 7-5: Since $\angle XYZ \ne 60^\circ$, we have $D\ne A$.","(Partial)
 1. Constructed $D$ and found $\theta$ correctly.
 2. Found the necessary and sufficient conditions for the existence of isogonal conjugate in quadrilateral.
(Almost)
 1. Verification contains minor mistakes only.",Geometry,IMO-medium, ,Novel Problem,"import Mathlib

/-
Let $\angle XYZ$ be an acute angle with $\angle XYZ \ne 60^\circ$, and let $A$ be a point inside $\angle XYZ$. Prove that there exists $D\ne A$ inside $\angle XYZ$ and $\theta\in (0,2\pi )$ satisfying the following condition:

For points $B$ and $C$ on the rays $\overrightarrow{YX}$ and $\overrightarrow{YZ}$ respectively, then
\[
\angle BAC = \angle XYZ \quad \implies \quad \angle BDC = \theta.
\]
-/

open EuclideanGeometry Real

local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)
local notation degrees ""ᵒ"" => (degrees * π / 180)

-- Points involved in the problem are now of type `Point`
variable (X Y Z A : ℝ²)

/-- The set of points forming the ray $\overrightarrow{p₁ p₂}$. -/
def raySet (p₁ p₂ : ℝ²) : Set ℝ² :=
  { AffineMap.lineMap p₁ p₂ t | (t : ℝ) (_ : t ≥ 0) }

/-- A point `A` is in the interior of $\angle XYZ$ if it satisfies the angle addition property
and is not on the boundary. -/
def InsideAngle (A X Y Z : ℝ²) : Prop :=
  A ∉ raySet Y X ∧ A ∉ raySet Y Z ∧
  ∠ X Y A + ∠ A Y Z = ∠ X Y Z

/--
The formal statement of the geometric problem (PB-Advanced-005).
-/
theorem PBAdvanced005
    (hxy : X ≠ Y) (hzy : Z ≠ Y)
    (h_acute : ∠ X Y Z < 90ᵒ)
    (h_not_60 : ∠ X Y Z ≠ 60ᵒ)
    (hA_interior : InsideAngle A X Y Z) :
    ∃ D : ℝ², D ≠ A ∧ InsideAngle D X Y Z ∧
    ∃ θ : ℝ, 0 < θ ∧ θ < 2 * π ∧
    ∀ B C : ℝ²,
      (B ∈ raySet Y X) →
      (C ∈ raySet Y Z) →
      (∠ B A C = ∠ X Y Z) →
      (∠ B D C = θ) := by sorry"
PB-Advanced-006,"Find all functions$f:\mathbb{Z}\rightarrow\mathbb{Z}$
 such that the equation
 \[
 f(x-f(xy))=f(x)f(1-y)
 \]
 holds for all $x,y\in\mathbb{Z}$.","Let $P(x,y)$ be the assertion $f(x-f(xy))=f(x)f(1-y)$.

 If $f$ is constant, $f\equiv0$ or $f\equiv1$. From now on we will
 only consider non-constant functions.

 Claim 1 $f(f(x))=f(x)$ for all $x\in\mathbb{Z}$

 Proof. $P(0,y):f(-f(0))=f(0)f(1-y)\Longrightarrow f(0)=0$

 $P(x,0):f(x)=f(x)f(1)\Longrightarrow f(1)=1$

 $P(1,y):f(1-f(y))=f(1-y)$

 $P(1,1-y):f(1-f(1-y))=f(y)$

 $P(1,1-f(y)):f(y)=f(1-f(1-y))=f(1-f(1-f(y)))=f(f(y))\Longrightarrow f(f(x))=f(x)\forall x\in\mathbb{Z}\blacksquare$

 $\Longrightarrow P(x,1):f(x-f(x))=0$

 Now consider $f(\mathbb{Z})$.

 Case 1 $f(\mathbb{Z})\subset\{-1,0,1\}$

 Case 1.1. $f(\mathbb{Z})=\{0,1\}$

 We have $f(x)=1\textbackslash \Longrightarrow f(x-1)=0\textbackslash \Longrightarrow
 f(2-x)=1 \textbackslash \Longrightarrow f(1-x)=0$ and $f(1-x)=0
 \textbackslash \Longrightarrow f(x)=1$

 \[
 \therefore f(x)=1\Longleftrightarrow f(x-1)=0\Longleftrightarrow f(2-x)=1\Longleftrightarrow f(1-x)=0
 \]
 We can inductively prove that $f(x)=\begin{cases}
 0 & 2|x

 1 & 2\nmid x
 \end{cases}$

 Case 1.2. $f(\mathbb{Z})=\{-1,0,1\}\Longrightarrow f(-1)=-1$

 $P(1,x)$ and $P(-1,x)$ gives

 \begin{align*}
 f(x)=1 & \Longrightarrow f(1-x)=0,f(x+1)=-f(-2)

 f(x)=0 & \Longrightarrow f(1-x)=1,f(x+1)=1

 f(x)=-1 & \Longrightarrow f(1-x)=f(2),f(x+1)=0
 \end{align*}
 It's easy to see that $f(-2)=1,f(2)=-1$, and using this, we can inductively
 prove that $f(x)=\begin{cases}
 0 & 3|x

 1 & 3|x-1

 -1 & 3|x+1
 \end{cases}$

 Case 2. $f(\mathbb{Z})$ is not a subset of $\{-1,0,1\}$

 Case 2.1 $f(t)=1$ for some $2\mid t$

 \[
 P(2,\frac{t}{2}):1=f(1)=f(2)f(1-\frac{t}{2})
 \]

 Case 2.1.1 : $f(2)=1\Longrightarrow P(2,1):1=f(2)f(0)=0$. Contradiction!

 Case 2.1.2 : $f(2)=-1\Longrightarrow f(-1)=-1,f(1-\frac{t}{2})=-1$

 $P(-1,-x):f(-1-f(x))=-f(1+x)$

 $P(-1,-f(x)):f(-1-f(x))=f(-1-f(f(x)))=-f(1+f(x))\Longrightarrow f(1+x)=f(1+f(x))...(*)$

 Since $f(2)=-1$, so plugging this in $(*)$, we have

 \[
 f(2)=-1\Longrightarrow f(3)=0\Longrightarrow f(4)=1\Longrightarrow f(5)=-1\cdots
 \]
 We can prove inductively that$f(x)=\begin{cases}
 0 & 3|x

 1 & 3|x-1

 -1 & 3|x+1
 \end{cases}$ for all $x\in\mathbb{N}$.

 Plugging in $x=-1,y=-n(n\in\mathbb{N})$ gives $f(-1-n)=f(-1-f(n))=-f(1+n)$,
 so we have $f(x)=\begin{cases}
 0 & 3|x

 1 & 3|x-1

 -1 & 3|x+1
 \end{cases}$for all $x$, so contradiction since $f(\mathbb{Z})\in\{-1,0,1\}$.

 Case 2.2 $f(t)=1\Longrightarrow2\nmid t$

 Claim 2 $f(-1)=-1$ or $f(2)=0$

 Proof. Assume that $f(2)\neq0$

 We will prove that $f(t)=1\Longrightarrow f(1-t)=0$. $P(2,\frac{1-t}{2})$
 gives

 \[
 f(2)=f(2)f(\frac{t-3}{2})\Longrightarrow f(\frac{t-3}{2})=1
 \]
 We repeat this, then we get $t\equiv1(mod2^{n})$ for all $n\Longrightarrow t=1$.
 Note that

 \[
 f(-1)=c\Longrightarrow f(-1-c)=0\Longrightarrow f(c+2)=1
 \]
 Hence we conclude that $c=-1$.

 Case 2.2.1. $f(2)=0\Longrightarrow f(-1)=1$

 Define $n$ as the element of $f(\mathbb{Z})-\{-1,0,1\}$ which has
 the smallest absolute value. Then we have

 \[
 f(x-f(xy))=n\Longrightarrow\{f(x),f(1-y)\}=\{1,n\}\text{ or }\{f(x),f(1-y)\}=\{-1,-n\}...(\star)
 \]
 . Also note that $P(-n,- 1):f(-n-f(n))=0$, where $f(n)=n$ by Claim 1. Thus by $x=n,y=-2$ $(\star)\Longrightarrow f(3)=1$.
 Case 2.2.1. now becomes a repetition of Case 1.1, so contradiction!

 Case 2.2.2. $f(2)\neq0,f(-1)=-1,f(-2)=-f(2)$

 Claim 3 $f(n)=t\Longrightarrow n=t\forall t\leq1$

 Proof. We use induction on $t$. $t=1$ is already proven above. Assume
 that $t$ works.

 If $f(n)=t-1\Longrightarrow f(1+n)=f(1+f(n))=f(t)\Longrightarrow n=t-1$,
 so we're done.

 Use the same method to prove that $f(n)=n\forall n\leq1$, and $P(-1,n)(n>0)$
 gives $f(n+1)=-f(-n-1)=n+1$, so $f(n)=n$ for all $n\in\mathbb{Z}$.","(Partial)
 1. Proved $f(f(x))=f(x)$.
 2. Handled the case where $f\subseteq \{-1, 0, 1\}$.
 3. Handled the case where $f\not \subset \{-1, 0, 1\}$.
(Almost)
 1. Verification contains minor mistakes only.",Algebra,IMO-hard,"$f_{1}(x)\equiv0$, $f_{2}(x)\equiv1$,
$f_{3}(x)\equiv x$, $f_{4}(x)=\begin{cases}
0, & x=2n\\
1, & x=2n+1
\end{cases}$, where $n\in\mathbb{Z}$, $f_{5}(x)=\begin{cases}
0, & x=3n\\
1, & x=3n+1\\
-1, & x=3n+2
\end{cases}$, where $n\in\mathbb{Z}$",Novel Problem,"import Mathlib

/-
Find all functions$f:\mathbb{Z}\rightarrow\mathbb{Z}$
such that the equation
\[
f(x-f(xy))=f(x)f(1-y)
\]
holds for all $x,y\in\mathbb{Z}$.

Solution: $f_{1}(x)\equiv0$, $f_{2}(x)\equiv1$,
$f_{3}(x)\equiv x$, $f_{4}(x)=\begin{cases}
0, & x=2n\\
1, & x=2n+1
\end{cases}$, where $n\in\mathbb{Z}$, $f_{5}(x)=\begin{cases}
0, & x=3n\\
1, & x=3n+1\\
-1, & x=3n+2
\end{cases}$, where $n\in\mathbb{Z}$
-/
theorem PBAdvanced006 :
    {f : ℤ → ℤ | ∀ x y, f (x - f (x * y)) = f x * f (1 - y)}
      = {(fun _ => 0), (fun _ => 1), id, (fun x => x % 2),
          (fun x => if 3 ∣ x then 0 else if 3 ∣ x - 1 then 1
            else -1)} := by sorry"
PB-Advanced-007,"Prove that there exist two real-coefficient polynomials $P(x)$ and $Q(x)$ satisfying the following condition:

 (Condition) The degree of the polynomial $P(x)$ is at least 2024, the degree of $Q(x)$ is at least 2, and for any real number $x$, the following holds:

 \[
 P(Q(x)-x-1)=Q(P(x))
 \]","<Step 1> Checking the possibility of recursive construction using induction

 <Step 1.1> Let's assume that two polynomials $P_{0}$ and $Q_{0}$ satisfy the polynomial equation given in the problem as follows:

 \[
 P_{0}\left(Q_{0}(x)-x-1\right)=Q_{0}\left(P_{0}(x)\right)
 \]

 <Step 1.2> For a positive integer $n$, let's define the polynomial $P_{n}(x)$ recursively as follows:

 \[
 P_{n}(x)=P_{n-1}\left(Q_{0}(x)-x-1\right); n \geq 1
 \]

 <Step 1.3> Let's prove by mathematical induction that for any non-negative integer $n$, $(P_{n}, Q_{0})$ also satisfies the given polynomial equation.

 *(1)* First, we have already confirmed that the case $n=0$ holds.

 *(2)* Next, let's assume that the case $n=k$ holds, i.e., assume that $(P_{k}, Q_{0})$ is also a solution to the polynomial equation. Then, by the induction hypothesis and the recursive formula for $P_{n}$ defined above, we have

 \[
 P_{k+1}(x)=P_{k}\left(Q_{0}(x)-x-1\right)=Q_{0}\left(P_{k}(x)\right)
 \]

 Now, substituting $x=Q_{0}(y)-y-1$, we get

 \[
 P_{k+1}\left(Q_{0}(y)-y-1\right)=Q_{0}\left(P_{k}\left(Q_{0}(y)-y-1\right)\right)=Q_{0}\left(P_{k+1}(y)\right)
 \]

 Therefore, we can see that the proposition holds for the case $n=k+1$.

 <Step 2> Initialization

 <Step 2.1> As we saw in <Step 1>, if we find an initial solution $(P_{0}, Q_{0})$, we can continuously generate solutions. In particular, looking at the recursive formula for $P_{n}$, if the degree of $Q_{0}$ is at least 2, the degree of $P_{n}$ is twice the degree of $P_{n-1}$, which continuously increases. Therefore, by repeating this process, we can eventually make the degree of $P_{n}$ greater than 2024.

 <Step 2.2> Therefore, to find an initial solution, let's first try $P_{0}$ as a linear polynomial and $Q_{0}$ as a quadratic polynomial. It is easy to see that there are no possible cases. If both $P_{0}$ and $Q_{0}$ are quadratic polynomials, substituting $P_{0}(x)=ax^{2}+bx+c$, $Q_{0}(x)=ux^{2}+vx+w$ and using the method of undetermined coefficients, we find that

 \[
 P_{0}(x)=x^{2}+\frac{5}{2} x, \quad Q_{0}(x)=x^{2}+\frac{7}{2} x+\frac{3}{2}
 \]

 is a solution to the polynomial equation $P_{0}\left(Q_{0}(x)-x-1\right)=Q_{0}\left(P_{0}(x)\right)$.

 <Step 2.3> Now, repeating <Step 1> based on this initial solution, $(P_{n}, Q_{0})$ is also a solution for any $n$, and in particular, the degree of $P_{11}$ is 2048, which is greater than 2024.","(Partial)
 1. Found a non-constant solution $P$, $Q$ satisfies the equation.
(Almost)
 1. Described a contruction that enables us to increase the degree of the solutions, but did not fully verify.",Algebra,IMO-easy, ,Novel Problem,"import Mathlib

open Polynomial

/--
Prove that there exist two real-coefficient polynomials $P(x)$ and $Q(x)$ satisfying the following condition:

(Condition) The degree of the polynomial $P(x)$ is at least 2024, the degree of $Q(x)$ is at least 2, and for any real number $x$, the following holds:

\[
P(Q(x)-x-1)=Q(P(x))
\]
-/
theorem PBAdvanced007 : ∃ P Q : ℝ[X],
    2024 ≤ P.natDegree ∧ 2 ≤ Q.natDegree ∧ P.comp (Q - X - 1) = Q.comp P := by sorry"
PB-Advanced-008,"Let $\left\{F_{n}\right\}_{n \geq 1}$ be a sequence of integers satisfying $F_{1}=1$ and for $n \geq 2$,
 \[
 F_{n}=n^{F_{n-1}}.
 \]
 For example, $F_3 = 3^2= 9$ and $F_4 = 4^9$.

 Prove that for any positive integers $a, c$ and integer $b$, there exists a positive integer $n$ such that the following expression is an integer:

 \[
 \frac{a^{F_{n}}+n-b}{c}.
 \]","First, let's prove the following Lemma, which is an extension of Euler's theorem.

 <Lemma 1> Given positive integers $n$ and $a$, there exists a positive integer $N$ such that for any positive integer $k \geq N$, $a^{k+\phi(n)} \equiv a^{k}(\bmod n)$ holds.

 <Proof of Lemma 1> Let $p_1, p_2, \cdots, p_t$ the common prime divisors of $n$ and $a$, and let $\max_{1\le i\le t} v_{p_i} (n) =M$. Then, if we set $N=M$, the conclusion holds. Let $n=p_n \times q_n$, $a=p_a \times q_a$, where $p_n$ and $p_a$ represents the $p_1, p_2, \cdots, p_t$ part of the prime factorization of $n, a$, respectively. Note that $p_n | (p_1 p_2 \cdots p_t )^M$, and that $(a, q_n) =1$.

 First, if $k \geq M$, then $a^{k+\phi(n)} \equiv a^{k} \equiv 0\left(\bmod p_n \right)$. Also, $\phi(n)=\phi\left(p_n \right) \cdot \phi(q_n)$, so

 \[
 a^{k+\phi(n)} \equiv a^{k+\phi\left(p_n\right) \cdot \phi(q_n)} \equiv a^{k}(\bmod q_m)
 \]

 holds. Combining the two equations, we obtain $a^{k+\phi(n)} \equiv a^{k}(\bmod n)$, which completes the proof. $\qed$

 Let's also prove the following Lemma about $F_{n}$.

 <Lemma 2> Given positive integers $n$ and $a$, there exists a positive integer $N$ such that for any positive integer $k \geq N$, $F_{k} \equiv F_{k+\Phi(n)}(\bmod n)$ holds.

 Here, $\Phi(n)$ is the least common multiple of $n, \phi(n), \phi^{(2)}(n), \ldots$.

 <Proof of Lemma 2>

 We will prove this by induction on $n$.

 (1) (Base case) If $n=1$, the statement holds trivially.

 (2) (Induction step) Assume that the statement holds for all positive integers less than $n$.

 By the induction hypothesis, there exists a positive integer $N$ such that for $k \geq N$,

 \[
 F_{k} \equiv F_{k+\Phi(\phi(n))}(\bmod \phi(n))
 \]

 holds.

 Now, consider the least common multiple of $\Phi(\phi(n))$ and $n$, which is $\Phi(n)$. Since $F_{k+1}=(k+1)^{F_{k}}$, by <Lemma $1>$,

 \[
 (k+\Phi(n)+1)^{F_{k+\Phi(n)}} \equiv(k+1)^{F_{k+\Phi(n)}} \equiv(k+1)^{F_{k}}(\bmod n)
 \]

 holds for all sufficiently large $k$. Therefore, the statement also holds for $n$. $\qed$

 Now, let's prove the original problem.
 <Step 1>$ \operatorname{gcd}(\Phi(\phi(c)), c)<c$

 <Step 1.1> Factorize $c$ into primes as $c=p_{1}^{e_{1}} \cdots p_{k}^{e_{k}}$. Also, let $p_{1}<\cdots<p_{k}, e_{i}>0$.

 <Step 1.2> Then, $\phi(c)=p_{1}^{e_{1}-1} \cdots p_{k}^{e_{k}-1}\left(p_{1}-1\right) \cdots\left(p_{k}-1\right)$, and since $p_{i}-1$ is not a multiple of $p_{k}$,

 \[
 v_{p_{k}}(\phi(c))<v_{p_{k}}(c)
 \]

 holds. Also, the largest prime factor of $\phi(c)$ is less than or equal to $p_{k}$.

 <Step 1.3> Therefore, by induction, for all positive integers $r$, $v_{p_{k}}\left(\phi^{(r)}(c)\right)<v_{p_{k}}(c)$ holds.

 <Step 2> Applying strong induction on $c$

 We will prove the problem by strong mathematical induction on $c$.

 <Step 2.1> If $c=1$, the statement holds for any positive integer $n$.

 <Step 2.2> Assume that the statement holds for $c=1, \ldots c-1$, and let's prove the statement for $c$.

 First, choose $n$ large enough so that

 \[
 a^{F_{n}} \equiv a^{F_n+\phi(c)}(\bmod c)
 \]

 holds. By <Step 1>, $\operatorname{gcd}(c, \Phi(\phi(c)))<c$, so by the induction hypothesis,

 \[
 a^{F_{n}}+n \equiv b(\bmod \operatorname{gcd}(c, \Phi(\phi(c)))
 \]

 has infinitely many solutions for $n$.

 <Step 2.3> By <Lemma 2>, among these solutions, for sufficiently large $n$,

 \[
 a^{F_{n}} \equiv a^{F_{n+\Phi(\phi(c))}}(\bmod c)
 \]

 holds. Let this $n$ be $n_{0}$.

 <Step 2.4> Consider the following system of congruences for $n$:

 \[
 n \equiv n_{0}(\bmod \Phi(\phi(c))), a^{F_{n_{0}}}+n \equiv b(\bmod c)
 \]

 This system of congruences has $n_{0}$ as a solution modulo $\operatorname{gcd}(c, \Phi(\phi(c)))$, so by the Chinese Remainder Theorem, there exists an $n$ that satisfies both congruences modulo $\operatorname{lcm}(\Phi(\phi(c)), c)$. For this $n$,

 \[
 a^{F_{n}}+n \equiv a^{F_{n_{0}}}+n \equiv b(\bmod c)
 \]

 holds, so the desired equation holds, and therefore the induction hypothesis holds for $c$.

 By induction, for all $c$, there exist infinitely many $n$ that satisfy $a^{F_{n}}+n \equiv b(\bmod c)$.","(Partial)
 1. Obtained a period of $F_n \pmod c$.
 2. Showed that the period is not a multiple of $n$.
(Almost)
 1. Verification contains minor mistakes only.",Number theory,IMO-medium, ,Novel Problem,"import Mathlib

/-
Let $\left\{F_{n}\right\}_{n \geq 1}$ be a sequence of integers satisfying $F_{1}=1$ and for $n \geq 2$,
\[
F_{n}=n^{F_{n-1}}.
\]
For example, $F_3 = 3^2= 9$ and $F_4 = 4^9$.

Prove that for any positive integers $a, c$ and integer $b$, there exists a positive integer $n$ such that the following expression is an integer:

\[
\frac{a^{F_{n}}+n-b}{c}.
\]
-/
theorem PBAdvanced008
    (F : ℕ → ℕ) (hF1 : F 1 = 1) (hF_rec : ∀ n ≥ 1, F (n+1) = (n+1) ^ F n)
    (a b c : ℤ) (ha : 0 < a) (hc : 0 < c) :
    ∃ᵉ (n ≥ 1) (m : ℤ), (a ^ F n + n - b : ℚ) / c = m := by sorry"
PB-Advanced-009,"Let $H$ be the orthocenter of an acute-angled triangle $A B C$, and let $D, E, F$ be the feet of the altitudes from vertices $A, B, C$ to the opposite sides, respectively. Let $G$ be the midpoint of $B C$. Let $I, J$ be the feet of the perpendiculars from $B, C$ to $AG$, respectively. Let $K (\neq D)$ be the second intersection of the circumcircles of triangle $D I F$ and triangle $D J E$. Let $M$ be the midpoint of segment $A H$. Let $L$ be the foot of the perpendicular from $M$ to $A G$. Let $R (\neq G)$ be the second intersection of the circumcircle of triangle $A H G$ with $B C$. Let $S$ be the intersection of line $A H$ and $E F$. Let $N$ be the foot of the perpendicular from point $D$ to $R S$. Let $O$ be the midpoint of segment $D N$. Let line $D N$ intersect the circumcircle of triangle $D K L$ again at point $P (\neq D)$. Let $Q (\neq C)$ be the second intersection of the circumcircle of triangle $O C P$ and line $B C$. Prove that $A B=A Q$. "," First, let $Z$ be the intersection of lines $E F$ and $B C$. From the harmonic relation $(Z D, B C)=-1$, we have $$D G \times D R=D H \times D A=D B \times D C=D Z \times D G,$$ which implies $D Z=D R$. Therefore $R S$ is the reflection of $E F$ with respect to $A H$. Now, let's invert the figure with respect to the circle centered at $D$ with radius $\sqrt{D B \times D C}$, and then reflect it across $A D$. Let the image of a point or figure $X$ after this transformation be denoted by $X^{\prime}$. It can be easily seen that $(A, H),(B, C),(Z, G),(E, F),(S, M)$ are pairs of corresponding points under this transformation. $I$ and $J$ are the intersections of $A G$ with $\odot(A D B)$ and $\odot(A D C)$ respectively, so $I^{\prime}=C H \cap \odot(D H Z)$ and $J^{\prime}=BH\cap \odot(D H Z)$. Also, $L=A G \cap \odot(D E F)$, so $L^{\prime}=\odot(D H Z) \cap E F$, which means $L'$ is the foot of the perpendicular from $H$ to $E F$. Let this point be $X$. Also, if we consider $A_{1}$, the reflection of $A$ across $G$, then $$A F \times A B=A H \times A D=2 A M \times A D=2 A L \times A G=A L \times A A_{1}$$ so $\left(B F L A_{1}\right)$ and $\left(C E L A_{1}\right)$ are sets of concyclic points. Therefore $A_{1}^{\prime}$ maps to $T \equiv \odot(C E X) \cap \odot(B F X)$. Meanwhile, since $A$ is the radical center of $\odot\left(B F L A_{1}\right), \odot\left(C E L A_{1}\right), \odot(B F E C)$, points $A, X, T$ are collinear. Furthermore, by the collinearity of $A, G, A_1$, $(H, Z, D, T)$ is a set of concyclic points.  Therefore, $$ \angle X T C=\angle A E F=\angle A H F=\angle E H X=\angle J^{\prime} H X=\angle J^{\prime} T X $$ so $J^{\prime}, T, C$ are collinear. For similar reasons, $\left(I^{\prime}, T, B\right)$ are also collinear. Therefore, by Pappus's theorem, $B F \cap C E=A, C J^{\prime} \cap B I^{\prime}=T, F J^{\prime} \cap E I^{\prime}=Y$ are collinear. At this point, $K$ is mapped to $Y$ by the transformation, so the four points $H, K, L, D$ are concyclic. $H, K, L, D, A_1, P$ all lie on this circle. Therefore, after the transformation (inversion then reflection), $A, T, Y, X, P^{\prime}$ are all collinear. Now, let $U$ be the foot of the perpendicular from $D$ to $E F$, and let $V$ be the midpoint of $D U$. Then from the harmonic cross-ratio $$X(D U, V \infty)=(A H, S D)=-1$$ it can be seen that $V, X, A$ must be collinear. Therefore, $A, T, X, Y, V, P^{\prime}$ are all collinear. Therefore $V^{\prime}$ must lie on the circle passing through $H, K, L, D, A^{\prime}, P$. Since $V$ is the reflection of $O$ about $A H$, $D, O, V^{\prime}$ is collinear and so is $D, O, P$. Therefore, $P \equiv V^{\prime}$. Thus, from $$ D B \times D C=D O \times D P=D Q \times D C $$ we have $D B=D Q$. This means $A B=A Q$.","(Partial)
 1. Observed $R$ and $Z$ are symmetric about $AD$, and  $(A, H),(B, C),(Z, G),(E, F),(S, M)$ have inverse relationships respect to $D$.
 2. Constructs the point $T$, or a similar point on the line $AT$.
(Almost)
 1. Observes $O, P$ are the corresponding points of inversion w.r.t. $D$ with radius $\sqrt{D B \times D C}$.",Geometry,IMO-hard, ,Novel Problem,"import Mathlib

/-
Let $H$ be the orthocenter of an acute-angled triangle $A B C$, and let $D, E, F$ be the feet of the altitudes from vertices $A, B, C$ to the opposite sides, respectively. Let $G$ be the midpoint of $B C$. Let $I, J$ be the feet of the perpendiculars from $B, C$ to $AG$, respectively. Let $K (\neq D)$ be the second intersection of the circumcircles of triangle $D I F$ and triangle $D J E$. Let $M$ be the midpoint of segment $A H$. Let $L$ be the foot of the perpendicular from $M$ to $A G$. Let $R (\neq G)$ be the second intersection of the circumcircle of triangle $A H G$ with $B C$. Let $S$ be the intersection of line $A H$ and $E F$. Let $N$ be the foot of the perpendicular from point $D$ to $R S$. Let $O$ be the midpoint of segment $D N$. Let line $D N$ intersect the circumcircle of triangle $D K L$ again at point $P (\neq D)$. Let $Q (\neq C)$ be the second intersection of the circumcircle of triangle $O C P$ and line $B C$. Prove that $A B=A Q$.
-/

open Affine Simplex EuclideanGeometry
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

theorem PBAdvanced009
    -- Acute-angled triangle $A, B, C$
    (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
    (acute : AcuteAngled ⟨![A, B, C], tri⟩)
    (D : ℝ²) (hD : D = altitudeFoot ⟨![A, B, C], tri⟩ 0) -- D is the foot from A
    (E : ℝ²) (hE : E = altitudeFoot ⟨![A, B, C], tri⟩ 1) -- E is the foot from B
    (F : ℝ²) (hF : F = altitudeFoot ⟨![A, B, C], tri⟩ 2) -- F is the foot from C
    (H : ℝ²) (hH : H = Triangle.orthocenter ⟨![A, B, C], tri⟩) -- H is the orthocenter
    (G : ℝ²) (hG : G = midpoint ℝ B C) -- $G$ is the midpoint of $B C$
    -- $I, J$ are the feet of the perpendiculars from $B, C$ to $AG$
    (I : ℝ²) (hI : I = orthogonalProjection (affineSpan ℝ {A, G}) B)
    (J : ℝ²) (hJ : J = orthogonalProjection (affineSpan ℝ {A, G}) C)
    -- K ≠ D is the second intersection of the circumcircles of triangles $D I F$ and $D J E$
    (K : ℝ²) (hK : K ≠ D ∧ Cospherical {K, D, I, F} ∧ Cospherical {K, D, J, E})
    (M : ℝ²) (hM : M = midpoint ℝ A H) -- M is the midpoint of segment $A H$
    -- Let $L$ be the foot of the perpendicular from $M$ to $A G$.
    (L : ℝ²) (hL : L = orthogonalProjection (affineSpan ℝ {A, G}) M)
    -- Let R ≠ G be the second intersection of the circumcircle of triangle $A H G$ with $B C$.
    (R : ℝ²) (hR : (R ≠ G ∧ Cospherical {R, A, H, G} ∧ Collinear ℝ {R, B, C}))
    -- Let $S$ be the intersection of line $A H$ and $E F$.
    (S : ℝ²) (hS : Collinear ℝ {S, A, H} ∧ Collinear ℝ {S, E, F})
    -- Let $N$ be the foot of the perpendicular from point $D$ to $R S$.
    (N : ℝ²) (hN : N = orthogonalProjection (affineSpan ℝ {R, S}) D)
    -- Let $O$ be the midpoint of segment $D N$.
    (O : ℝ²) (hO : O = midpoint ℝ D N)
    -- Let line $D N$ intersect the circumcircle of triangle $D K L$ again at point $P (\neq D)$.
    (P : ℝ²) (hP : P ≠ D ∧ Collinear ℝ {P, D, N} ∧ Cospherical {P, D, K, L})
    -- Let $Q (\neq C)$ be the second intersection of the circumcircle of triangle $O C P$ and line $B C$.
    (Q : ℝ²) (hQ : Q ≠ C ∧ Cospherical {Q, O, C, P} ∧ Collinear ℝ {Q, B, C}) :
    -- Prove that $A B=A Q$.
    dist A B = dist A Q := by sorry"
PB-Advanced-010,"Let $O$ and $G$ be the circumcenter and centroid of a non-isosceles triangle $ABC$, respectively. Let $H$ be the foot of the perpendicular from $A$ to $BC$, and let $M$ be the midpoint of $BC$. For a point $X$ on the line $OG$, let the line $BX$ intersect $AC$ at $P$, and let the line $CX$ intersect $AB$ at $Q$. Let $H_1$ be the foot of the perpendicular from $P$ to the line $AB$, and let $K$ be the reflection of $A$ about $H_1$. Let $T$ be the intersection of the circumcircle of triangle $KPQ$ and the circumcircle of triangle $PHM$. Prove that as $X$ moves along the line $OG$, $T$ moves along a fixed circle.","The most difficult part of this problem is to observe that $(B, T, P, C)$ are concyclic. If this holds, let $Y$ be the intersection of $TP$ and $BC$. Then $YH \cdot YM = YT \cdot YP = YB \cdot YC$,
 which means that $Y$ is the point such that $(B, H, C, Y)$ is a harmonic division. This is a fixed point. Thus, $T$ lies on the inversion of $AC$ with respect to the circle centered at $Y$ with radius $\sqrt{YB \cdot YC}$. This is a fixed circle.

 \textbf{Claim:} $(B, T, P, C)$ are concyclic. This means that the circumcircles of $\triangle BPC$, $\triangle PHM$, and $\triangle KPQ$ are coaxial. We will use the following well-known Lemma: For two circles, the locus of points where the ratio of the powers with respect to the two circles is constant is a circle coaxial with the two circles.
 Now, using the Lemma, we see that it suffices to show that: The ratio of the powers of $B$ and $C$ with respect to the circumcircles of $\triangle PHM$ and $\triangle KPQ$ are the same.

 1) The ratio of powers of $B$ and $C$ with respect to the circumcircle of $\triangle PHM$ is $\frac{BH \cdot BM}{CH \cdot CM} = \frac{BH}{CH}$.

 2) Let's compute the ratio of powers of $B$ and $C$ with respect to the circumcircle of $\triangle KPQ$. Let $H_2$ be the foot of the perpendicular from $Q$ to $AC$, and let $L$ be the reflection of $A$ across $H_2$. Since $\angle QKP = \angle QLP = 180^\circ - \angle A$, $(K, L, P, Q)$ are concyclic. The power of $B$ is $BK \cdot BQ$, and the power of $C$ is $CL \cdot CP$. We want to show that $\frac{CL \cdot CP}{BK \cdot BQ} = \frac{CH}{BH}$.

 Let $D, E$ be the intersection of the Euler line of $\triangle ABC$ with $AB, AC$ respectively. Let $R$ be the intersection of $AX$ and $BC$.
 By Ceva's theorem, $\frac{CP}{BQ} = \frac{AP}{AQ} \times \frac{CR}{BR}$. $BK = BA - KA = BA - 2AP \cos A$, $CL = CA - LA = CA - 2AQ \cos A$.
 Thus $\frac{CL \cdot CP}{BK \cdot BQ} = \frac{CA - 2AQ \cos A}{BA - 2AP \cos A} \times \frac{AP}{AQ} \times \frac{CR}{BR} = \frac{\frac{CA}{AQ} - 2 \cos A}{\frac{BA}{AP} - 2 \cos A} \times \frac{CR}{BR}$.

 Let $\frac{XE}{DX} = \lambda$.
 $\frac{CR}{BR} = \frac{AC \sin \angle CAR}{AB \sin \angle BAR} = \frac{AC}{AB} \cdot \frac{AD}{AE} \cdot \frac{XE}{DX} = \frac{AC \cdot AD}{AB \cdot AE} \lambda$.
 By Menelaus' theorem on $\triangle ADE$ and line $CXQ$,
 $\frac{AQ}{QD} \cdot \frac{CE}{AC} \cdot \frac{DX}{EX} = 1$. Thus, $\frac{AQ}{QD} = \frac{AC}{CE} \lambda$, so $AQ = \frac{1}{1 + \frac{AC}{CE} \lambda} AD = \frac{AD}{1+\lambda \frac{AC}{CE}}.$
 Similarly, $AP = \frac{1}{1 + \frac{1}{\lambda} \frac{AB}{BD}} AE$.
 Thus $\frac{CL \cdot CP}{BK \cdot BQ} = \frac{\frac{CA}{AQ} - 2 \cos A}{\frac{BA}{AP} - 2 \cos A} \times \frac{CR}{BR}$ has the form $\frac{(\text{linear in } \lambda)}{(\text{linear in } \lambda)}$. That is, if it is $\frac{CH}{BH}$ for two values of $\lambda$, then it is $\frac{CH}{BH}$ for all $\lambda$.

 Case 1) $X = H$.
 In this case, $P, Q$ are the feet of the altitudes from $B, C$ respectively. Thus $KL \parallel BC$. Thus $\frac{CL \cdot CP}{BK \cdot BQ} = \frac{CH \cdot CB}{BH \cdot BC} = \frac{CH}{BH}$.

 Case 2) $X = G$.
 In this case, $P, Q$ are the midpoints of $AC, AB$, and $K, L$ are the feet of altitudes from $C, B$ respectively. Thus $\frac{CL \cdot CP}{BK \cdot BQ} = \frac{CH \cdot \frac{CB}{2}}{BH \cdot \frac{BC}{2}} = \frac{CH}{BH}$.
 Since $H \neq G$ due to scalene condition, we have two different values of $\lambda$, completing the proof.","(Partial)
 1. Proved that $B, T, P, C$ lie on a circle.
 2. Observed that it suffices to show that the ratio of powers of $B$ and $C$ respect to the circumcircles of $\triangle PHM$ and $\triangle KPQ$ are the same.
(Almost)
 1. In addition to items 1 and 2 of (Partial), observed that it is enough to prove $2$ of (Partial) only for two different $X$.
 2. Verification contains minor mistakes only.",Geometry,IMO-medium, ,Novel Problem,"import Mathlib

/-
Let $O$ and $G$ be the circumcenter and centroid of a non-isosceles triangle $ABC$, respectively. Let $H$ be the foot of the perpendicular from $A$ to $BC$, and let $M$ be the midpoint of $BC$. For a point $X$ on the line $OG$, let the line $BX$ intersect $AC$ at $P$, and let the line $CX$ intersect $AB$ at $Q$. Let $H_1$ be the foot of the perpendicular from $P$ to the line $AB$, and let $K$ be the reflection of $A$ about $H_1$. Let $T$ be the intersection of the circumcircle of triangle $KPQ$ and the circumcircle of triangle $PHM$. Prove that as $X$ moves along the line $OG$, $T$ moves along a fixed circle.
-/
open Affine.Simplex EuclideanGeometry
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

theorem PBAdvanced010
    -- a triangle ABC
    (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
    -- ABC is non-isosceles
    (n_isosceles : [dist A B, dist A C, dist B C].Nodup )
    (O : ℝ²) (hO : O = circumcenter ⟨![A, B, C], tri⟩) -- O is the circumcenter
    (G : ℝ²) (hG : G = Finset.centroid ℝ .univ ![A, B, C]) -- G is the centroid
    (H : ℝ²) (hH : H = altitudeFoot ⟨![A, B, C], tri⟩ 0) -- H is the foot from A
    (M : ℝ²) (hM : M = midpoint ℝ B C) : -- $M$ is the midpoint of $B C$
    -- There is a sphere ω independent of the point `X`
    ∃ ω : Sphere ℝ²,
    -- For a point $X$ on the line $OG$,
    ∀ (X : ℝ²) (hX : Collinear ℝ {X, O, G})
    -- let the line $BX$ intersect $AC$ at $P$,
    (P : ℝ²) (hP : Collinear ℝ {P, B, X} ∧ Collinear ℝ {P, A, C})
    -- let the line $CX$ intersect $AB$ at $Q$.
    (Q : ℝ²) (hQ : Collinear ℝ {Q, C, X} ∧ Collinear ℝ {Q, A, B})
    -- Let $H_1$ be the foot of the perpendicular from $P$ to the line $AB$,
    (H₁ : ℝ²) (hH₁ : H₁ = orthogonalProjection (affineSpan ℝ {A, B}) P)
    -- let $K$ be the reflection of $A$ about $H_1$.
    (K : ℝ²) (hK : K = reflection (affineSpan ℝ {H₁}) A)
    -- Let $T$ be the intersection of the circumcircle of triangle $KPQ$ and the circumcircle of triangle $PHM$.
    (T : ℝ²) (hT : T ≠ P ∧ Cospherical {T, K, P, Q} ∧ Cospherical {T, P, H, M}),
    -- Prove that as $X$ moves along the line $OG$, $T$ moves along a fixed circle.
    T ∈ ω := by sorry"
PB-Advanced-011,"Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $$yf(yf(x)+1) = f(\frac{1}{x} + f(y))$$ for all $x, y \in \mathbb{R}^+$","
Let $P(x,y)$ be the given assertion.
$$P(c+1, \frac{c}{f(c+1)}): \quad c= \frac{c}{f(c+1)} f(\frac{c}{f(c+1)} \times f(c+1)+1)= f(\frac{1}{c+1} + f(\frac{c}{f(c+1)}))$$, hence $f$ is surjective. \\ Now assume that $f(x) = f(b)$ for some $a>b>0$. Since $f$ is surjective, there exists some $t\in \mathbb{R}^+$ such that $f(t)=1$. comparing $P(a,t)$ and $f(b,t)$ gives $f(1+\frac{1}{a}) = f(1+\frac{1}{b})$. Again since $f$ is surjective, there exists some $t\in \mathbb{R}^+$ such that $f(t)=\frac{1}{ab}$. Comparing $P(s,a)$ and $P(s,b)$ gives $af(\frac{1}{b} +1) = b f(\frac{1}{a}+1)$, which is a contradiction. Therefore $f$ is injective. \\
Now $P(x,1)$ implies $f(x) = \frac{1}{x}+c$, and by surjectivity, $c=0$, thus $f(x) = \frac{1}{x}$, which is indeed a solution.
","(Partial)
 1. Considered $(x,y) = (c+1, \frac{c}{f(c+1)})$.
 2. Proved that $f$ is subjective.
(Almost)
 1. Proved that $f$ is injective.",Algebra,IMO-medium,$f(x) = \frac{1}{x}$,Novel Problem,"import Mathlib

/-
Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that $$yf(yf(x)+1) = f(\frac{1}{x} + f(y))$$ for all $x, y \in \mathbb{R}^+$
Solution: ${fun x => 1/x}$
-/
open Classical

theorem PBAdvanced011 :
    {f : {x : ℝ // 0 < x} → {x : ℝ // 0 < x} | ∀ x y, y * f (y * f x + 1) = f (1 / x + f y)}
      = {fun x ↦ 1 / x} := by sorry"
PB-Advanced-012,"Let $p$ be a prime number and $a, b$ be positive integers such that

 \[
 p^{n}=a^{4}+b^{4}
 \]

 for some integer $n \geq 2$. Prove that $n \geq 5$.","First, let's summarize the facts that are frequently used in Diophantine equations. We will omit the proofs of well-known facts. The following lemma characterizes Pythagorean triples and is used repeatedly.

 <Lemma 1> If $a, b, c$ are positive integers with $ab \neq 0$ satisfying $a^{2}+b^{2}=c^{2}$, then there exist positive integers $m \neq n$ and $k$ such that

 \[
 (a, b, c)=k\left(2 m n, m^{2}-n^{2}, m^{2}+n^{2}\right) \text { or } k\left(m^{2}-n^{2}, 2 m n, m^{2}+n^{2}\right).
 \]

 In particular, if $a, b$ are relatively prime, then $k=1$ and $m$ and $n$ have different parity, and one of $a, b$ is odd and the other is even. The odd one is expressed as $m^{2}-n^{2}$ and the even one is expressed as $2mn$. Also, in this case, $c$ is always odd.
 <Proof of Lemma 1> Omitted. \qed
 <Lemma 2> If $p$ is a prime number or 1, and positive integers $x$ and relatively prime positive integers $a, b$ satisfy

 \[
 p a b=x^{2},
 \]

 then there exist positive integers $u, v$ such that $(a, b, x)=\left(p u^{2}, v^{2}, p u v\right)$ or $\left(u^{2}, p v^{2}, p u v\right)$.
 <Proof of Lemma 2> Omitted. \qed

 <Lemma 3> If $a^{2}+b^{2}$ is a multiple of 4 for integers $a, b$, then both $a$ and $b$ are even.

 <Proof of Lemma 3> $a$ and $b$ must have the same parity, but the square of an odd number has a remainder of 1 when divided by 4, so if $a, b$ have the same parity as odd numbers, then $a^{2}+b^{2}$ has a remainder of 2 when divided by 4, so it cannot be a multiple of 4. \qed

 Now, let's proceed with the proof of the main problem using these facts.
 <Step 1> There are no positive integers $x, y, z$ satisfying $x^{4}+y^{4}=z^{2}$. Therefore, $n$ is odd.

 <Step 1.1> Using proof by contradiction, assume that there exists a positive integer solution $x, y, z$. Among these solutions, choose the solution $(x, y, z)$ with the smallest $z$. If there is a prime number $q$ that divides two of $x, y, z$, then it must also divide the remaining number, and in this case, $\left(\frac{x}{q}, \frac{y}{q}, \frac{z}{q^{2}}\right)$ becomes a solution, contradicting the minimality of $z$. Therefore, this solution is pairwise relatively prime.

 <Step 1.2> Since $\left(x^{2}, y^{2}, z\right)$ is a Pythagorean triple, by <Lemma 1>, there exist relatively prime positive integers $u, v$ with different parity such that


 \begin{equation*}
 x^{2}=u^{2}-v^{2}, y^{2}=2 u v, z=u^{2}+v^{2} \tag{1.1}
 \end{equation*}


 We can assume that $u > v$ without loss of generality. Since $y^{2}=2 u v$ and $u, v$ are relatively prime, by <Lemma 2>, there exist positive integers $s, t$ such that $u=2 s^{2}, v=t^{2}$ or $u=s^{2}, v=2 t^{2}$.

 (Case 1) $u=2 s^{2}, v=t^{2}$

 $x^{2}+t^{4}=4 s^{4}$. By <Lemma 3>, both $x$ and $t$ must be even, but from $x^{2}=u^{2}-v^{2}$, $x$ is odd, which is a contradiction.

 (Case 2) $u=s^{2}, v=2 t^{2}$

 $x^{2}+4 t^{4}=s^{4}$ and since $t|v, v| y^{2}$,

 \[
 \operatorname{gcd}(x, t)|\operatorname{gcd}(x, v)| \operatorname{gcd}\left(x, y^{2}\right)=1.
 \]

 Therefore, $x$ and $t^{2}$ are also relatively prime, and by <Lemma 1>, there exist relatively prime positive integers $q, r$ with different parity (since $x$ is odd by (1.1)) such that

 \[
 x=q^{2}-r^{2}, 2 t^{2}=2 q r, s^{2}=q^{2}+r^{2}.
 \]

 Since $t^{2}=q r$, by <Lemma 2>, there exist positive integers $x', y'$ such that $q=x^{\prime 2}, r=y^{\prime 2}$, and thus by the above equation, $x^{\prime 4}+y^{\prime 4}=s^{2}$. Since

 \[
 s=\sqrt{u} \leq u<u^{2}+v^{2}=z,
 \]

 this contradicts the minimality of $(x, y, z)$. Therefore, the proof is complete.

 <Step 2> If $p^{3}=a^{4}+b^{4}$, then there exist positive integers $x, y, z$ such that $x^{2}+3 y^{4}=z^{4}$.

 <Step 2.1> If $p \mid a$, then clearly $p \mid b$ must also hold, and vice versa. Therefore, $a, b$ are either both multiples of $p$ or not. However, if the former is the case, the right-hand side becomes a multiple of $p^{4}$, which is a contradiction, so $p$ is relatively prime to both $a$ and $b$.

 <Step 2.2> Therefore, $p$ is a divisor of the sum of squares, and since $(p, a)=(p, b)=1$, $p$ is a prime number of the form $4 k+1$, so it can be expressed in the form $p=c^{2}+d^{2}$. By Jacobi's theorem, there are only two ways (considering order) to express $p^{3}$ as the sum of perfect squares:

 \[
 (p c)^{2}+(p d)^{2} \text { and } \left(c\left(c^{2}-3 d^{2}\right)\right)^{2}+\left(d\left(3 c^{2}-d^{2}\right)\right)^{2}.
 \]

 By (1), the first case cannot be expressed as $a^{4}+b^{4}$, so

 \[
 a^{2}=\left|c\left(c^{2}-3 d^{2}\right)\right|, \quad b^{2}=\left|d\left(3 c^{2}-d^{2}\right)\right|
 \]

 (or vice versa, but we can ignore this case due to symmetry) must hold. Letting $C=|c|, D=|d|$, we have


 \begin{equation*}
 a^{2} b^{2}=C D \times\left|C^{2}-3 D^{2}\right| \times\left|3 C^{2}-D^{2}\right| \tag{1.2}
 \end{equation*}


 and $\operatorname{gcd}(C, D)=1$. Now, we divide into cases depending on whether $C D$ is a multiple of 3.

 (Case 1) If $3 \mid C D$

 Without loss of generality, assume $3 \mid C$. Then, from

 \[
 \left(\frac{a b}{3}\right)^{2}=\frac{C}{3} \cdot D \cdot\left|3\left(\frac{C}{3}\right)^{2}-D^{2}\right| \cdot\left|3 C^{2}-D^{2}\right|,
 \]

 since the four terms are pairwise relatively prime, there exist positive integers $e, f, g, h$ such that

 \[
 C=3 e^{2}, \quad D=f^{2},\left|3 e^{4}-f^{4}\right|=g^{2},\left|27 e^{4}-f^{4}\right|=h^{2}.
 \]

 Then, by $\bmod 3$, we have $g^{2}+3 e^{4}=f^{4}$.

 (Case 2) If $3 \nmid CD$

 From (1.2), since the four terms are pairwise relatively prime, there exist positive integers $e, f, g, h$ such that

 \[
 C=e^{2}, \quad D=f^{2},\left|e^{4}-3 f^{4}\right|=g^{2},\left|3 e^{4}-f^{4}\right|=h^{2}.
 \]

 Similarly, by $\bmod 3$, we have $e^{4}-3 f^{4}=g^{2}$.

 Therefore, in both cases, there exists a positive integer solution to $x^{2}+3 y^{4}=z^{4}$.

 <Step 3> There are no positive integers $x, y, z$ satisfying $x^{2}+3 y^{4}=z^{4}$.

 Using proof by contradiction, assume that there exists a positive integer solution $x, y, z$. Among these solutions, choose the solution $(x, y, z)$ with the smallest $z$. If there is a prime number $q$ that divides two of $x, y, z$, then it must also divide the remaining number, and in this case, $\left(\frac{x}{q^{2}}, \frac{y}{q}, \frac{z}{q}\right)$ becomes a solution, contradicting the minimality of $z$. Therefore, this solution is pairwise relatively prime.

 Now, rewriting the equation as


 \begin{equation*}
 3 y^{4}=\left(z^{2}-x\right)\left(z^{2}+x\right) \tag{1.3}
 \end{equation*}


 we can divide into two cases since $\operatorname{gcd}\left(z^{2}-x, z^{2}+x\right)=1$ or 2.

 (Case 1) If $x$ and $z$ have the same parity

 By <Lemma 2>, there exist relatively prime positive integers $u, v$ such that

 \[
 z^{2}-x=2 u^{2}, z^{2}+x=6 v^{2}, y^{2}=2 u v \quad \text { or } \quad z^{2}+x=2 u^{2}, z^{2}-x=6 v^{2}, y^{2}=2 u v.
 \]

 In both cases, we have


 \begin{equation*}
 z^{2}=u^{2}+3 v^{2}, y^{2}=2 u v. \tag{1.4}
 \end{equation*}


 Since $y$ is even, one of $u, v$ must be even.

 <Step 3.1> If $u$ is even, then $v$ is odd, and then $z^{2}=u^{2}+3 v^{2} \equiv 3(\bmod 4)$, which is a contradiction.

 <Step 3.2> If $v$ is even, then by <Lemma 2>, there exist positive integers $s, t$ such that

 \[
 u=s^{2}, v=2 t^{2},
 \]

 and substituting this back into (1.4) gives $s^{4}+12 t^{4}=z^{2}$. Since both $s$ and $z$ are odd, this equation can be written as

 \[
 \frac{z+s^{2}}{2} \cdot \frac{z-s^{2}}{2}=3 t^{4},
 \]

 and since $\operatorname{gcd}\left(\frac{z+s^{2}}{2}, \frac{z-s^{2}}{2}\right)=1$, by <Lemma 2>, there exist relatively prime positive integers $q, r$ such that

 \[
 \frac{z+s^{2}}{2}=q^{4}, \frac{z-s^{2}}{2}=3 r^{4} \quad \text { or } \quad \frac{z+s^{2}}{2}=3 r^{4}, \frac{z-s^{2}}{2}=q^{4}.
 \]

 In the second case, we have $s^{2}=3 r^{4}-q^{4}$, but since $q$ and $s$ cannot both be multiples of 3 (in this case, $r$ would also be a multiple of 3, leading to a contradiction), $s^{2}+q^{4}$ cannot be a multiple of 3, which is a contradiction. In the remaining case, we have $s^{2}=q^{4}-3 r^{4}$, but since $q \leq q^{4} = \frac{z+s^2}{2} < z$, this contradicts the minimality of $z$.

 (Case 2) If $x$ and $z$ have different parity

 In this case, by (1.3) and <Lemma 2>, there exist relatively prime positive integers $u, v$ such that

 \[
 z^{2}+x=u^{4}, z^{2}-x=3 v^{4}, y^{2}=u v \quad \text { or } \quad z^{2}-x=3 v^{4}, z^{2}+x=u^{4}, y^{2}=u v.
 \]

 In both cases, we have $2 z^{2}=u^{4}+3 v^{4}$, but since $z$ and $u$ cannot both be multiples of 3, this leads to a contradiction modulo 3.

 Therefore, by <Step 1>, $n \neq 2,4$, and by <Step 2>, $n \neq 3$, so the proof is complete.","(Partial)
 1. Observed and verified that $x^2=y^4+z^4$ has no nontrivial solution
 2. Transformed the equation $p^3=a^4+b^4$ into $p=c^2+d^2, a^2=c(c^2-3d^2), b^2=d(3c^2-d^2)$
(Almost)
 1. Transformed the given equation to $x^2+3y^4=z^4$ or its equivalent form, but failed to prove that it has no nontrivial solution",Number theory,IMO-hard, ,Novel Problem,"import Mathlib

/-
Let $p$ be a prime number and $a, b$ be positive integers such that

\[
p^{n}=a^{4}+b^{4}
\]

for some integer $n \geq 2$. Prove that $n \geq 5$.
-/
theorem PBAdvanced012
    (p a b n : ℕ) (hp : p.Prime) (ha : a ≠ 0) (hb : b ≠ 0)
    (hp' : p ^ n = a ^ 4 + b ^ 4) (hn : 2 ≤ n) : 5 ≤ n := by sorry"
PB-Advanced-013,"For an integer $n \geq 2$, let $a_{1} \leq a_{2} \leq \cdots \leq a_{n}$ be positive real numbers satisfying $a_{1} a_{2} \cdots a_{n}=1$. For each $k=1,2, \cdots, n$, define $b_{k}=2^{k}\left(1+a_{k}^{2^{k}}\right)$. Prove that the following inequality holds:

 \[
 \frac{1}{2}-\frac{1}{2^{n+1}} \leq \frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{n}}
 \]","The following Lemma is often used in problems involving multiple variables or sequences.

 <Lemma> For positive real numbers $x, y$ satisfying $xy \geq 1$, the following inequality holds:

 \[
 \frac{1}{1+x}+\frac{1}{1+y} \geq \frac{2}{1+\sqrt{x y}}
 \]

 <Proof of Lemma> The proof is a simple calculation. Multiplying both sides by $(1+\sqrt{x y})(1+x)(1+y)$, the given inequality simplifies to $(\sqrt{x y}-1)(\sqrt{x}-\sqrt{y})^{2} \geq 0$, which holds trivially under the condition $x y \geq 1$. Therefore, the inequality is proven.\qed

 Now, let's prove the main problem. We will prove by induction that the following inequality holds for each $k=n, n-1, \cdots, 1$:

 \begin{equation*}
 \frac{1}{2^{n+1}}+\frac{1}{b_{n}}+\frac{1}{b_{n-1}}+\cdots+\frac{1}{b_{k}} \geq \frac{1}{2^{k-1}} \frac{1}{1+\left(a_{n} a_{n-1} \cdots a_{k}\right)^{2^{k-1}}} \tag{1.1}
 \end{equation*}

 In particular, if this inequality holds for $k=1$, then the inequality in the problem is proven due to the condition $a_{1} a_{2} \cdots a_{n}=1$.
 <Step 1> First, for $k=n$, since $a_{n} \geq 1$, we obtain the following by <Lemma>:

 \[
 \frac{1}{2^{n+1}}+\frac{1}{b_{n}}=\frac{1}{2^{n}}\left(\frac{1}{1+1}+\frac{1}{1+a_{n}^{2^{n}}}\right) \geq \frac{1}{2^{n-1}} \frac{1}{1+a_{n}^{2^{n-1}}}
 \]

 Therefore, (1.1) holds.

 <Step 2> Next, let's prove that if (1.1) holds for $k=m$, then it also holds for $k=m-1$ (where $2 \leq m \leq n$).

 <Step 2.1> First, by the induction hypothesis, we obtain the following inequality:

 \[
 \frac{1}{2^{n+1}}+\frac{1}{b_{n}}+\frac{1}{b_{n-1}}+\cdots+\frac{1}{b_{m}}+\frac{1}{b_{m-1}} \geq \frac{1}{2^{m-1}}\left[\frac{1}{1+\left(a_{n} a_{n-1} \cdots a_{m}\right)^{2^{m-1}}}+\frac{1}{1+a_{m-1}^{2_{m-1}^{m}}}\right]
 \]

 <Step 2.2> By the condition of the problem,

 \[
 \left(a_{n} a_{n-1} \cdots a_{m}\right)^{2^{m-1}} \cdot a_{m-1}^{2^{m-1}} \geq 1
 \]

 Therefore, by <Lemma>,

 \[
 \frac{1}{2^{m-1}}\left[\frac{1}{1+\left(a_{n} a_{n-1} \cdots a_{m}\right)^{2^{m-1}}}+\frac{1}{1+a_{m-1}^{2^{m-1}}}\right] \geq \frac{1}{2^{m-2}} \frac{1}{1+\left(a_{n} a_{n-1} \cdots a_{m} a_{m-1}\right)^{2^{m-2}}}
 \]

 This confirms that inequality (1.1) also holds for $k=m-1$.

 In conclusion, by mathematical induction, (1.1) holds for $k=1$, completing the proof.","(Partial)
 1. Tried to apply induction in an appropriate direction.
 2. Found any localization inequality, e.g., the one equivalent to the Lemma of solution.
 (Almost)
 1. Verification contains minor mistakes only.",Algebra,IMO-easy, ,Novel Problem,"import Mathlib

/-
For an integer $n \geq 2$, let $a_{1} \leq a_{2} \leq \cdots \leq a_{n}$ be positive real numbers satisfying $a_{1} a_{2} \cdots a_{n}=1$. For each $k=1,2, \cdots, n$, define $b_{k}=2^{k}\left(1+a_{k}^{2^{k}}\right)$. Prove that the following inequality holds:

\[
\frac{1}{2}-\frac{1}{2^{n+1}} \leq \frac{1}{b_{1}}+\frac{1}{b_{2}}+\cdots+\frac{1}{b_{n}}
\]
-/
theorem PBAdvanced013
    (n : ℕ) (hn : 2 ≤ n) (a b : ℕ → ℝ)
    (ha : ∀ i ∈ Finset.Icc 1 n, 0 < a i)
    (ha' : ∏ i ∈ Finset.Icc 1 n, a i = 1)
    (ha'' : MonotoneOn a (Finset.Icc 1 n))
    (hb : ∀ i, b i = 2 ^ i * (1 + a i ^ (2 ^ i))) :
    1 / 2 - 1 / 2 ^ (n + 1) ≤ ∑ i ∈ Finset.Icc 1 n, 1 / b i := by sorry"
PB-Advanced-014,"Two distinct positive integers \( a \) and \( b \) are written on a blackboard. At each step, you must choose one of the two operations — either adding 2 or multiplying by 3 — for each number separately, and apply the chosen operation to both numbers (possibly using different operations). You must apply an operation to both numbers every step. Find all pairs $(a, b)$ such that we can make the two numbers on the blackboard equal after a finite number of steps.","The answer is that $a$ and $b$ are both even, or $a$ and $b$ are both odd and have the same remainder when divided by 4.
 <Step 1> Finding an invariant

 Let's observe how the remainder when divided by 4 changes under the operations. First, for odd numbers, both operations change the remainder as follows:

 $1 \rightarrow 3,3 \rightarrow 1$

 For even numbers, adding 2 changes the remainder as follows:

 \[
 0 \rightarrow 2,2 \rightarrow 0
 \]

 and multiplying by 3 changes the remainder as follows:
 \[
 0 \rightarrow 0,2 \rightarrow 2
 \]


 <Step 2> Finding impossible cases

 <Step 2.1> From the observation in <Step 1>, we can find conditions on $a$ and $b$ for which the conclusion of the problem does not hold. Let's consider the following two cases:

 <Step 2.2> (Case 1) $a$ and $b$ have different parity.

 Since the parity of the two numbers does not change under the operations, the two numbers cannot become equal. Therefore, the answer is impossible in this case.

 <Step 2.3> (Case 2) $a$ and $b$ are both odd and have different remainders when divided by 4.

 Since odd numbers change their remainders when divided by 4 under each operation, the remainders of the two numbers when divided by 4 will always be different, regardless of the operations. Therefore, the answer is impossible in this case.

 Now, let's prove that the answer is possible for $a$ and $b$ that do not satisfy (Case 1) and (Case 2).

 <Step 3> Showing that the operations exist for the remaining cases.

 <Step 3.1> The remaining cases are as follows:

 (Case 3) $a, b$ are both odd and have the same remainder when divided by 4.

 (Case 4) $a, b$ are both even.

 <Step 3.2> First, let's show that we can make $a \equiv b(\bmod 4)$ for both cases by applying operations to $a$ and $b$. In (Case 3), $a \equiv b(\bmod 4)$ is already satisfied.

 In (Case 4), if $a \neq b(\bmod 4)$, then by applying the operation of changing $a$ to $3a$ and $b$ to $b+2$, we can see that $3a \equiv b+2(\bmod 4)$. Therefore, we can make the two numbers congruent modulo 4.

 <Step 3.3> Now, let's write the two numbers with the same remainder when divided by 4 as $x$ and $x+4n$. Let's add 2 to $x$ for $k$ times and then multiply it by 3. On the other hand, let's multiply $x+4n$ by 3 and then add 2 to it for $k$ times. As a result of the operations, $x$ becomes $3(x+2k)$ and $x+4n$ becomes $3x+12n+2k$.

 Therefore, if we set $k=3n$, then the two numbers become equal.

 [Short answer] $a$ and $b$ are both even, or $a$ and $b$ are both odd and have the same remainder when divided by 4.","(Partial)
 1. Found a correct invariant $\pmod 4$.
(Almost)
 1. Verification contains minor mistakes only.",Combinatorics,IMO-medium,"$a$ and $b$ are both even, or $a$ and $b$ are both odd and have the same remainder when divided by 4.",Novel Problem,"import Mathlib

open Relation

/--
The single-step operation in the problem: either adding 2 or multiplying by 3.
-/
def operation (x x' : ℕ) : Prop :=
  x' = x + 2 ∨ x' = 3 * x

/--
A single step in the blackboard game, applied simultaneously to both numbers.
We define the relation on pairs of natural numbers.
-/
def single_step (p q : ℕ × ℕ) : Prop :=
  operation p.fst q.fst ∧ operation p.snd q.snd

/--
The pair `q` is reachable from `p` if it can be obtained by a finite sequence of steps.
This is the reflexive and transitive closure of the `single_step` relation.
-/
def reachable (p q : ℕ × ℕ) : Prop :=
  ReflTransGen single_step p q

/--
A pair of numbers `(a, b)` can eventually be made equal if there exists some number `x`
such that the state `(x, x)` is reachable from the initial state `(a, b)`.
-/
def eventually_equal (a b : ℕ) : Prop :=
  ∃ x : ℕ, reachable (a, b) (x, x)

/--
The property characterizing the pairs (a, b) that can eventually be made equal,
as given in the answer: Either both are even, or both are odd and congruent modulo 4.
-/
def good_pair (a b : ℕ) : Prop :=
  (a % 2 = 0 ∧ b % 2 = 0) ∨ (a % 2 = 1 ∧ b % 2 = 1 ∧ a % 4 = b % 4)

/--
The main theorem: Two distinct positive integers a and b can be made equal
if and only if they satisfy the `good_pair` property.
The hypothesis requires a and b to be positive and distinct, as stated in the problem.
Note: If a and b were allowed to be equal initially, `eventually_equal a a` would be trivally true
via zero steps, and the `good_pair` condition would still hold for any positive `a`.
-/
theorem PBAdvanced014 (a b : ℕ) (h_pos : a > 0 ∧ b > 0) (h_ne : a ≠ b) :
    eventually_equal a b ↔ good_pair a b := by sorry"
PB-Advanced-015,"Consider an acute triangle $ABC$ that is not isosceles. Let $H_0$, $E$, and $F$ be the feet of the perpendiculars dropped from vertices $A$, $B$, and $C$ to their opposite sides, respectively. Let $D$ be the point where the incircle of $\triangle ABC$ is tangent to side $ BC $. Denote the incenter and circumcenter of $\triangle ABC$ as $I$ and $O$, respectively. Let $K$ be the intersection of line $IO$ and line $BC$. Let $Q$ be the point where the ray $IH_0$ intersects the circumcircle of $\triangle ABC$ again. Let $X$ be the point where the line $ QD $ intersects the circumcircle of $\triangle ABC$ at a point other than $Q$.
 Let $Y$ be the point where the circle that touches rays $AB$, $AC$, and is also externally tangent to the circumcircle of $\triangle ABC$, touches the circumcircle of $ \triangle ABC$. Prove that if segment $EF$ is tangent to the incircle of $ \triangle ABC$, then $X$, $Y$, and $K$ are collinear.","Let $W, V$ be the points where the incircle of $\triangle ABC$ tangent to $AB,AC$, respectively.

 <Step 1> The circumcircle of $\triangle A B C$, the circumcircle of $\triangle A V W$, and the circumcircle of $\triangle A E F$ meet at a point $P$ other than $A$.

 (1) Let $P(\neq A)$ be the intersection of the circumcircle of $\triangle A B C$ and the circumcircle of $\triangle A E F$. Then, $\triangle P E C \sim \triangle P F B(A A)$.

 (2) From the perspective of $\triangle A E F$, the incircle is the excircle, and since $\triangle A B C \sim \triangle A E F$, we have $\frac{F W}{E V}=\frac{B W}{C V}$.

 (3) Therefore, $\frac{B W}{F W}=\frac{C V}{E V}$, so $\triangle P E V \sim \triangle P F W(S A S)$, and by the inscribed angle theorem, we know that $A, V, W, P$ must lie on a circle.


 <Step 2> Using radical axis

 (1) By considering the radical axis of the circumcircle of $\triangle A B C$, the circumcircle of $\triangle A E F$, and $(E, F, B, C)$, we see that $E F, A P, B C$ must be concurrent. Let this point be $R$.

 (2) By considering the radical axis of the circumcircle of $\triangle A B C$, the circumcircle of $\triangle I B C$, and the circumcircle of $\triangle A V W$, we see that $I R$ must be tangent to the circumcircle of $\triangle I B C$.


 <Step 3> $Q$ is the point of tangency between the $A$-inmixtilinear circle of $\triangle A B C$ and the circumcircle.

 (1) $B, H_{0}, C, R$ are harmonic conjugates, so for the midpoint $M$ of $B C, R M \times R H_{0}=R B \times R C$. This is equal to $R I^{2}$ since $R I$ is tangent to the circumcircle of $\triangle I B C$.

 (2) Therefore, $I H_{0}$ is the symmedian of $\triangle I B C$.

 (3) Therefore, $Q$ must be the point of tangency between the $A$-inmixtilinear circle of $\triangle A B C$ and the circumcircle.


 <Step 4> $V, H, W$ are collinear and $H V$ bisects $\angle E H C$.

 Since $\frac{H E}{H C}=\frac{E F}{B C}=\frac{E V}{C V}$(because $\triangle A B C \sim \triangle A E F$) $H V$ bisects $\angle E H C$. Similarly, $H W$ bisects $\angle F H B$, and therefore $V, H, W$ are collinear.


 <Step 5> $K$ is the reflection of $D$ with respect to $M$.

 (1) Since $\frac{H B}{H C}=\frac{B W}{C V}=\frac{B D}{C D}, H D$ is the angle bisector of $\angle B H C$, and therefore $H D$ is perpendicular to $V W$. That is, $H D$ is parallel to $A I$.

 (2) Therefore, quadrilateral $A I D H$ is a parallelogram, and $A H=I D$. Therefore, $O M$ is half the length of the inradius. Therefore, $K$ is the reflection of $D$ with respect to $M$.


 <Step 6> Finishing the proof

 (1) Let $l$ be the perpendicular bisector of $B C$.

 (2) By Step $3$, $X$ is the reflection of $A$ with respect to $l$.

 (3) By Step $5$ $K$ is the reflection of $D$ with respect to $l$.

 (4) Therefore, the intersection $Y$ of $X K$ and the circumcircle of $\triangle A B C$ is the reflection of the intersection of $A D$ and the circumcircle of $\triangle A B C$ with respect to $l$. This holds because this is the point of tangency between the $A$-exmixtilinear circle and the circumcircle of $\triangle A B C$.
 [0pt]

 [Marking Scheme]

 (1) Circumcircles of $A B C, A V W$, and $A V W$ meet simultaneosly at a point $P$ other than $A$ : 2 points

 (2) $Q$ is the point of tangency between the circumcircle of $A B C$ and $A$-mixtilinear circle: 2 points

 (3) $K$ is the reflection of $D$ with respect to the midpoint $M$ of $B C: 2$ points","(Partial)
 1. Let $V, W$ be the points where the incircle of $\triangle ABC$ tangent to $AB,AC$, respectively. Then, proved that circumcircles of $ABC$, $AVW$, and $AEF$ meet simultaneosly at a point $P$ other than $A$.
(Almost)
 1. Proved that $Q$ is the point of tangency between the circumcircle of $ABC$ and $A$-mixtilinear circle.
 2. Proved that $K$ is the reflection of $D$ with respect to the midpoint $M$ of $BC$.",Geometry,IMO-hard, ,Novel Problem,"import Mathlib

/-
Consider an acute triangle $ABC$ that is not isosceles. Let $H_0$, $E$, and $F$ be the feet of the perpendiculars dropped from vertices $A$, $B$, and $C$ to their opposite sides, respectively. Let $D$ be the point where the incircle of $\triangle ABC$ is tangent to side $ BC $. Denote the incenter and circumcenter of $\triangle ABC$ as $I$ and $O$, respectively. Let $K$ be the intersection of line $IO$ and line $BC$. Let $Q$ be the point where the ray $IH_0$ intersects the circumcircle of $\triangle ABC$ again. Let $X$ be the point where the line $ QD $ intersects the circumcircle of $\triangle ABC$ at a point other than $Q$.
Let $Y$ be the point where the circle that touches rays $AB$, $AC$, and is also externally tangent to the circumcircle of $\triangle ABC$, touches the circumcircle of $ \triangle ABC$. Prove that if segment $EF$ is tangent to the incircle of $ \triangle ABC$, then $X$, $Y$, and $K$ are collinear.
-/
open Affine.Simplex EuclideanGeometry
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

/-- The set of points forming the ray $\overrightarrow{p₁ p₂}$. -/
def ray (p₁ p₂ : ℝ²) : Set ℝ² :=
  { AffineMap.lineMap p₁ p₂ t | (t : ℝ) (_ : t ≥ 0) }

theorem PBAdvanced015
    -- a triangle ABC
    (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
    -- ABC is acute
    (h_acute : AcuteAngled ⟨![A, B, C], tri⟩)
    -- ABC is non-isosceles
    (n_isosceles : [dist A B, dist A C, dist B C].Nodup )
    (H₀ : ℝ²) (hH₀ : H₀ = altitudeFoot ⟨![A, B, C], tri⟩ 0) -- H₀ is the foot from A
    (E : ℝ²) (hE : E = altitudeFoot ⟨![A, B, C], tri⟩ 1) -- E is the foot from B
    (F : ℝ²) (hF : F = altitudeFoot ⟨![A, B, C], tri⟩ 2) -- F is the foot from C
    -- Let $D$ be the point where the incircle of $\triangle ABC$ is tangent to side $ BC $.
    (D : ℝ²) (hD : D = touchpoint ⟨![A, B, C], tri⟩ ∅ 0)
    (I : ℝ²) (hI : I = incenter ⟨![A, B, C], tri⟩) -- I is the incenter
    (O : ℝ²) (hO : O = circumcenter ⟨![A, B, C], tri⟩) -- O is the circumcenter
    -- Let $K$ be the intersection of line $IO$ and line $BC$.
    (K : ℝ²) (hK : Collinear ℝ {K, I, O} ∧ Collinear ℝ {K, B, C})
    -- We denote the circumcircle of ABC as ω
    (ω : Sphere ℝ²) (hω : ω = circumsphere ⟨![A, B, C], tri⟩)
    -- Let $Q$ be the point where the ray $IH_0$ intersects the circumcircle of $\triangle ABC$ again.
    (Q : ℝ²) (hQ : Q ∈ ray I H₀ ∩ ω)
    -- Let $X$ be the point where the line $ QD $ intersects the circumcircle of $\triangle ABC$ at a point other than $Q$.
    (X : ℝ²) (hX : X ≠ Q ∧ X ∈ (affineSpan ℝ {Q, D} ∩ ω : Set ℝ²))
    -- Let $Y$ be the point where the circle that touches rays $AB$, $AC$, and is also externally tangent to the circumcircle of $\triangle ABC$, touches the circumcircle of $ \triangle ABC$.
    (Y : ℝ²) (ωY : Sphere ℝ²) (hY :
      ωY.IsTangent (affineSpan ℝ {A, B}) ∧
      ωY.IsTangent (affineSpan ℝ {A, C}) ∧
      ωY.IsExtTangentAt ω Y ∧
      (affineSpan ℝ {A, ωY.center}).SOppSide B C) -- ensure ωY is in the angle BAC
    -- $EF$ is tangent to the incircle of $ \triangle ABC$
    (h : (insphere ⟨![A, B, C], tri⟩).IsTangent (affineSpan ℝ {E, F}))
    -- then $X$, $Y$, and $K$ are collinear.
    : Collinear ℝ {X, Y, K} := by sorry"
PB-Advanced-016,"Let $ABC$ be a non-isosceles triangle with incenter $I$. Let line $BI$ intersect $AC$ at $E$, and line $CI$ intersect $AB$ at $F$. Two Points $U$ and $V$ are on segments $AB$ and $AC$ respectively, such that $AU = AE$ and $AV = AF$. Let the line passing through $I$ and perpendicular to $AI$ intersect line $BC$ at $L$. The circumcircle of $\triangle ILC$ intersects line $LU$ at $X$ (other than $L$), and the circumcircle of triangle $\triangle ILB$ intersects line $LV$ at $Y$ (other than $L$). Prove that if $P$ is the intersection of lines $YB$ and $XC$, then line $IP$ is parallel to line $XY$.","By symmetry, we assume that $AC>BC$.

 <Step 1> $U, V, L$ lie on a same line

 (1) Let $B^{\prime}$ be the point on $A C$ such that $A B=A B^{\prime}$ and $C^{\prime}$ be the point on $A B$ such that $A C=A C^{\prime}$. These points lie on the circumcircle of $\triangle I B C$.

 (2) By symmetry, $\left(B^{\prime}, I, U\right)$ and $\left(C^{\prime}, I, V\right)$ are collinear.

 (3) By Pascal's theorem on $B^{\prime} C B C^{\prime} I I, U, V, L$ lie on a same line.


 <Step 2> $IPXY$ is an isosceles trapezoid

 Since $U, V, L$ are collinear,

 \[
 \angle I X Y=\angle I C B=\angle B I L=\angle B Y L=\angle P Y X
 \]

 and similarly, $\angle I Y X=\angle P X Y$, so the statement holds.


 <Step 3> Therefore, by <Step 2>, $I P$ and $X Y$ are parallel.","(Partial)
 1. Proved that $U, V$, and $L$ lie on a same line.
(Almost)
 1. Made a minor mistake at the proof, or made a small gap.",Geometry,IMO-easy, ,Novel Problem,"import Mathlib

/-
Let $ABC$ be a non-isosceles triangle with incenter $I$. Let line $BI$ intersect $AC$ at $E$, and line $CI$ intersect $AB$ at $F$. Two Points $U$ and $V$ are on segments $AB$ and $AC$ respectively, such that $AU = AE$ and $AV = AF$. Let the line passing through $I$ and perpendicular to $AI$ intersect line $BC$ at $L$. The circumcircle of $\triangle ILC$ intersects line $LU$ at $X$ (other than $L$), and the circumcircle of triangle $\triangle ILB$ intersects line $LV$ at $Y$ (other than $L$). Prove that if $P$ is the intersection of lines $YB$ and $XC$, then line $IP$ is parallel to line $XY$.
-/
open Affine.Simplex EuclideanGeometry Real
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

theorem PBAdvanced016
    -- a triangle ABC
    (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
    -- ABC is non-isosceles
    (n_isosceles : [dist A B, dist A C, dist B C].Nodup )
    (I : ℝ²) (hI : I = incenter ⟨![A, B, C], tri⟩) -- I is the incenter
    (E : ℝ²) (hE : Collinear ℝ {E, B, I} ∧ Collinear ℝ {E, A, C}) -- BI intersect AC at E
    (F : ℝ²) (hF : Collinear ℝ {F, C, I} ∧ Collinear ℝ {F, A, B}) -- CI intersect AB at E
    -- Two Points $U$ and $V$ are on segments $AB$ and $AC$ respectively, such that $AU = AE$ and $AV = AF$.
    (U : ℝ²) (hU : Sbtw ℝ A U B ∧ dist A U = dist A E)
    (V : ℝ²) (hV : Sbtw ℝ A V C ∧ dist A V = dist A F)
    -- the line passing through $I$ and perpendicular to $AI$ intersect line $BC$ at $L$.
    (L : ℝ²) (hL : ∠ A I L = π / 2 ∧ Collinear ℝ {L, B, C})
    -- The circumcircle of ILC intersects line LU at X (other than L),
    (X : ℝ²) (hX : X ≠ L ∧ Cospherical {X, I, L, C} ∧ Collinear ℝ {X, L, U})
    -- the circumcircle of ILB intersects line LV at Y (other than L).
    (Y : ℝ²) (hY : Y ≠ L ∧ Cospherical {Y, I, L, B} ∧ Collinear ℝ {Y, L, V})
    -- P is the intersection of lines YB and XC
    (P : ℝ²) (hP : Collinear ℝ {P, Y, B} ∧ Collinear ℝ {P, X, C}) :
    -- Prove that line $IP$ is parallel to line $XY$.
    (affineSpan ℝ {I, P}).Parallel (affineSpan ℝ {X, Y}) := by sorry"
PB-Advanced-017,"Let $n$ be a positive integer that has a remainder of 6 when divided by 7. Let $d$ be any divisor of $n$.

Define $c$ such that when the expression $\left(d+\frac{n}{d}\right)^{2}$ is divided by $n$, the remainder is $n-c$.

What is the smallest possible value of $c$ among all $d$ and $n$ satisfying the conditions above?

(Note that the remainder when a positive integer $a$ is divided by a positive integer $b$ is the value of $r$ in the expression $a=b q+r, 0 \leq r \leq b-1$.)","<Step 1> Understanding the problem situation

 <Step 1.1> Let $d=a, \frac{n}{d}=b$, then $n=a b$. So if the remainder when $(a+b)^{2}$ is divided by $n$ is $n-c$ (where $c \geq 1$), then

 \[
 -c \equiv(a+b)^{2}=a^{2}+b^{2}+2 a b \equiv a^{2}+b^{2}(\bmod n)
 \]

 which implies that $\frac{a^{2}+b^{2}+c}{a b}$ must be a positive integer.

 <Step 1.2> First, if $c=3$, then setting $a=4, b=19$ gives

 \[
 \frac{4^{2}+19^{2}+3}{4 \cdot 19}=5
 \]

 which is an integer. In fact, this corresponds to the case when $n=76, d=4$, and $\left(4+\frac{76}{4}\right)^{2}=529 \equiv -3(\bmod 76)$.

 <Step 1.3> Therefore, if $c \leq 2$, it is sufficient to prove that when $\frac{a^{2}+b^{2}+c}{a b}$ is a positive integer, $n$ cannot have remainder 6 when divided by $7$. Now, let's assume that $c$ is 1 or 2.

 <Step 2> Vieta jumping

 <Step 2.1> First, let's find the solutions when $a=b$. Then $\frac{2 a^{2}+c}{a^{2}}$ must be a positive integer, so $a^{2} \mid c$. Since $c$ is 1 or 2, we know that $a=1$. Therefore, $(a, b)=(1,1)$ is the only solution in this case. Now, let's find the solutions when $a \neq b$. By symmetry, we can assume $a<b$, so let's find all such pairs $(a, b)$.

 <Step 2.2> We need to find pairs $(a, b)$ that satisfy the following for some positive integer $k$:


 \begin{equation*}
 \frac{a^{2}+b^{2}+c}{a b}=k \tag{1.1}
 \end{equation*}


 The above equation (1.1) can be rewritten as:

 \[
 a^{2}-k a b+b^{2}+c=0
 \]

 Now, if we view this equation as a quadratic equation in $b$, we can observe the following facts from the relationship between roots and coefficients:

 (1) The sum of the two roots is $k a$, so if a positive integer $b$ is a root, then the other root is $k a-b$, which is also an integer.

 (2) The product of the two roots is $a^{2}+c$, so if a positive integer $b$ is a root, then $\frac{a^{2}+c}{b}$ is also a root, and by (1), it is a positive integer solution.

 (2) Therefore, if $(a, b)$ is a solution to (1.1), then


 \begin{equation*}
 \left(a^{\prime}, b^{\prime}\right)=\left(\frac{a^{2}+c}{b}, a\right)=(k a-b, a) \tag{1.2}
 \end{equation*}


 defined as above is also a solution to (1.1).

 This process gives us a way to find another solution if we have a solution $(a, b)$.

 <Step 3> Applying the method of infinite descent

 Suppose $(a, b)$ is a solution to (1.1) with $a<b$. Now, let's apply infinite descent to the recursive construction process of solutions described in <Step 2> .

 <Step 3.1> Since we assumed $a<b$,

 \[
 b^{2} \geq(a+1)^{2}>a^{2}+2 \geq a^{2}+c
 \]

 so $\frac{a^{2}+c}{b}<b$. In other words, from (1.2),

 \[
 a^{\prime}+b^{\prime}=\frac{a^{2}+c}{b}+a<a+b
 \]

 holds. That is, when performing the recursive construction in <Step 1>, the sum of the two numbers forming the solution always decreases by at least 1.

 <Step 3.2> If $\frac{a^{2}+c}{b}<a$, we can apply the process in <Step 1> again to the new solution $\left(a^{\prime}, b^{\prime}\right)=\left(\frac{a^{2}+c}{b}, a\right)$ to obtain another solution. However, by the result in (1) and the principle of infinite descent, this process cannot be repeated infinitely, and we will eventually reach a solution $(a, b)$ that satisfies $\frac{a^{2}+c}{b} \geq a$.

 <Step 3.3> Let's find what the final solution we must reach in (2) is. If $a<b$, then $b \geq a+1$ must hold. In this situation, if $\frac{a^{2}+c}{b} \geq a$, then

 \[
 a^{2}+c \geq a b \geq a(a+1)=a^{2}+a
 \]

 so $a \leq c$ must hold. Since $a$ is a positive integer, only $a=1,2$ are possible. Here, if $a=2$, then all equalities in the above inequality must hold, so $b=a+1=3$ and $c=2$, which gives

 \[
 \frac{a^{2}+b^{2}+c}{a b}=\frac{2^{2}+3^{2}+3}{2 \cdot 3}=\frac{8}{3}
 \]

 which is not an integer. Therefore, $a$ must be 1 in any case, and $\frac{a^{2}+c}{b}=\frac{1+c}{b}$ must be an integer.$b>a$ and $c+1=2or3$ so $b=c+1$.

 Therefore, this process ends when $(1, c+1)$ becomes $\left(\frac{1^{2}+c}{c+1}, 1\right)=(1,1)$.

 <Step 3.4> Summarizing the above discussion, if $(a, b)$ is a solution to (1.1) with $a<b$, then $\left(a^{\prime}, b^{\prime}\right)$ obtained in <Step 1> is a new solution satisfying $a^{\prime}<b^{\prime}$, and this process of generating solutions ends when the solution reaches $(1,1)$.

 <Step 4> Reversing the descent process

 <Step 4.1> First, the value of $k$ in (1.1) does not change during the process described in <Step 3> . Therefore, the value of $k$ must be the same as the value at the endpoint $(a, b)=(1,1)$ of this process, so

 \[
 k=\frac{1^{2}+1^{2}+1}{c \cdot 1}=c+2
 \]

 must hold.

 <Step 4.2> Now, we know that any solution $(a, b)$ to (1.1) with $a<b$ reaches the endpoint $(a, b)=(1,1)$ by going through the process described in <Step 3> . Therefore, by reversing this process, we can find all solutions $(a, b)$ to (1.1) with $a<b$.

 <Step 4.3> On the other hand, from (1), $k=c+2$, so looking at (1.2), we see that $\left(a^{\prime}, b^{\prime}\right)=((c+2) a-b, a)$. Therefore,

 \[
 (a, b)=\left(b^{\prime},(c+2) b^{\prime}-a^{\prime}\right)
 \]

 holds. In other words, the process of mapping $(x, y) \mapsto(y,(c+2) y-x)$ is the reverse of the process where the solution descends in <Step 2> .

 <Step 4.4> Therefore, by (3), if we define the sequence $\left(x_{n}, y_{n}\right)_{n=1}^{\infty}$ as $\left(x_{1}, y_{1}\right)=(1,1)$ and

 \begin{equation} \tag{1.3}
 \left\{\begin{array}{l}
 x_{n+1}=y_{n}

 y_{n+1}=(c+2) y_{n}-x_{n}
 \end{array}\right.
 \end{equation}

 then this sequence describes the reverse of the process where the solution descends in <Step 2>, and by (2), these $\left(x_{n}, y_{n}\right)$ with $n \geq 2$ are all solutions to (1.1) with $a<b$.

 Note that the descent process described in <Step 2> is

 \[
 \left(x_{k}, y_{k}\right) \rightarrow\left(x_{k-1}, y_{k-1}\right) \rightarrow \cdots \rightarrow\left(x_{1}, y_{1}\right)=(1,1)
 \]

 <Step 4.5>

 (Case 1) When $c=1$, substituting the first equation of (1.3) into the second equation gives the recurrence relation

 \[
 y_{n+1}=3 y_{n}-y_{n-1}
 \]

 and using $y_{1}=1, y_{2}=2$, we can see that this sequence repeats as $1,2,5,6,6,5,2,1,1,2$ modulo 7. Therefore, $x_n y_{n} = y_n y_{n-1}$ repeats as $2,3,2,1,2,3,2,1$ and thus cannot have remainder 6 when devided by $7$.

 (Case 2) When $c=2$, substituting the first equation of (1.3) into the second equation gives the recurrence relation

 \[
 y_{n+1}=4 y_{n}-y_{n-1}
 \]

 and using $y_{1}=1, y_{2}=3$, we can see that this sequence repeats as $1,3,4,6,6,4,3,1,1,3$ modulo 7. Therefore, $x_n y_{n} = y_n y_{n-1}$ repeats as $3,5,3,1,3,5,3,1$ and thus cannot have remainder 6 when divided by $7$.

 In conclusion, when $c=1,2$, there is no case where $ab$ has remainder $6$ when divided by $7$ and $\frac{a^{2}+b^{2}+c}{a b}$ is an integer. Therefore, $c=3$ is the smallest possible value.","(Partial)
 1. Proved $c\ge 3$.
(Almost)
 1. Applied infinite descent or equivalent strategy to prove $c=3$, but not completed.",Number theory,IMO-easy,3,Novel Problem,"import Mathlib

/-
Let $n$ be a positive integer that has a remainder of 6 when divided by 7. Let $d$ be any divisor of $n$.
Define $c$ such that when the expression $\left(d+\frac{n}{d}\right)^{2}$ is divided by $n$, the remainder is $n-c$.
What is the smallest possible value of $c$ among all $d$ and $n$ satisfying the conditions above?
(Note that the remainder when a positive integer $a$ is divided by a positive integer $b$ is the value of $r$ in the expression $a=b q+r, 0 \leq r \leq b-1$.)
Solution: 3
-/
theorem PBAdvanced017
    (r : ℕ → ℕ → ℕ) (hr : ∀ n d, r n d = (d + (n / d : ℕ)) ^ 2 % n) :
    IsLeast {(n - r n d : ℕ) | (n : ℕ) (d : ℕ) (_ : n % 7 = 6) (_ : d ∣ n)} 3 := by sorry"
PB-Advanced-018,"For given integers $n \ge 5$ and $k \ge 1$, we color each of the $n^2$ cells of an $n \times n$ grid using one of $k$ colors. If $q$ is the largest integer which is not larger than $\frac{n^2}{k}$, then, each of the $k$ colors must be used to color exactly $q$ or $q+1$ cells. A sequence of $n$ different cells $c_1, c_2, \ldots, c_n$ in the grid is called a \textit{snake} if it satisfies the following conditions simultaneously:

 \begin{enumerate}
 \item[(a)] For each $1 \le i \le n-1$, two cells $c_i$ and $c_{i+1}$ are adjacent to their sides,
 \item[(b)] For each $1 \le i \le n-1$, cell $c_i$ and cell $c_{i+1}$ are colored with different colors.
 \end{enumerate}
 Let $a(n)$ be the minimum $k$ such that a snake exists regardless of the method of coloring. Find a constant $L$ that satisfies the following inequality and prove it:

 \[
 |La(n)- n^2 | \le n +2 \sqrt n + 3 \;.
 \]","Let $(i, j)$ denote the cell in the $i$-th column from the left and the $j$-th row from the bottom.


 <Step 1> For any $n \geq 5$, we can color the grid with $q=3$ and $r:= n^2 - kq \leq n+\lceil 2 \sqrt{n}\rceil+2$ so that it does not satisfy the conditions. From this, we can see that $a(n) \geq \frac{n^{2}- (n+2 \sqrt{n}+3)}{3}$.


 (1) Let $m=\lceil\sqrt{n}\rceil-1$.


 (2) For the following cells, color the $2 \times 2$ square with that cell as its bottom left corner with the same color.

 \[
 \{(i, j): i, j \in \mathbb{Z}, m|i, \quad m| j, \quad 1 \leq i, j \leq n-1\}
 \]

 Color each $2 \times 2$ square with a different color. Let $A$ be the area colored in this way.


 (3) In the remaining area excluding $A$, we can color the cells so that each color is used 3 times. For any $i$ such that $m \mid i$, color the cells $(x, y)$ with $x=i$ that do not belong to $A$ with the same color as the cell above it, and color the cells $(x, y)$ with $y=i$ that do not belong to $A$ with the same color as the cell to its right. Color the remaining cells arbitrarily, but make sure that each color appears 3 or 4 times.


 (4) Since we can do this with $k=\left\lfloor\frac{n^{2}-|A|}{3}\right\rfloor, r \leq|A|+2$.


 (5) From this,
 \[
 |A|=\left\lfloor\frac{n-1}{m}\right\rfloor^{2}<\left(\frac{n-1}{\sqrt{n}-1}\right)^{2}<n+2 \sqrt{n}+1
 \]
 and $r \leq n+\lceil 2 \sqrt{n}\rceil+2$ holds.


 (6) Show that for this arrangement, there do not exist $n$ cells that satisfy the conditions. Group the cells into $m \times m$ squares starting from the bottom left of the grid. Treat any remaining cells at the top or right as if the grid were extended and group them in the same way. Each grouped section contains at most $m^{2}$ cells.


 (7) In order to start from within a grouped section and exit it, one must pass through consecutive cells of the same color due to the construction in (3) and (4). Therefore, $n \leq m^{2}$ must hold. This contradicts the definition of $m$. That is, this coloring does not satisfy the conditions.


 <Step 2> If $k \geq \frac{n^{2}}{3}$, the conditions are satisfied. Therefore, $a(n) \leq\left\lceil\frac{n^{2}}{3}\right\rceil \leq \frac{n^{2}+2}{3}$.


 (1) Consider the following graph: each cell is a vertex, and each cell is connected to the vertices of cells adjacent to it that have a different color.


 (2) Consider the $4 n-4$ cells on the border of the given $n \times n$ grid. Now, let's prove the following claim.

 Claim. For any two vertices $x, y$ belonging to the same connected component $\Phi$ on the border, if there is no vertex in $\Phi$ when moving clockwise along the border from $x$ to $y$, then the vertex one step clockwise from $x$ and the vertex one step counterclockwise from $y$ belong to the same connected component.


 (3) Let $z$ be the vertex one step clockwise from $x$ and $w$ be the vertex one step counterclockwise from $y$. Consider the boundary of the polygon formed by the cells in $\Phi$. This boundary consists of the perimeter edges of the given $n \times n$ grid or the edges shared by two adjacent cells within the $n \times n$ grid. The edge $l$ shared by $x$ and $z$ and the edge $m$ shared by $y$ and $w$ are both included in the boundary of this polygon. By the condition, on the boundary of this polygon, the line segments between $l$ and $m$ are all inside the $n \times n$ grid.


 (4) Let these line segments between $l$ and $m$ be $a_{0}, a_{1}, \cdots, a_{k}$ in order ( $a_{0}=l, a_{k}=m$ ). Each $a_{i}$ lies on the boundary between a cell belonging to $\Phi$ and a cell not belonging to $\Phi$. That is, the two cells sharing $a_{i}$ must have the same color. We will show that the cells not belonging to $\Phi$ that share $a_{i}$ all belong to the same connected component. We will use induction to show that for $i=0, \cdots, s$, the cells not belonging to $\Phi$ that share $a_{i}$ all belong to the same connected component.


 (5) Now, if $s=0$, it is trivial since there is only one cell. Therefore, using the inductive hypothesis, assume that the statement holds for $s-1$ and consider the case for $s$.


 (Case 1) If $a_{s-1}$ and $a_{s}$ lie on the same line

 This is the case where for some $2 \times 2$ square and the line connecting the midpoints of two opposite sides, one side belongs to $\Phi$ and the opposite side does not. In this case, the coloring must be symmetric with respect to this line. Therefore, the two cells not belonging to $\Phi$ must also be connected by a line segment.


 (Case 2) If $a_{s-1}$ and $a_{s}$ form a $90^{\circ}$ angle

 This means that for some $2 \times 2$ square, the two lines connecting its center to the midpoints of two consecutive sides are $a_{s-1}$ and $a_{s}$. These lines divide the $2 \times 2$ square into 1 cell and 3 cells. Place the centers of these cells on a new coordinate plane. Suppose they are divided into $(0,0)$ and $(1,0),(0,1),(1,1)$. Then $(0,0)$ and $(1,0),(0,1)$ must have the same color. Since each color is used at most 3 times, these are different from the color of $(1,1)$. Therefore, $(1,1)$ is connected to $(1,0)$ and $(0,1)$. That is, both sides of the division belong to the same connected component.


 (6) Therefore, by mathematical induction, all cells not belonging to $\Phi$ that share $a_{i}$ are in the same connected component. This means that $z$ and $w$ are in the same connected component. This completes the proof of the Claim.


 (7) Returning to the main proof, we will consider the vertices on the border. For two points $(x_1, y_1)$ and $(x_2, y_2)$, define the 'grid distance' as $|x_1 -x_2| +|y_1 - y_2|$. Let points $a$ and $b$ be the pair of points on the boarder that have the largest grid distance between them while being on the same connected component.

(8) If the grid distance between $a$ and $b$ is larger or equal to $n-1$, then the path on the connected component that connects $a$ and $b$ is a desired \textit{snake} and we are done.

(9) If the grid distance is smaller than $n-1$, then let the squares one step clockwise from $a$ and $b$ be $a_1$ and $b_1$, respectively, and the squares one step counterclockwise from $a$ and $b$ be  $a_2$ and $b_2$, respectively. Now without loss of generality, we can say the grid distance between $a_1$ and $b_2$ is smaller than $n-1$. Now think of the path along the boundary connecting $a_2$ and $b_1$, not passing $a$ and $b$. The length of this path is at least $4n-4 -(n-1)-2 = 3n-5$, so if there is a square $c$ that is in the same connected component with $a$ and $b$, the grid distance between $(a,c)$ or $(b,c)$ is larger than $n-1$, thus a contradiction to maximality. However, if there are no squares in the path from $a_2$ to $b_1$ , by Claim $a_2$ and $b_1$ must be in the same connected component, which is agian a contradiction to the maximality of $(a,b)$.","(Partial)
 1. Guessed $L=3$ and then explained the strategy of the proof (namely, explained how one can obtain lower and upper bound of $a(n)$ ).
 2. Partitioned the given grid into squares of size of order $\sqrt{n} \times \sqrt{n}$ or used a similar logic to determine a lower bound for $N a(n)$, but failed to meet the required lower bound as specified by the problem.
(Almost)
 1.  Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,IMO-hard,3,Novel Problem,"import Mathlib

/-
For given integers $n \ge 5$ and $k \ge 1$, we color each of the $n^2$ cells of an $n \times n$ grid  using one of $k$ colors. If $q$ is the largest integer which is not larger than $\frac{n^2}{k}$, then, each of the $k$ colors must be used to color exactly $q$ or $q+1$ cells. A sequence of $n$ different cells $c_1, c_2, \ldots, c_n$ in the grid is called a \textit{snake} if it satisfies the following conditions simultaneously:

\begin{enumerate}
\item[(a)] For each $1 \le i \le n-1$, two cells $c_i$ and $c_{i+1}$ are adjacent to their sides,
\item[(b)] For each $1 \le i \le n-1$, cell $c_i$ and cell $c_{i+1}$ are colored with different colors.
\end{enumerate}
Let $a(n)$ be the minimum $k$ such that a snake exists regardless of the method of coloring. Find a constant $L$ that satisfies the following inequality and prove it:

\[
|La(n)- n^2 | \le n +2 \sqrt n + 3 \;.
\]

Answer: 3
-/

-- The set of valid coloring: each of the $k$ colors must be used to color exactly $q$ or $q+1$ cells
def validColorings (n k : ℕ) : Set (Matrix (Fin n) (Fin n) (Fin k)) :=
  let q : ℕ := ⌊ n^2 / k ⌋₊
  { M |
    ∀ c : Fin k, let cn := Finset.card {coor : Fin n × Fin n | M.uncurry coor = c}
    cn = q ∨ cn = q+1
  }

-- Definition of a snake on a fixed grid
structure Snake {n k : ℕ} (M : Matrix (Fin n) (Fin n) (Fin k)) where
  c : List (Fin n × Fin n)
  c_nodup : c.Nodup -- the cells must be different
  c_len : c.length = n
  c_adj : c.Chain' (fun (y₁,x₁) (y₂,x₂) ↦
    abs (x₁ - x₂ : ℤ) + abs (y₁ - y₂ : ℤ) = 1 ∧ -- the cells are adjacent
    M y₁ x₁ ≠ M y₂ x₂ -- and they have a different color
  )

-- Let $a(n)$ be the minimum $k$ such that a snake exists regardless of the method of coloring.
-- We are given k ≥ 1 among the main assumptions, so we will force k ≥ 1 here.
noncomputable
def a (n : ℕ) := sInf { k : ℕ | k ≥ 1 ∧ ∀ M ∈ validColorings n k, Nonempty (Snake M) }

theorem PBAdvanced018 (n : ℕ) (hn : n ≥ 5) : |(3 * a n - n^2 : ℤ)| ≤ n + 2 * √ n + 3 := by sorry"
PB-Advanced-019,"For a real number $r$, let $A(r)$ denote the fractional part of $2r$ in its decimal representation. For a real number $r$ and a positive integer $n$, define $B(n,r)$ as
 $$
 B(n,r)=\sum_{k=1}^n A(kr).
 $$
 Find all positive real numbers $r$ such that $n(n+1)r - B(n,r)$ is a multiple of $n$ for all positive integers $n$.","Let $r$ be a positive real number. Let $A(r)$ denote the fractional part of $2r$, so $A(r) = 2r - \lfloor 2r \rfloor$. We are given that $B(n,r) = \sum_{k=1}^n A(kr)$. We want to find all positive real numbers $r$ such that $n(n+1)r - B(n,r)$ is a multiple of $n$ for all positive integers $n$.

 We can rewrite the condition as
 $$
 n(n+1)r - \sum_{k=1}^n (2kr - \lfloor 2kr \rfloor) = \sum_{k=1}^n \lfloor 2kr \rfloor \quad \text{is a multiple of } n.
 $$
 Let $\alpha = 2r$. Then the condition becomes: $\sum_{k=1}^n \lfloor k\alpha \rfloor$ is a multiple of $n$ for all positive integers $n$.

 We prove that all even integers $\alpha$ satisfy the condition, and no other real number $\alpha$ does so. First we will show that
 even integers satisfy the condition. If $\alpha=2m$ where $m$ is
 an integer then

 \[
 \lfloor\alpha\rfloor+\lfloor2\alpha\rfloor+\cdots+\lfloor n\alpha\rfloor=2m+4m+\cdots+2mn=mn(n+1)
 \]

 which is a multiple of $n$. Since $r > 0$, we must have $\alpha = 2r > 0$. So $\alpha$ must be a positive even integer. If $\alpha = 2m$ for some positive integer $m$, then $r = m$, which is a positive integer.

 Now we will show that they are the only real numbers satisfying the
 conditions of the problem. Let $\alpha=k+\epsilon$ where $k$ is
 an integer and $0\leqslant\epsilon<1$. Then the number
 \[
 \begin{aligned}\lfloor\alpha\rfloor+\lfloor2\alpha\rfloor+\cdots+\lfloor n\alpha\rfloor & = \lfloor k+\epsilon \rfloor+\lfloor2(k+\epsilon)\rfloor+\cdots+\lfloor n(k+\epsilon)\rfloor

 & = k+\lfloor\epsilon\rfloor+2k+\lfloor2\epsilon\rfloor+\cdots+nk+\lfloor n\epsilon\rfloor

 & =\frac{kn(n+1)}{2}+\lfloor\epsilon\rfloor+\lfloor2\epsilon\rfloor+\cdots+\lfloor n\epsilon\rfloor
 \end{aligned}
 \]
 has to be a multiple of $n$. We consider two cases based on the parity
 of $k$.


 \begin{itemize}
 \item Case 1: \textbf{$k$ is even.} Then $\frac{kn(n+1)}{2}$ is always
 a multiple of $n$. Thus
 \[
 \lfloor\epsilon\rfloor+\lfloor2\epsilon\rfloor+\cdots+\lfloor n\epsilon\rfloor
 \]
 also has to be a multiple of $n$. We will prove that $\lfloor n\epsilon\rfloor=0$
 for every positive integer $n$ by strong induction. The base case
 $n=1$ follows from the fact that $0\leqslant\epsilon<1$. Let us
 suppose that $\lfloor m\epsilon\rfloor=0$ for every $1\leqslant m<n$.
 Then the number
 \[
 \lfloor\epsilon\rfloor+\lfloor2\epsilon\rfloor+\cdots+\lfloor n\epsilon\rfloor=\lfloor n\epsilon\rfloor
 \]
 has to be a multiple of $n$. As $0\leqslant\epsilon<1$ then $0\leqslant n\epsilon<n$,
 which means that the number $\lfloor n\epsilon\rfloor$ has to be
 equal to 0. The equality $\lfloor n\epsilon\rfloor=0$ implies $0\leqslant\epsilon<1/n$.
 Since this has to happen for all $n$, we conclude that $\epsilon=0$.
 Thus $\alpha$ is an even integer. Since $r > 0$, $\alpha = 2r > 0$. So $\alpha$ must be a positive even integer. This means $2r = 2m$ for some positive integer $m$. Therefore, $r=m$ must be a positive integer.
 \item Case 2: \textbf{$k$ is odd.} We will prove that $\lfloor n\epsilon\rfloor=n-1$
 for every natural number $n$ by strong induction. The base case $n=1$
 again follows from the fact that $0\leqslant\epsilon<1$. Let us suppose
 that $\lfloor m\epsilon\rfloor=m-1$ for every $1\leqslant m<n$.
 We need the number
 \[
 \begin{aligned}\frac{kn(n+1)}{2}+\lfloor\epsilon\rfloor+\lfloor2\epsilon\rfloor+\cdots+\lfloor n\epsilon\rfloor & =\frac{kn(n+1)}{2}+0+1+\cdots+(n-2)+\lfloor n\epsilon\rfloor

 & =\frac{kn(n+1)}{2}+\frac{(n-2)(n-1)}{2}+\lfloor n\epsilon\rfloor

 & =\frac{k+1}{2}n^{2}+\frac{k-3}{2}n+1+\lfloor n\epsilon\rfloor
 \end{aligned}
 \]
 to be a multiple of $n$. As $k$ is odd, we need $1+\lfloor n\epsilon\rfloor$
 to be a multiple of $n$. Again, as $0\leqslant\epsilon<1$ then $0\leqslant n\epsilon<n$,
 so $\lfloor n\epsilon\rfloor=n-1$ as we wanted. This implies that
 $1-\frac{1}{n}\leqslant\epsilon<1$ for all $n$. Taking the limit as $n \to \infty$, we get $\epsilon=1$, which contradicts $0 \le \epsilon < 1$.
 So there are no other solutions in this case.

 \end{itemize}
 Therefore, $\alpha = 2r$ must be a positive even integer. This implies $r$ must be a positive integer.","(Partial)
 1. Verified that integer $r$ works.
 2. Proved that $r$ should be integer if $r$ is a rational number.
(Almost)
 1. Proved that $r$ should be an integer, but did not verified that integers indeed work.
 2. Solution is almost complete, but made minor mistakes which are not negligible.",Algebra,IMO-easy,all integers,(Modified) IMO 2024 P1,"import Mathlib

/-
For a real number $r$, let $A(r)$ denote the fractional part of $2r$ in its decimal representation. For a real number $r$ and a positive integer $n$, define $B(n,r)$ as
$$
B(n,r)=\sum_{k=1}^n A(kr).
$$
Find all positive real numbers $r$ such that $n(n+1)r - B(n,r)$ is a multiple of $n$ for all positive integers $n$.

Solution: $r$ is a positive integer.
-/
theorem PBAdvanced019
    (A : ℝ → ℝ) (hA : ∀ r, A r = Int.fract (2 * r))
    (B : ℕ → ℝ → ℝ) (hB : ∀ n r, B n r = ∑ k ∈ Finset.Icc 1 n, A (k * r)) :
    {r : ℝ | 0 < r ∧ ∀ n > 0, ∃ (m : ℤ), n * (n + 1) * r - B n r = n * m}
      = {(n : ℝ) | (n : ℕ) (_ : 0 < n)} := by sorry"
PB-Advanced-020,"For given positive integers $x$ and $y$, we define a sequence $(a_n)_{n \ge 1}$ as
 $$
 a_n = \gcd \left( x^n +y , \, (y-x)\left(\sum_{i=0}^{n-1} y^i x^{n-i-1} - 1\right) \right)
 $$
 for all $n\in \mathbb{N}$. Find all pairs $(x,y)$ of positive integers such that the limit of the sequence $(a_n)$ exists.","To begin with, we can change the expression in the problem to
 $\gcd \left( x^n +y , \, (y-x)\left(\sum_{i=0}^{n-1} y^i x^{n-i-1} - 1\right) \right) = \gcd \left( x^n +y , \, y^n - x^n -(y-x) \right) = \gcd(x^n +y , y^n +x)$.
 Let the limit of the sequence $(a_n)$ exist and be equal to $g$. Then, for sufficiently large $n$,

 Lemma. If the limit of $a_n$ as $n \to \infty$ exists and is equal to $g$, then $g$ divides $2 \gcd(x, y)$.

 Proof. For sufficiently large $n$, we have $a_n = \gcd \left( x^n +y , \, y^n +x \right) = g$.
 This implies that $g$ divides $x^n + y$ and $g$ divides $y^n +x$ for all $n \ge N$, for some positive integer $N$.

 Consider $n \ge N$. We have $x^n + y \equiv 0 \pmod{g}$ and $x^{n+1} + y \equiv 0 \pmod{g}$.
 Multiplying the first congruence by $x$, we get $x^{n+1} + xy \equiv 0 \pmod{g}$.
 Subtracting the second congruence from this, we have $(x^{n+1} + xy) - (x^{n+1} + y) \equiv 0 - 0 \pmod{g}$, which simplifies to $xy - y = y(x-1) \equiv 0 \pmod{g}$.

 Analogously, $x(y-1)$ is divisible by $g$.
 Their difference $x-y$ is then divisible by $g$, so $g$ also divides
 $x(y-1)+x(x-y)=x^2 -x$. All powers of $x$ are then congruent modulo
 $g$, so $x+y\equiv x^{N}+y\equiv0(\bmod g)$. Then $2x=(x+y)+(x-y)$
 and $2y=(x+y)-(x-y)$ are both divisible by $g$, so $g\mid2\operatorname{gcd}(x,y)$.
 On the other hand, it is clear that $\operatorname{gcd}(x,y)\mid g$,
 thus proving the Lemma.

 Let $d=\operatorname{gcd}(x,y)$, and write $x=da$ and $y=db$ for
 coprime positive integers $a$ and $b$. We have that

 \[
 \operatorname{gcd}\left((da)^{n}+db,(db)^{n}+da\right)=d\operatorname{gcd}\left(d^{n-1}a^{n}+b,d^{n-1}b^{n}+a\right)
 \]

 so the Lemma tells us that

 \[
 \operatorname{gcd}\left(d^{n-1}a^{n}+b,d^{n-1}b^{n}+a\right)\leqslant2
 \]

 for all $n\geqslant N$. Defining $K=d^{2}ab+1$, note that $K$ is
 coprime to each of $d,a$, and $b$. By Euler's theorem, for $n\equiv-1(\bmod\varphi(K))$
 we have that

 \[
 d^{n-1}a^{n}+b\equiv d^{-2}a^{-1}+b\equiv d^{-2}a^{-1}\left(1+d^{2}ab\right)\equiv 0\quad(\bmod K)
 \]

 so $K\mid d^{n-1}a^{n}+b$. Analogously, we have that $K\mid d^{n-1}b^{n}+a$.
 Taking such an $n$ which also satisfies $n\geqslant N$ gives us
 that

 \[
 K\mid\operatorname{gcd}\left(d^{n-1}a^{n}+b,d^{n-1}b^{n}+a\right)\leqslant2
 \]

 This is only possible when $d=a=b=1$, which yields the only solution
 $(x,y)=(1,1)$. Therefore, the only solution is $(x,y)=(1,1)$.","(Partial)
 1. Considered a prime $p|xy+1$.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.
 2. Omitted the case when $xy+1$ doesn't have an odd prime factor.",Number theory,IMO-medium,"(1,1)",(Modified) IMO 2024 P2,"import Mathlib

open scoped Topology

/-
For given positive integers $x$ and $y$, we define a sequence $(a_n)_{n \ge 1}$ as
$$
a_n = \gcd \left( x^n +y , \, (y-x)\left(\sum_{i=0}^{n-1} y^i x^{n-i-1} - 1\right) \right)
$$
for all $n\in \mathbb{N}$. Find all pairs $(x,y)$ of positive integers such that the limit of the sequence $(a_n)$ exists.
-/

/-- Recursive definition of `a x y n` -/
def a (x y : ℤ) (n : ℕ) : ℤ :=
  (x^n + y).gcd ((y - x) * (∑ i ∈ Finset.range n, y^i * x^(n - i - 1 : ℕ) - 1))

/-- All pairs of positive integers x, y where a limit exists -/
def satisfyingPairs : Set (ℤ × ℤ) := {
  (x, y) | (x : ℤ) (_ : 0 < x) (y : ℤ) (_ : 0 < y) (l : ℝ)
  (_ : Filter.atTop.Tendsto (fun n ↦ (a x y n : ℝ)) (𝓝 l))
}

theorem PBAdvanced020 : satisfyingPairs = {(1, 1)} := by sorry"
PB-Advanced-021,"Infinitely many boys and girls stand in a line alternatingly. For
 a given positive integer $N$, they choose positive integers according
 to the following rules:

 (1) The first $N$ people in the line choose any positive integers
 they want.

 (2) For each $m>N$, the $m$-th person chooses the number that is
 one more than the number of appearances of the number chosen by the
 $(m-1)$-th person appears among the numbers chosen by the first $m-2$
 people.

 Let $b_{i}$ be the number chosen by the $i$-th boy, and $g_{i}$
 be the number chosen by the $i$-th girl. Prove that at least one
 of the following holds:

 (i) There exists a sufficiently large integer $M$ such that $\{b_{n}\}_{n\ge M}$
 is a periodic sequence.

 (ii) There exists a sufficiently large integer $M$ such that $\{g_{n}\}_{n\ge M}$
 is a periodic sequence.","Let $N$ be the given positive integer. Let $s_{m}$ be the number
 chosen by the $m$-th person in the line. For $m\ge N+1$, $s_{m}$
 is one more than the number of appearances of $s_{m-1}$ in the sequence
 $s_{1},s_{2},\dots,s_{m-2}$. Let $b_{i}=s_{2i-1}$ be the number
 chosen by the $i$-th boy, and $g_{i}=s_{2i}$ be the number chosen
 by the $i$-th girl. We want to show that at least one of the sequences
 $\{b_{n}\}_{n\ge M}$ or $\{g_{n}\}_{n\ge M}$ is periodic for some
 sufficiently large integer $M$.


 Let $M>\max\left(s_{1},\ldots,s_{2N}\right)$. We first prove that
 some integer appears infinitely many times. If not, then the sequence
 contains arbitrarily large integers. The first time each integer larger
 than $M$ appears, it is followed by a 1 . So 1 appears infinitely
 many times, which is a contradiction.


 Now we prove that every integer $x\geq M$ appears at most $M-1$
 times. If not, consider the first time that any $x\geq M$ appears
 for the $M^{\text{th }}$ time. Up to this point, each appearance
 of $x$ is preceded by an integer which has appeared $x\geq M$ times.
 So there must have been at least $M$ numbers that have already appeared
 at least $M$ times before $x$ does, which is a contradiction.


 Thus there are only finitely many numbers that appear infinitely many
 times. Let the largest of these be $k$. Since $k$ appears infinitely
 many times there must be infinitely many integers greater than $M$
 which appear at least $k$ times in the sequence, so each integer
 $1,2,\ldots,k-1$ also appears infinitely many times. Since $k+1$
 doesn't appear infinitely often there must only be finitely many numbers
 which appear more than $k$ times. Let the largest such number be
 $l\geq k$. From here on we call an integer $x$ big if $x>l$, medium
 if $l\geq x>k$ and small if $x\leq k$. To summarise, each small
 number appears infinitely many times in the sequence, while each big
 number appears at most $k$ times in the sequence.


 Choose a large enough $N^{\prime}>2N$ such that $s_{N^{\prime}}$
 is small, and in $s_{1},\ldots,s_{N^{\prime}}$ :
 \begin{enumerate}
 \item every medium number has already made all of its appearances;
 \item every small number has made more than $\max(k,N)$ appearances.
 \end{enumerate}
 Since every small number has appeared more than $k$ times, past this
 point each small number must be followed by a big number. Also, by
 definition each big number appears at most $k$ times, so it must
 be followed by a small number. Hence the sequence alternates between
 big and small numbers after $s_{N^{\prime}}$.


 \textbf{Lemma 1. }Let $g$ be a big number that appears after $s_{N^{\prime}}$.
 If $g$ is followed by the small number $h$, then $h$ equals the
 amount of small numbers which have appeared at least $g$ times before
 that point.


 \emph{Proof. }By the definition of $N^{\prime}$, the small number
 immediately preceding $g$ has appeared more than $\max(k,2N)$ times,
 so $g>\max(k,2N)$. And since $g>2N$, the $g^{\text{th }}$ appearance
 of every small number must occur after $s_{2N}$ and hence is followed
 by $g$. Since there are $k$ small numbers and $g$ appears at most
 $k$ times, $g$ must appear exactly $k$ times, always following
 a small number after $s_{2N}$. Hence on the $h^{\text{th }}$ appearance
 of $g$, exactly $h$ small numbers have appeared at least $g$ times
 before that point.


 Denote by $s_{[i,j]}$ the subsequence $s_{i},s_{i+1},\ldots,s_{j}$.


 \textbf{Lemma 2. }Suppose that $i$ and $j$ satisfy the following
 conditions:

 (a) $j>i>N^{\prime}+2$,

 (b) $s_{i}$ is small and $s_{i}=s_{j}$,

 (c) no small value appears more than once in $s_{[i,j-1]}$.

 Then $s_{i-2}$ is equal to some small number in $s_{[i,j-1]}$.


 \emph{Proof. }Let $\mathcal{I}$ be the set of small numbers that
 appear at least $s_{i-1}$ times in $s_{[1,i-1]}$. By Lemma 1, $s_{i}=|\mathcal{I}|$.
 Similarly, let $\mathcal{J}$ be the set of small numbers that appear
 at least $s_{j-1}$ times in $s_{[1,j-1]}$. Then by Lemma $1,s_{j}=|\mathcal{J}|$
 and hence by (b), $|\mathcal{I}|=|\mathcal{J}|$. Also by definition,
 $s_{i-2}\in\mathcal{I}$ and $s_{j-2}\in\mathcal{J}$.


 Suppose the small number $s_{j-2}$ is not in $\mathcal{I}$. This
 means $s_{j-2}$ has appeared less than $s_{i-1}$ times in $s_{[1,i-1]}$.
 By (c), $s_{j-2}$ has appeared at most $s_{i-1}$ times in $s_{[1,j-1]}$,
 hence $s_{j-1}\leq s_{i-1}$. Combining with $s_{[1,i-1]}\subset s_{[1,j-1]}$,
 this implies $\mathcal{I}\subseteq\mathcal{J}$. But since $s_{j-2}\in\mathcal{J}\backslash\mathcal{I}$,
 this contradicts $|\mathcal{I}|=|\mathcal{J}|$. So $s_{j-2}\in\mathcal{I}$,
 which means it has appeared at least $s_{i-1}$ times in $s_{[1,i-1]}$
 and one more time in $s_{[i,j-1]}$. Therefore $s_{j-1}>s_{i-1}$.


 By (c), any small number appearing at least $s_{j-1}$ times in $s_{[1,j-1]}$
 has also appeared $s_{j-1}-1\geq$ $s_{i-1}$ times in $s_{[1,i-1]}$.
 So $\mathcal{J}\subseteq\mathcal{I}$ and hence $\mathcal{I}=\mathcal{J}$.
 Therefore, $s_{i-2}\in\mathcal{J}$, so it must appear at least $s_{j-1}-s_{i-1}=1$
 more time in $s_{[i,j-1]}$.


 For each small number $s_{n}$ with $n>N^{\prime}+2$, let $p_{n}$
 be the smallest number such that $s_{n+p_{n}}=s_{i}$ is also small
 for some $i$ with $n\leq i<n+p_{n}$. In other words, $s_{n+p_{n}}=s_{i}$
 is the first small number to occur twice after $s_{n-1}$. If $i>n$,
 Lemma 2 (with $j=n+p_{n}$ ) implies that $s_{i-2}$ appears again
 before $s_{n+p_{n}}$, contradicting the minimality of $p_{n}$. So
 $i=n$. Lemma 2 also implies that $p_{n}\geq p_{n-2}$. So $p_{n},p_{n+2},p_{n+4},\ldots$
 is a nondecreasing sequence bounded above by $2k$ (as there are only
 $k$ small numbers). Therefore, $p_{n},p_{n+2},p_{n+4},\ldots$ is
 eventually constant and the subsequence of small numbers is eventually
 periodic with period at most $k$.","(Partial)
 1. Proved that there are only finitely many integers that appear infinitely many times in $\{b_{n}\}\cup\{g_{n}\}$.
 2. Came up with multiple observations that may lead to the proof, but failed to rigorously combine them.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,IMO-hard, ,(Modified) IMO 2024 P3,"import Mathlib

/-- The property that a sequence `c` of positive integers satisfies the recursive rule
starting from index `N + 1`. Indices start at 1. -/
def SatisfiesRule (c : ℕ → ℕ) (N : ℕ) : Prop :=
  (N > 0) ∧
  -- (1) All chosen numbers are positive integers (for i ≥ 1).
  (∀ i : ℕ, i > 0 → c i > 0) ∧
  -- (2) For m > N, the rule applies.
  (∀ m : ℕ, m > N →
    -- The set of indices {k | 1 ≤ k ≤ m - 2} is Finset.Icc 1 (m - 2)
    c m = 1 + Finset.card (
      (Finset.Icc 1 (m - 2)).filter (fun k => c k = c (m - 1)))
    )

variable (c : ℕ → ℕ)

/-- The sequence of numbers chosen by the boys (at odd positions 1, 3, 5, ...). -/
def boys_sequence : ℕ → ℕ :=
  fun i => c (2 * i - 1)

/-- The sequence of numbers chosen by the girls (at even positions 2, 4, 6, ...). -/
def girls_sequence : ℕ → ℕ :=
  fun i => c (2 * i)

/-- A sequence `a : ℕ → ℕ` is eventually periodic. We require the starting index M and period P to be positive. -/
def IsEventuallyPeriodic (a : ℕ → ℕ) : Prop :=
  -- M is the index after which periodicity starts, P is the period.
  ∃ M P : ℕ, M > 0 ∧ P > 0 ∧ ∀ n : ℕ, n ≥ M → a n = a (n + P)

-- The main claim of the problem: At least one of the gender subsequences is eventually periodic.
theorem PBAdvanced021 {N : ℕ} (h_valid : SatisfiesRule c N) :
  IsEventuallyPeriodic (boys_sequence c) ∨ IsEventuallyPeriodic (girls_sequence c) := by sorry"
PB-Advanced-022,"Given a triangle $ABC$ with $AB<AC<BC$, let $I$ be the incenter
 of triangle $ABC$, and let $M$ and $N$ be the midpoints of sides
 $CA$ and $AB$, respectively. Let $K$ be the midpoint of the arc
 $BC$ of the circumcircle of triangle $ABC$ which does not contain
 $A$. Let $B'\neq C$ be the point where the line parallel to $AC$
 and tangent to the incircle of triangle $ABC$ intersects side $BC$,
 and similarly, let $C'\neq B$ be the point where the line parallel
 to $AB$ and tangent to the incircle of triangle $ABC$ intersects
 side $BC$. Find the value of $\angle NIM+\angle B'KC'$ in terms
 of degree.","Let $IB'$ intersect $CK$ at $Q$.


 \textbf{Claim. }Triangle $CB'Q$ is similar to $\triangle AIB$ (and
 $Q$ lies on circle $BIC$).


 \emph{Proof.} Since the tangent at $B'$ to the incircle, distinct
 from $BC$, is parallel to $AC$, and $B'I$ bisects the two tangents
 from $B'$,
 \[
 \angle CB'I=\frac{1}{2}\left(180^{\circ}-\angle ACB\right)=90^{\circ}-\frac{1}{2}\angle ACB.
 \]
 Using this, we get
 \[
 \angle CQB'=\angle CB'I-\angle BCK=90^{\circ}-\frac{1}{2}\angle ACB-\angle BAK=90^{\circ}-\frac{1}{2}(\angle ACB+\angle BAC)=\frac{1}{2}\angle CBA
 \]
 Which is enough to show $Q$ lies on circle $BIC$. Combining this
 with $\angle B'CQ=\angle BCK=$ $\angle BAI$ gives the similarity.

 Note that $K$ is the centre of circle $BIC$ so $K$ is the midpoint
 of $CQ$. Since $N$ is the midpoint of $AB$, $K$ and $N$ are similar
 points in the two triangles in the Claim so
 \[
 \angle KB'C'=\angle KB'C=\angle AIN.
 \]

 Similarly $\angle B'C'K=\angle MIA$ so
 \[
 \angle MIN+\angle C'KB'=\angle MIA+\angle AIN+\angle C'KB'=\angle B'C'K+\angle KB'C'+\angle C'KB'=180^{\circ}.
 \]","(Partial)
 1. Constructed the reflection of $A$ over $I$, or any auxiliary point that helps angle computation regarding the midpoints.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Geometry,IMO-easy,180,(Modified) IMO 2024 P4,"import Mathlib

/-
Given a triangle $ABC$ with $AB < AC < BC$, let $I$ be the incenter
of triangle $ABC$, and let $M$ and $N$ be the midpoints of sides
$CA$ and $AB$, respectively. Let $K$ be the midpoint of the arc
$BC$ of the circumcircle of triangle $ABC$ which does not contain
$A$. Let $B' \neq C$ be the point where the line parallel to $AC$
and tangent to the incircle of triangle $ABC$ intersects side $BC$,
and similarly, let $C' \neq B$ be the point where the line parallel
to $AB$ and tangent to the incircle of triangle $ABC$ intersects
side $BC$. Find the value of $\angle NIM+\angle B'KC'$ in terms
of degree.

Answer: 180
-/
open Real EuclideanGeometry Affine.Simplex

local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)
local notation degrees ""ᵒ"" => (degrees * π / 180)

/--
Auxiliary definition for constructing the points B' & C'.
Assuming ω is the incircle of ABC, then `IsParaTangentFoot A B C B' ω` is true
if `B'` is the point where the line parallel to AC and tangent to ω intersects side BC, and not C.
-/
def IsParaTangentFoot (A B C B' : ℝ²) (ω : Sphere ℝ²) : Prop :=
  B' ≠ C ∧ Collinear ℝ {B', B, C} ∧
  ∃ t : AffineSubspace ℝ ℝ², t.Parallel (affineSpan ℝ {A, C}) ∧ ω.IsTangent t ∧ B' ∈ t

theorem PBAdvanced022
    -- a triangle ABC
    (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
    -- ABC satisfies the given side inequalities
    (h_side_len : dist A B < dist A C ∧ dist A C < dist B C)
    (I : ℝ²) (hI : I = incenter ⟨![A, B, C], tri⟩) -- I is the incenter
    -- M and N are the midpoints of sides CA and AB
    (M : ℝ²) (hM : M = midpoint ℝ C A)
    (N : ℝ²) (hN : N = midpoint ℝ A B)
    -- We will denote the incircle ω, and circumcircle Ω
    (ω : Sphere ℝ²) (hω : ω = insphere ⟨![A, B, C], tri⟩)
    (Ω : Sphere ℝ²) (hΩ : Ω = circumsphere ⟨![A, B, C], tri⟩)
    -- K is the midpoint of the arc BC of the circumcircle of triangle ABC which does not contain A
    (K : ℝ²) (hK : K ∈ Ω ∧ dist K B = dist K C ∧ (affineSpan ℝ {B, C}).SOppSide A K)
    -- B' ≠ B is the point where the line parallel to AC and tangent to the incircle of triangle ABC intersects side BC
    (B' : ℝ²) (hB : IsParaTangentFoot A B C B' ω)
    -- C' ≠ C be the point where the line parallel to AB and tangent to the incircle of triangle ABC intersects side BC
    (C' : ℝ²) (hC : IsParaTangentFoot A C B C' ω) :
    --  The value of $\angle NIM+\angle B'KC'$ in terms of degree is 180
    ∠ N I M + ∠ B' K C' = 180ᵒ := by sorry"
PB-Advanced-023,"On a table of size $3002\times3001$, a stone is placed on the leftmost cell of the first row. James and Peter play a game on this table. Peter selects $3000$ cells, under the rule that he must choose one from each row except the first and last rows (i.e., the $1$st and $3002$th row), and there must be at most one selected cell in each column. James knows this rule too, but he doesn't know which cells Peter selected. The goal of James is to move the stone to the last row, avoiding the cells selected by Peter. The stone can only move to adjacent cells on the table. If the stone enters a cell selected by Peter, James receives a penalty of 1 point, and the stone returns to its initial position (i.e., the leftmost cell). Find the smallest positive integer $n$ such that there exists a method for James to achieve his goal before receiving a penalty of $n$ points.","First we demonstrate that there is no winning strategy if James has
 2 attempts.


 Suppose that $(2,i)$ is the first cell in the second row that James
 reaches on his first attempt. Peter could have selected this cell,
 in which case James receives a penalty and returns to the initial
 position, and he cannot have reached any other cells past the first
 row.


 Next, suppose that $(3,j)$ is the first cell in the third row that
 James reaches on his second attempt. James must have moved to this
 cell from $(2,j)$, so we know $j\neq i$. So it is possible that
 Peter selected the cell $(3,j)$, in which case James also receives
 a penalty and returns to the initial position on his second attempt.
 Therefore James cannot guarantee to reach the last row in 2 attempts.


 Next, we exhibit a strategy for $n=3$. On the first attempt, James
 travels along the path

 \[
 (1,1)\rightarrow(2,1)\rightarrow(2,2)\rightarrow\cdots\rightarrow(2,3001).
 \]

 This path meets every cell in the second row, so James will find the
 selected cell in row 2, receive a penalty, and return to the initial
 position.


 If the selected cell in the second row is not on the edge of the board
 (that is, it is in cell $(2,i)$ with $2\leq i\leq3000$ ), then James
 takes the following two paths in his second and third attempts:

 \[
 \begin{aligned} & (1,1)\rightarrow\dots\rightarrow(1,i-1)\rightarrow(2,i-1)\rightarrow(3,i-1)\rightarrow(3,i)\rightarrow(4,i)\rightarrow\cdots\rightarrow(3002,i).

  & (1,1)\rightarrow\dots\rightarrow(1,i+1)\rightarrow(2,i+1)\rightarrow(3,i+1)\rightarrow(3,i)\rightarrow(4,i)\rightarrow\cdots\rightarrow(3002,i).
 \end{aligned}
 \]

 The only cells that Peter might have selected in either of these paths
 are $(3,i-1)$ and $(3,i+1)$. At most one of these can have been
 selected by Peter, so at least one of the two paths will be successful.
 James starts each path from the initial position $(1,1)$, but he
 knows where the selected cell in row 2 is, so he can navigate to $(1,i-1)$
 or $(1,i+1)$ directly.


 If the selected cell in the second row is on the edge of the board,
 without loss of generality we may assume it is in $(2,1)$. Then,
 on the second attempt, James takes the following path:

 \[
 (1,1)\rightarrow(1,2)\rightarrow(2,2)\rightarrow(2,3)\rightarrow(3,3)\rightarrow\cdots\rightarrow(3000,3001)\rightarrow(3001,3001)\rightarrow(3002,3001).
 \]

 If none of the cells on this path were selected by Peter, then James
 wins. Otherwise, let $(i,j)$ be the first cell on which James encounters
 a selected cell. We have that $j=i$ or $j=i+1$. Then, on the third
 attempt, James takes the following path:

 \[
 \begin{aligned}(1,1)\rightarrow\dots\rightarrow(1,2) & \rightarrow(2,2)\rightarrow(2,3)\rightarrow(3,3)\rightarrow\cdots\rightarrow(i-2,i-1)\rightarrow(i-1,i-1)

  & \rightarrow(i,i-1)\rightarrow(i,i-2)\rightarrow\cdots\rightarrow(i,2)\rightarrow(i,1)

  & \rightarrow(i+1,1)\rightarrow\cdots\rightarrow(3001,1)\rightarrow(3002,1).
 \end{aligned}
 \]

 Now note that:
 \begin{itemize}
 \item The cells from $(1,2)$ to $(i-1,i-1)$ were not selected because
 they were reached earlier than $(i,j)$ on the previous attempt.
 \item The cells $(i,k)$ for $1\leq k\leq i-1$ were not selected because
 there is only one selected cell in row $i$, and it lies in $(i,i)$
 or $(i,i+1)$.
 \item The cells $(k,1)$ for $i\leq k\leq3002$ were not selected because
 Peter selected at most one cell in column 1 (since there is one selected
 cell per row from 2 to 3001), and it lies in $(2,1)$.
 \end{itemize}
 Therefore James will win on the third attempt. Therefore, the smallest
 positive integer $n$ is $3$.","(Partial)
 1. Proved that the answer is at least $3$.
 2. Solved the case where the stone in the second row is not on the edge.
(Almost)
 1. Solved the case where the stone in the second row is on the edge.
 2. Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,IMO-medium,3,(Modified) IMO 2024 P5,"import Mathlib

/-
On a table of size $3002\times3001$, a stone is placed on the leftmost cell of the first row. James and Peter play a game on this table. Peter selects $3000$ cells, under the rule that he must choose one from each row except the first and last rows (i.e., the $1$st and $3002$th row), and there must be at most one selected cell in each column. James knows this rule too, but he doesn't know which cells Peter selected. The goal of James is to move the stone to the last row, avoiding the cells selected by Peter. The stone can only move to adjacent cells on the table. If the stone enters a cell selected by Peter, James receives a penalty of 1 point, and the stone returns to its initial position (i.e., the leftmost cell). Find the smallest positive integer $n$ such that there exists a method for James to achieve his goal before receiving a penalty of $n$ points.

Answer: 3
-/
open Classical

-- Game Constants
def R_rows : ℕ := 3002
def C_cols : ℕ := 3001
abbrev Cell : Type := ℕ × ℕ

-- Location definitions
/-- A cell is on the table if its indices are within [1, R_rows] x [1, C_cols]. -/
def IsOnTable (c : Cell) : Prop :=
  c.fst > 0 ∧ c.fst ≤ R_rows ∧ c.snd > 0 ∧ c.snd ≤ C_cols

def start_cell : Cell := (1, 1)
/-- The goal is to reach any cell in the last row. -/
def target_row : Set Cell := { (r, c) | r = R_rows ∧ IsOnTable (r, c) }

-- Adjacency: orthogonal movement within bounds
/-- A move from c₁ to c₂ is adjacent if they are orthogoanlly next to each other and c₂ is on the table. -/
def IsAdjacentMove (c₁ c₂ : Cell) : Prop :=
  IsOnTable c₂ ∧
  (abs ((c₁.fst : ℤ) - (c₂.fst : ℤ)) + abs ((c₁.snd : ℤ) - (c₂.snd : ℤ)) = 1)

-- Peter's Constraints
/-- The set of rows Peter must choose one blocking cell from: {2, 3, ..., R_rows - 1}. -/
def PeterRows : Set ℕ := {r : ℕ | 2 ≤ r ∧ r ≤ R_rows - 1}
/-- The set of valid columns. -/
def Columns : Set ℕ := {c : ℕ | 1 ≤ c ∧ c ≤ C_cols}

/-- A set of selected cells S that satisfies Peter's rules: one cell per row \{2, ..., R-1\}, at most one cell per column {1, ..., C}. -/
def IsPeterSelection (S : Set Cell) : Prop :=
  -- 1. Cells are valid and within Peter's designated rows/columns
  (∀ (r c : ℕ), (r, c) ∈ S → r ∈ PeterRows ∧ c ∈ Columns) ∧
  -- 2. Exactly one selected cell per Peter row
  (∀ r ∈ PeterRows, ∃! c : ℕ, (r, c) ∈ S) ∧
  -- 3. At most one selected cell per column (injectivity)
  (∀ (r₁ c₁ r₂ c₂ : ℕ), (r₁, c₁) ∈ S → (r₂, c₂) ∈ S → c₁ = c₂ → r₁ = r₂)

def PeterSelections : Set (Set Cell) := {S : Set Cell | IsPeterSelection S}

/--
A valid path is a list of cells that starts at the start cell,
finishes at the target row, and consists of adjacent, on-table moves. -/
def IsValidPath (P : List Cell) : Prop :=
  P.head? = some start_cell ∧
  (∃ c ∈ target_row, P.getLast? = some c) ∧
  (∀ c : Cell, c ∈ P → IsOnTable c) ∧
  P.Chain' IsAdjacentMove

/-- A path is winning against a selection S if it is valid, avoids S entirely, and ends on the target row. -/
noncomputable
def pathFirstHit (P : List Cell) (S : Set Cell) : Option (Fin P.length) :=
  P.findFinIdx? (fun c ↦ c ∈ S)

def PathIsWinning (P : List Cell) (S : Set Cell) : Prop :=
  IsValidPath P ∧
  (∀ c : Cell, c ∈ P → c ∉ S)

/--
James' strategy is a finite tree of choosing paths.
Every time James tries a path, this path either succeeds, or James gets an information about
at which step it failed. This equals to the maximum possible penalty a strategy can get.
-/
inductive JamesStrategy
| giveUp
| tryPath (P : List Cell) (hP : IsValidPath P)
  (next : (i : ℕ) → i < P.length → JamesStrategy)

/-- The maximal number of steps a strategy can take -/
noncomputable
def JamesStrategy.depth : JamesStrategy → ℕ
| giveUp => 0
| tryPath P _ next => sSup { (next i.val i.isLt).depth | (i : Fin P.length) } + 1

def JamesStrategy.winsOn (S : Set Cell) : JamesStrategy → Prop
| giveUp => False
| tryPath P _ next =>
  match pathFirstHit P S with
  | some i => (next i.val i.isLt).winsOn S
  | none => True

/-- A given strategy always wins regardless of Peter's selection of cells. -/
def JamesStrategy.wins (strategy : JamesStrategy) : Prop :=
  ∀ S ∈ PeterSelections, strategy.winsOn S

/-- The property that there exists a strategy of length n that guarantees a win before n penalties. -/
def HasWinningStrategyBeforeNPenalties (n : ℕ) : Prop :=
  n > 0 ∧
  ∃ strategy : JamesStrategy, strategy.wins ∧ strategy.depth ≤ n

/-- The smallest positive integer n such that there exists a method for James to achieve his goal before receiving a penalty of n points. -/
noncomputable
def SmallestWinningPenaltyBound : ℕ :=
  sInf {n : ℕ | HasWinningStrategyBeforeNPenalties n}

-- Formalization of the Claim provided in the problem statement
theorem PBAdvanced023 : SmallestWinningPenaltyBound = 3 := by sorry"
PB-Advanced-024,"Let $P$ be a function from the set $\mathbb{Q}$ of rational numbers
 to itself, and suppose that $P$ satisfies
 \[
 (P(b-P(a))+a-P(b))(P(a+P(b-P(a)))-b)=0
 \]
 for all rational numbers $a,b$. Prove that the set $\{P(a)+P(-a):a\in\mathbb{Q}\}$
 is a finite set, and find the maximum possible number of elements
 in this set.","Let the given condition be
 \[
 (P(b-P(a))+a-P(b))(P(a+P(b-P(a)))-b)=0\quad(*)
 \]
 for all $a,b\in\mathbb{Q}$. Let $y=b-P(a)$. Then $b=y+P(a)$. Substituting
 this into $(*)$, we get
 \[
 (P(y)+a-P(y+P(a)))(P(a+P(y))-(y+P(a)))=0.
 \]
 This must hold for all $a\in\mathbb{Q}$ and all $y$ in the image
 of $b\mapsto b-P(a)$. Since $b$ can be any rational number, $y$
 can be any rational number (assuming $P(a)\in\mathbb{Q}$). Thus,
 for all $a,y\in\mathbb{Q}$, we have
 \[
 P(y+P(a))=P(y)+a\quad\text{or}\quad P(a+P(y))=y+P(a).
 \]
 Let's rewrite this by swapping variable names $a\to y$ and $y\to x$:
 For all $x,y\in\mathbb{Q}$,
 \[
 P(x+P(y))=P(x)+y\quad\text{or}\quad P(P(x)+y)=x+P(y).\quad(**)
 \]
 We want to find the maximum possible number of elements in the set
 $S=\{P(a)+P(-a):a\in\mathbb{Q}\}$. Let $g(a)=P(a)+P(-a)$. We want
 to find the maximum possible size of the image of $g$.


 We begin by providing an example of a function $P$ for which there
 are two values of $g(a)$. We take the function $P(x)=2\lfloor x\rfloor-x=\lfloor x\rfloor-\{x\}$.
 First, we show that $P$ satisfies $(**)$. Given $x,y\in\mathbb{Q}$,
 writing $f$ instead of $P$: If $\{x\}<\{y\}$, then $f(f(x)+y)=x+f(y)$.
 This is the second case of $(**)$. If $\{x\}>\{y\}$, then $f(x+f(y))=f(x)+y$.
 This is the first case of $(**)$. If $\{x\}=\{y\}$, then $f(x)+y=x+f(y)=N$
 (an integer). $f(N)=N$. So $f(x+f(y))=f(N)=N=f(x)+y$ and $f(f(x)+y)=f(N)=N=x+f(y)$.
 Both cases of $(**)$ hold. In all cases, the relation $(**)$ is
 satisfied, which is equivalent to the original condition $(*)$.


 Finally, we compute $g(a)=P(a)+P(-a)$ for this function. $g(a)=(2\lfloor a\rfloor-a)+(2\lfloor-a\rfloor-(-a))=2(\lfloor a\rfloor+\lfloor-a\rfloor)$.
 If $a$ is an integer, $\lfloor a\rfloor=a$ and $\lfloor-a\rfloor=-a$,
 so $g(a)=2(a-a)=0$. If $a$ is not an integer, $\lfloor-a\rfloor=-\lfloor a\rfloor-1$,
 so $g(a)=2(\lfloor a\rfloor-\lfloor a\rfloor-1)=-2$. The set of values
 for $g(a)$ is $\{0,-2\}$. There are two values for $g(a)$.

 Now, we prove that there cannot be more than two values of $g(a).$
 Applying $(**)$ with $y=x$ tells us that $P(x+P(x))=P(x)+x$ or
 $P(P(x)+x)=x+P(x)$. In either case, if we let $z=x+P(x)$, then $P(z)=z$.
 So
 \[
 P(x+P(x))=x+P(x)\qquad\tag{(1)}
 \]
 for all $x\in\mathbb{Q}$.


 We begin with the following lemma.


 \textbf{Lemma 1.} $P$ is a bijection, and satisfies
 \[
 P(-P(-x))=x\qquad\tag{(2)}
 \]

 \emph{Proof.} We first prove that $P$ is injective. Suppose that
 $P(x_{1})=P(x_{2})=Y$. The condition $(**)$ applied to $(x_{1},x_{2})$
 gives $P(x_{1}+P(x_{2}))=P(x_{1})+x_{2}$ or $P(P(x_{1})+x_{2})=x_{1}+P(x_{2})$.
 Substituting $P(x_{1})=P(x_{2})=Y$, this becomes $P(x_{1}+Y)=Y+x_{2}$
 or $P(Y+x_{2})=x_{1}+Y$.

 From (1), we know $P(x_{1}+P(x_{1}))=x_{1}+P(x_{1})$, i.e., $P(x_{1}+Y)=x_{1}+Y$.
 Also from (1), $P(x_{2}+P(x_{2}))=x_{2}+P(x_{2})$, i.e., $P(x_{2}+Y)=x_{2}+Y$.

 Case A: $P(x_{1}+Y)=Y+x_{2}$ holds. Using $P(x_{1}+Y)=x_{1}+Y$,
 we get $x_{1}+Y=Y+x_{2}$, which implies $x_{1}=x_{2}$.

 Case B: $P(Y+x_{2})=x_{1}+Y$ holds. Using $P(x_{2}+Y)=x_{2}+Y$,
 we get $x_{2}+Y=x_{1}+Y$, which implies $x_{1}=x_{2}$.

 In both cases, $x_{1}=x_{2}$. So $P$ is injective.


 Now, (1) with $x=0$ tells us that $P(0+P(0))=0+P(0)$, so $P(P(0))=P(0)$.
 By injectivity, $P(0)=0$.

 Let $C(x,y)$: $P(x+P(y))=P(x)+y$ or $P(P(x)+y)=x+P(y)$.

 $C(x,-P(x))$: $P(x+P(-P(x)))=P(x)-P(x)=0$ or $P(P(x)-P(x))=x+P(-P(x))$.

 Case 1: $P(x+P(-P(x)))=0$. Since $P(0)=0$ and $P$ is injective,
 $x+P(-P(x))=0$, so $P(-P(x))=-x$.

 Case 2: $P(0)=x+P(-P(x))$. Since $P(0)=0$, this gives $0=x+P(-P(x))$,
 so $P(-P(x))=-x$.

 Both options yield $P(-P(x))=-x$. Replace $x$ with $-x$: $P(-P(-x))=-(-x)=x$.
 This proves (2).


 Finally, bijectivity follows immediately from (2). $P$ is injective.
 For any $y\in\mathbb{Q}$, let $x=-P(-y)$. Then $P(x)=P(-P(-y))=y$.
 So $P$ is surjective.


 Since $P$ is bijective, it has an inverse, which we denote $P^{-1}$.
 From (2), $P(-P(-x))=x$. Let $z=-P(-x)$. Then $P(z)=x$, so $z=P^{-1}(x)$.
 Thus, $P^{-1}(x)=-P(-x)$. Rearranging gives $P(-x)=-P^{-1}(x)$.
 We have $g(x)=P(x)+P(-x)=P(x)-P^{-1}(x)$.

 Suppose $g(x)=u$ and $g(y)=v$, where $u\neq v$ are both nonzero.
 Define $x^{\prime}=P^{-1}(x)$ and $y^{\prime}=P^{-1}(y)$. By definition,
 $P(x')=x$ and $P(y')=y$. We have $g(x)=P(x)-P^{-1}(x)=P(x)-x'=u$,
 so $P(x)=x'+u$. Similarly,
 \[
 g(y)=P(y)-P^{-1}(y)=P(y)-y'=v,
 \]
 so $P(y)=y'+v$.


 Apply $(**)$ to $(x',y)$:

 \[
 P(x'+P(y))=P(x')+y\text{ or }P(P(x')+y)=x'+P(y).
 \]
 $P(x'+y'+v)=x+y$ or $P(x+y)=x'+y'+v$. (Let $A=x+y,B=x'+y'+v$. $P(B)=A$
 or $P(A)=B$.)



  Apply $(**)$ to $(x,y')$:
 \[
 P(x+P(y'))=P(x)+y'\text{ or }P(P(x)+y')=x+P(y').
 \]
 $P(x+y)=P(x)+y'=(x'+u)+y'$ or $P(x'+u+y')=x+y$. (Let $C=x'+y'+u$.
 $P(A)=C$ or $P(C)=A$.)


 We have the possibilities:

 (1a) $P(B)=A$ and (2a) $P(A)=C$.

 (1a) $P(B)=A$ and (2b) $P(C)=A$. Since $P$ is injective, $B=C$,
 so $x'+y'+v=x'+y'+u$, which implies $u=v$. Contradiction.

 (1b) $P(A)=B$ and (2a) $P(A)=C$. Since $P(A)$ must be unique, $B=C$,
 which implies $u=v$. Contradiction.

 (1b) $P(A)=B$ and (2b) $P(C)=A$.

 So we must have either ($P(B)=A$ and $P(A)=C$) or ($P(A)=B$ and
 $P(C)=A$).


 Case 1: $P(x'+y'+v)=x+y$ and $P(x+y)=x'+y'+u$.

 Case 2: $P(x+y)=x'+y'+v$ and $P(x'+y'+u)=x+y$.


 Swapping $(x,u)$ with $(y,v)$ transforms Case 1 into Case 2 structure
 and vice versa. So, without loss of generality, assume Case 2 holds:
 $P(x+y)=x'+y'+v$ and $P(x'+y'+u)=x+y$.


 Now consider $(**)$ applied to $(x+y,-x'-u)$. Let $A=x+y$ and $Z=-x'-u$.
 Using $P(-z)=-P^{-1}(z)$, we have $P(-x'-u)=-P^{-1}(x'+u)$. From
 $P(x)=x'+u$ we have $x=P^{-1}(x'+u)$. Therefore, $P(Z)=P(-x'-u)=-x$.

 Now $C(A,Z)$ gives: $P(A+P(Z))=P(A)+Z$ or $P(P(A)+Z)=A+P(Z)$. Substitute
 $A=x+y$, $Z=-x'-u$, $P(Z)=-x$. And $P(A)=P(x+y)=x'+y'+v$ (using
 Case 2).


 Option 1: $P((x+y)+(-x))=P(x+y)+(-x'-u)$. $P(y)=(x'+y'+v)-x'-u=y'+v-u$.

 Option 2: $P(P(x+y)+Z)=A+P(Z)$. $P((x'+y'+v)+(-x'-u))=(x+y)+(-x)$.
 $P(y'+v-u)=y$.


 So we have $P(y)=y'+v-u$ or $P(y'+v-u)=y$. This means $y'+v-u$
 must be either $P(y)$ or $P^{-1}(y)$. We know $P(y)=y'+v$ and $P^{-1}(y)=y'$.
 If $y'+v-u=P(y)=y'+v$, then $v-u=v$, which means $u=0$. This contradicts
 our assumption $u\neq0$. If $y'+v-u=P^{-1}(y)=y'$, then $v-u=0$,
 which means $u=v$. This contradicts our assumption $u\neq v$.


 Both possibilities lead to a contradiction. Therefore, the initial
 assumption that there exist distinct non-zero values $u,v$ for $g(x)$
 must be false. The set $S=\{g(a):a\in\mathbb{Q}\}$ can contain $g(0)=P(0)+P(0)=0$.
 It can contain at most one non-zero value $u$. Thus, the set $S$
 must be a subset of $\{0,u\}$ for some $u\in\mathbb{Q}$. The maximum
 possible number of elements in $S$ is $2$.


 Since we found an example where the set of values is $\{0,-2\}$,
 which has 2 elements, the maximum possible number of elements is exactly
 2.","(Partial)
 1. Correctly guessed all the possible $P$.
 2. Proved that $P$ is bijective.
 3. Proved that $P(-P(x))=-x$.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Algebra,IMO-hard,2,(Modified) IMO 2024 P6,"import Mathlib

/-
Let $P$ be a function from the set $\mathbb{Q}$ of rational numbers to itself, and suppose that $P$ satisfies
$$(P(b-P(a))+a-P(b))(P(a+P(b-P(a)))-b)=0$$
for all rational numbers $a, b$. Prove that the set $\{P(a)+P(-a):a\in\mathbb{Q}\}$ is a finite set, and find the maximum possible number of elements in this set.

Solution: the maximum number of elements is 2.
-/
theorem PBAdvanced024
    (IsGood : (ℚ → ℚ) → Prop)
    (IsGood_def : ∀ P, IsGood P ↔ ∀ a b,
      (P (b - P a) + a - P b) * (P (a + P (b - P a)) - b) = 0) :
    (∀ P, IsGood P → {P a + P (- a) | (a : ℚ)}.encard < ⊤) ∧
      IsGreatest {{P a + P (-a) | (a : ℚ)}.encard | (P : ℚ → ℚ) (_ : IsGood P)} 2 := by sorry"
PB-Advanced-025,"Let $k$ and $d$ be positive integers. Prove that there exists a
 positive integer $N$ such that for every odd integer $n>N$, the
 digits in the base-$2n$ representation of $n^{k}$ are all greater
 than $d$.","The problem actually doesn't have much to do with digits: the idea
 is to pick any length $\ell\leq k$, and look at the rightmost $\ell$
 digits of $n^{k}$; that is, the remainder upon division by $(2n)^{\ell}$.
 We compute it exactly:

 Claim - Let $n\geq1$ be an odd integer, and $k\geq\ell\geq1$ integers.
 Then

 \[
 n^{k}\bmod(2n)^{l}=c(k,\ell)\cdot n^{l}
 \]

 for some odd integer $1\leq c(k,\ell)\leq2^{\ell}-1$.

 Proof. This follows directly by the Chinese remainder theorem, with
 $c(k,\ell)$ being the residue class of $n^{k-i}\left(\bmod2^{\ell}\right)$
 since $n$ is odd. In particular, for the $\ell$ th digit from the
 right to be greater than $d$, it would be enough that
 \[
 c(k,\ell)\cdot n^{\ell}\geq(d+1)\cdot(2n)^{\ell-1}
 \]
 But this inequality holds whenever $n\geq(d+1)\cdot2^{\ell-1}$.

 Putting this together by varying $\ell$, we find that for all odd

 \[
 n\geq(d+1)\cdot2^{k-1}
 \]

 we have that
 \begin{itemize}
 \item $n^{k}$ has $k$ digits in base- $2n$; and
 \item for each $\ell=1,\ldots,k$, the $\ell^{\text{th }}$ digit from the
 right is at least $d+1$

  so the problem is solved.

  Remark. Note it doesn't really matter that $c(k,i)$ is odd per se;
 we only need that $c(k,i)\geq1$.
 \end{itemize}","(Partial)
 1. Proved that $n^{k}$ has $k$ digits on base $2n$ representation for sufficiently large $n$.
 2. Expressed the $i$th digit such as $\lfloor\frac{n^{k}}{(2n)^{i}}\rfloor\pmod{2n}$.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Number theory,IMO-easy, ,USAMO 2025,"import Mathlib

/-
Let $k$ and $d$ be positive integers. Prove that there exists a positive integer $N$ such that for every odd integer $n > N$, the digits in the base-$2n$ representation of $n^k$ are all greater than $d$.
-/
theorem PBAdvanced025
    (k d : ℕ) (hk : 0 < k) (hd : 0 < d) :
    ∃ N > 0, ∀ n > N, Odd n → ∀ u ∈ Nat.digits (2*n) (n^k), d < u := by sorry"
PB-Advanced-026,"Let $n$ and $k$ be positive integers with $k<n$. Let $P(x)$ be
 a polynomial of degree $n$ with real coefficients, nonzero constant
 term, and no repeated roots. Suppose that for any real numbers $a_{0},a_{1},\dots,a_{k}$
 such that the polynomial $a_{k}x^{k}+\dots+a_{1}x+a_{0}$ divides
 $P(x)$, the product $a_{0}a_{1}\dots a_{k}$ is zero. Prove that
 $P(x)$ has a non-real root.","Let $n>k\geq1$ be integers. Let $P(x)\in\mathbb{R}[x]$ be a polynomial
 of degree $n$ with no repeated roots and $P(0)\neq0$. Suppose that
 for any real numbers $a_{0},\ldots,a_{k}$ such that the polynomial
 $a_{k}x^{k}+\cdots+a_{1}x+a_{0}$ divides $P(x)$, the product $a_{0}a_{1}\ldots a_{k}$
 is zero. Prove that $P(x)$ has a non-real root.


 By considering any $k+1$ of the roots of $P$, we may as well assume
 WLOG that $n=k+1$. Suppose that $P(x)=\left(x+r_{1}\right)\ldots\left(x+r_{n}\right)\in\mathbb{R}[x]$
 has $P(0)\neq0$. Then the problem hypothesis is that each of the
 $n$ polynomials (of degree $n-1$ ) given by

 \[
 \begin{aligned}P_{1}(x) & =\left(x+r_{2}\right)\left(x+r_{3}\right)\left(x+r_{4}\right)\ldots\left(x+r_{n}\right)

 P_{2}(x) & =\left(x+r_{1}\right)\left(x+r_{3}\right)\left(x+r_{4}\right)\ldots\left(x+r_{n}\right)

 P_{3}(x) & =\left(x+r_{1}\right)\left(x+r_{2}\right)\left(x+r_{4}\right)\ldots\left(x+r_{n}\right)

  & \vdots

 P_{n}(x) & =\left(x+r_{1}\right)\left(x+r_{2}\right)\left(x+r_{3}\right)\ldots\left(x+r_{n-1}\right)
 \end{aligned}
 \]

 has at least one coefficient equal to zero. (Explicitly, $P_{i}(x)=\frac{P(x)}{x+r_{i}}$.)
 We'll prove that at least one $r_{i}$ is not real.


 Obviously the leading and constant coefficients of each $P_{i}$ are
 nonzero, and there are $n-2$ other coefficients to choose between.
 So by pigeonhole principle, we may assume, say, that $P_{1}$ and
 $P_{2}$ share the position of a zero coefficient, say the $x^{k}$
 one, for some $1\leq k<n-1$.


 \textbf{Claim 1. }If $P_{1}$ and $P_{2}$ both have $x^{k}$ coefficient
 equal to zero, then the polynomial

 \[
 Q(x)=\left(x+r_{3}\right)\left(x+r_{4}\right)\ldots\left(x+r_{n}\right)
 \]

 has two consecutive zero coefficients.


 \emph{Proof.} Suppose that

 \[
 Q(x)=x^{n-2}+b_{n-3}x^{n-3}+\cdots+b_{0}.
 \]

 (And let $b_{n-2}=1$. ) Then the fact that the $x^{k}$ coefficient
 of $P_{1}$ and $P_{2}$ are both zero means

 \[
 r_{1}b_{k}+b_{k-1}=r_{2}b_{k}+b_{k-1}=0
 \]

 and hence that $b_{k}=b_{k-1}=0$ (since the $r_{i}$ are distinct).



  To solve the problem, it suffices to prove:


 \textbf{Claim 2.} If $F(x)\in\mathbb{R}[x]$ is a polynomial with
 two consecutive zero coefficients, it cannot have all distinct real
 roots.


 \emph{Proof. }Say $x^{t}$ and $x^{t+1}$ coefficients of $F$ are
 both zero. Assume for contradiction all the roots of $F$ are real
 and distinct. Then by Rolle's theorem, every higher-order derivative
 of $F$ should have this property too. However, the $t$th order derivative
 of $F$ has a double root of $0$, contradiction.","(Partial)
 1. (WLOG assumed $n=k+1$ and) proved that two different divisors of $P$ share the position of a zero coefficient.
 2. Found a polynomial with only distinct real roots and has two consecutive zero coefficients.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible.",Algebra,IMO-medium, ,USAMO 2025,"import Mathlib

/-
Let $n$ and $k$ be positive integers with $k < n$. Let $P(x)$ be a polynomial of degree $n$ with real coefficients, nonzero constant term, and no repeated roots. Suppose that for any real numbers $a_0, a_1, \dots, a_k$ such that the polynomial $a_k x^k + \dots + a_1 x + a_0$ divides $P(x)$, the product $a_0 a_1 \dots a_k$ is zero. Prove that $P(x)$ has a nonreal root.
-/
open scoped Polynomial

theorem PBAdvanced026
    (P : ℝ[X]) (hP : P.coeff 0 ≠ 0) (hP' : (P.aroots ℂ).Nodup)
    (k : ℕ) (hk : 0 < k) (hkn : k < P.natDegree)
    (H : ∀ Q, Q ∣ P → Q.natDegree ≤ k →
      ∏ i ∈ Finset.range k.succ, Q.coeff i = 0) :
    ∃ z, z.im ≠ 0 ∧ z ∈ P.aroots ℂ := by sorry"
PB-Advanced-027,"Alice the architect and Bob the builder play a game. First, Alice
 chooses two points $P$ and $Q$ in the plane and a subset $S$ of
 the plane, which are announced to Bob. Next, Bob marks infinitely
 many points in the plane, designating each a city. He may not place
 two cities within distance at most one unit of each other, and no
 three cities he places may be collinear. Finally, roads are constructed
 between the cities as follows: for each pair $A,B$ of cities, they
 are connected with a road along the line segment $AB$ if and only
 if the following condition holds: For every city $C$ distinct from
 $A$ and $B$, there exists $R\in S$ such that $\triangle PQR$ is
 directly similar to either $\triangle ABC$ or $\triangle BAC$. Alice
 wins the game if

 \noindent (i) the resulting roads allow for travel between any pair
 of cities via a finite sequence of roads and

 \noindent (ii) no two roads cross.

 \noindent Otherwise, Bob wins. Determine, with proof, which player
 has a winning strategy. (Note: $\triangle UVW$ is directly similar
 to $\triangle XYZ$ if there exists a sequence of rotations, translations,
 and dilations sending $U$ to $X$, $V$ to $Y$, and $W$ to $Z$.)","The answer is that Alice wins. Let's define a Bob-set $V$ to be
 a set of points in the plane with no three collinear and with all
 distances at least 1 . The point of the problem is to prove the following
 fact.

 Claim - Given a Bob-set $V\subseteq\mathbb{R}^{2}$, consider the
 Bob-graph with vertex set $V$ defined as follows: draw edge $ab$
 if and only if the disk with diameter $\overline{ab}$ contains no
 other points of $V$ on or inside it. Then the Bob-graph is connected
 and planar.

 Proving this claim shows that Alice wins since Alice can specify $\mathcal{S}$
 to be the set of points outside the disk of diameter $PQ$.

 We start with proving that every Bob-graph is connected. Assume for
 contradiction the graph is disconnected. Let $p$ and $q$ be two
 points in different connected components. Since $pq$ is not an edge,
 there exists a third point $r$ inside the disk with diameter $\overline{pq}$.

 Hence, $r$ is in a different connected component from at least one
 of $p$ or $q$ - let's say point $p$. Then we repeat the same argument
 on the disk with diameter $\overline{pr}$ to find a new point $s$,
 non-adjacent to either $p$ or $r$. See the figure below, where the
 X'ed out dashed edges indicate points which are not only non-adjacent
 but in different connected components.

 In this way we generate an infinite sequence of distances $\delta_{1},\delta_{2},\delta_{3},\ldots$
 among the non-edges in the picture above. By the ""Pythagorean theorem""
 (or really the inequality for it), we have

 \[
 \delta_{i}^{2}\leq\delta_{i-1}^{2}-1
 \]

 and this eventually generates a contradiction for large $i$, since
 we get $0\leq\delta_{i}^{2}\leq\delta_{1}^{2}-(i-$ $1)$.

 Now we prove that every Bob-graph is planar. Assume for contradiction
 that edges $ac$ and $bd$ meet, meaning $abcd$ is a convex quadrilateral.
 WLOG assume $\angle bad\geq90^{\circ}$ (each quadrilateral has an
 angle at least $90^{\circ}$ ). Then the disk with diameter $\overline{bd}$
 contains $a$, contradiction.","(Partial)
 1. Argued that Alice should take $S$ to be the exterior of the circle of diameter $PQ$.
 2. Constructed $S$ as above and proved that the resulting graph is planar.
(Almost)
 1. Constructed $S$ as above and proved that the resulting graph is connected.
 2. Solution is almost complete, but made minor mistakes which are not negligible.",Combinatorics,IMO-hard,Alice,USAMO 2025,"import Mathlib

/-
Alice the architect and Bob the builder play a game. First, Alice
chooses two points $P$ and $Q$ in the plane and a subset $S$ of
the plane, which are announced to Bob. Next, Bob marks infinitely
many points in the plane, designating each a city. He may not place
two cities within distance at most one unit of each other, and no
three cities he places may be collinear. Finally, roads are constructed
between the cities as follows: for each pair $A,B$ of cities, they
are connected with a road along the line segment $AB$ if and only
if the following condition holds: For every city $C$ distinct from
$A$ and $B$, there exists $R\in S$ such that $\triangle PQR$ is
directly similar to either $\triangle ABC$ or $\triangle BAC$. Alice
wins the game if

\noindent (i) the resulting roads allow for travel between any pair
of cities via a finite sequence of roads and

\noindent (ii) no two roads cross.

\noindent Otherwise, Bob wins. Determine, with proof, which player
has a winning strategy. (Note: $\triangle UVW$ is directly similar
to $\triangle XYZ$ if there exists a sequence of rotations, translations,
and dilations sending $U$ to $X$, $V$ to $Y$, and $W$ to $Z$.)

Answer: Alice
-/

open EuclideanGeometry Relation
local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

theorem finRank_eq2 : Module.finrank ℝ ℝ² = 2 := by simp

local instance : Fact (Module.finrank ℝ ℝ² = 2) := ⟨finRank_eq2⟩

noncomputable
local instance : Module.Oriented ℝ ℝ² (Fin 2) where
  positiveOrientation :=
    cast (a := (Module.finBasis (R := ℝ) (M := ℝ²)).orientation) (by rw [finRank_eq2])

def DirectlySimilar (A B C A' B' C' : ℝ²) : Prop :=
  oangle A B C = oangle A' B' C' ∧
  oangle B C A = oangle B' C' A'

-- Alice chooses two points $P$ and $Q$ in the plane and a subset $S$ of the plane
structure AliceData where
  (P Q : ℝ²)
  S : Set ℝ²

/-
Bob marks infinitely
many points in the plane, designating each a city. He may not place
two cities within distance at most one unit of each other, and no
three cities he places may be collinear.
-/
def ValidBobSet (S : Set ℝ²) : Prop :=
  S.Infinite ∧
  (∀ A ∈ S, ∀ B ∈ S, A ≠ B → dist A B > 1) ∧
  (∀ triple ⊆ S, triple.encard = 3 → ¬ Collinear ℝ triple)

/-
For each pair $A,B$ of cities, they are connected with a road along
the line segment $AB$ if and only if the following condition holds:
For every city $C$ distinct from $A$ and $B$, there exists $R\in S$
such that $\triangle PQR$ is directly similar to either $\triangle ABC$
or $\triangle BAC$.
-/
def hasRoad (alice : AliceData) (bob : Set ℝ²) (A B : ℝ²) : Prop :=
  ∀ C ∈ bob, C ≠ A → C ≠ B →
  ∃ R ∈ alice.S,
    DirectlySimilar A B C alice.P alice.Q R ∨
    DirectlySimilar B A C alice.P alice.Q R

-- Alice wins the game if
def AliceWins (alice : AliceData) (bob : Set ℝ²) : Prop :=
  /-
  (i) the resulting roads allow for travel between any pair
  of cities via a finite sequence of roads and
  -/
  (∀ A ∈ bob, ∀ B ∈ bob, ReflTransGen (hasRoad alice bob) A B) ∧
  -- (ii) no two roads cross.
  (∀ A ∈ bob, ∀ B ∈ bob, ∀ C ∈ bob, ∀ D ∈ bob,
    [A, B, C, D].Nodup → hasRoad alice bob A B → hasRoad alice bob C D →
    (openSegment ℝ A B) ∩ (openSegment ℝ C D) = ∅)

-- Proposition that Alice can win the game no matter of Bob's action
def AliceHasWinningStrategy : Prop :=
  ∃ alice : AliceData, ∀ bob : Set ℝ², ValidBobSet bob → AliceWins alice bob

-- We want to prove that Alice has a winning strategy
theorem PBAdvanced027 : AliceHasWinningStrategy := by sorry"
PB-Advanced-028,"Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be
 the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection
 of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$
 intersects line $BC$ at two distinct points $X$ and $Y$. Prove
 that $C$ is the midpoint of $XY$.","Let $Q$ be the antipode of $B$ in the circumcircle of $ABC$.


 \textbf{Claim.} $AHQC$ is a parallelogram, and $APCQ$ is an isosceles
 trapezoid.



  \emph{Proof.} The claim follows from $\overline{AH}\perp\overline{BC}\perp\overline{CQ}$
 and $\overline{CF}\perp\overline{AB}\perp\overline{AQ}$.


 Let $M$ be the midpoint of $\overline{QC}$.


 \textbf{Claim.} Point $M$ is the circumcenter of $\triangle AFP$.


 \emph{Proof.} It's clear that $MA=MP$ from the isosceles trapezoid
 $APCQ$. As for $MA=MF$, let $N$ denote the midpoint of $\overline{AF}$;
 then $\overline{MN}$ is a midline of the parallelogram, so $\overline{MN}\perp\overline{AF}$.


 Since $\overline{CM}\perp\overline{BC}$ and $M$ is the center of
 $(AFP)$, it follows $CX=CY$.","(Partial)
  1. Constructed an external point, including the antipode of $B$ w.r.t $(ABC)$ or the antipode of $A$ w.r.t $(AFP)$, that may lead to a solution.
(Almost)
  1. Solution is almost complete, but made minor mistakes which are not negligible.",Geometry,IMO-easy, ,USAMO 2025,"import Mathlib

/-
Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be
the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection
of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$
intersects line $BC$ at two distinct points $X$ and $Y$. Prove
that $C$ is the midpoint of $XY$.
-/
open Affine Simplex EuclideanGeometry

local notation ""ℝ²"" => EuclideanSpace ℝ (Fin 2)

theorem PBAdvanced028
    -- an acute-angled triangle ABC
    (A B C : ℝ²) (tri : AffineIndependent ℝ ![A, B, C])
    (h_acute : AcuteAngled ⟨![A, B, C], tri⟩)
    (H : ℝ²) (hH : H = Triangle.orthocenter ⟨![A, B, C], tri⟩) -- H is the orthocenter
    (F : ℝ²) (hF : F = altitudeFoot ⟨![A, B, C], tri⟩ 2) -- F is the foot from C
    -- P be the reflection of H across BC
    (P : ℝ²) (hP : P = reflection (affineSpan ℝ {B, C}) H)
    -- the circumcircle of triangle AFP intersects line BC at two distinct points X and Y
    (X Y : ℝ²) (hXY : X ≠ Y ∧ Collinear ℝ {X, Y, B, C} ∧ Cospherical {X, Y, A, F, P}) :
    C = midpoint ℝ X Y := by sorry"
PB-Advanced-029,"Determine, with proof, all positive integers $k$ such that
 \[
 \frac{1}{n+1}\sum_{i=0}^{n}\binom{n}{i}^{k}
 \]
 is an integer for every positive integer $n$.","The answer is all even $k$. Let's abbreviate $S(n):=\binom{n}{0}^{k}+\cdots+\binom{n}{n}^{k}$
 for the sum in the problem.


 \textbf{Proof that even $k$ is necessary.}

  Choose $n=2$. We need $3\mid S(2)=2+2^{k}$, which requires $k$
 to be even.


 \textbf{Proof that even $k$ is sufficient. }

 From now on we treat $k$ as fixed, and we let $p^{e}$ be a prime
 fully dividing $n+1$. It is sufficient to prove that $S(n)$ is a
 multiple of $p^{e}$ for each such $p^{e}$. The basic idea is to
 reduce from $n+1$ to $(n+1)/p$ by an induction. We prove the following
 lemma on each individual binomial coefficient.


 \textbf{Lemma. }Suppose $p^{e}$ is a prime power which fully divides
 $n+1$. Then

 \[
 \binom{n}{i}\equiv\pm\binom{\frac{n+1}{p}-1}{\lfloor i/p\rfloor}\quad\left(\bmod p^{e}\right)
 \]

 \emph{Proof.} If we write
 \[
 \binom{n}{i}=\prod_{1\leq j\leq i}\frac{n-(j-1)}{j}
 \]

 then the fractions for $p\nmid j$ are all $\pm1\left(\bmod p^{e}\right)$.
 So only considers those $j$ with $p\mid j$; in that case one obtains
 the claimed $\binom{\frac{n+1}{p}-1}{\lfloor i/p\rfloor}$ exactly
 (even without having to take modulo $p^{e}$).


 From the lemma, it follows if $p^{e}$ is a prime power which fully
 divides $n+1$, then
 \[
 S(n)\equiv p\cdot S\left(\frac{n+1}{p}-1\right)\quad\left(\bmod p^{e}\right)
 \]

 by grouping the $n+1$ terms (for $0\leq i\leq n$ ) into consecutive
 ranges of length $p$ (by the value of $\lfloor i/p\rfloor$ ) and
 the result follows by induction.","(Partial)
 1. Proved that $k$ should be even.
 2. Used the induction on $n+1$ and considered the equation for$\frac{n+1}{p}$.
 3. Stated the identity $n+1\mid\sum_{i=0}^{n}\left((-1)^{i}\binom{n}{i}\right)^{k}$ without proof.
(Almost)
 1. Solution is almost complete, but made minor mistakes which are not negligible, such as coming up with a proof that only works for odd primes.",Number theory,IMO-medium,All even integer $k$,USAMO 2025,"import Mathlib

/-
Determine, with proof, all positive integers $k$ such that
\[
\frac{1}{n+1} \sum_{i=0}^{n} \binom{n}{i}^k
\]
is an integer for every positive integer $n$.

Solution: all even positive integers $k$.
-/
theorem PBAdvanced029 :
    {(k : ℕ) | 0 < k ∧ ∀ n > (0 : ℕ), ∃ (m : ℕ),
      1 / (n + 1 : ℚ) * ∑ i ∈ Finset.range (n + 1), n.choose i ^ k = m }
    = {(k : ℕ) | 0 < k ∧ Even k} := by sorry"
PB-Advanced-030,"Let $m$ and $n$ be positive integers with $m\ge n$. There are
 $m$ cupcakes of different flavors arranged around a circle and $n$
 people who like cupcakes. Each person assigns a non-negative real
 number score to each cupcake, depending on how much they like the
 cupcake. Suppose that for each person $P$, it is possible to partition
 the circle of $m$ cupcakes into $n$ groups of consecutive cupcakes
 so that the sum of $P$'s scores of the cupcakes in each group is
 at least $1$. Prove that it is possible to distribute the $m$ cupcakes
 to the $n$ people so that each person $P$ receives cupcakes of total
 score at least $1$ with respect to $P$.","Arbitrarily pick any one person - call her Pip - and her $n$ arcs.
 The initial idea is to try to apply Hall's marriage lemma to match
 the $n$ people with Pip's arcs (such that each such person is happy
 with their matched arc). To that end, construct the obvious bipartite
 graph $\mathfrak{G}$ between the people and the arcs for Pip.

 We now consider the following algorithm, which takes several steps.
 \begin{itemize}
 \item If a perfect matching of $\mathfrak{G}$ exists, we're done!
 \item We're probably not that lucky. Per Hall's condition, this means there
 is a bad set $\mathcal{B}_{1}$ of people, who are compatible with
 fewer than $\left|\mathcal{B}_{1}\right|$ of the arcs. Then delete
 $\mathcal{B}_{1}$ and the neighbors of $\mathcal{B}_{1}$, then try
 to find a matching on the remaining graph.
 \item If a matching exists now, terminate the algorithm. Otherwise, that
 means there's another bad set $\mathcal{B}_{2}$ for the remaining
 graph. We again delete $\mathcal{B}_{2}$ and the fewer than $\mathcal{B}_{2}$
 neighbors.
 \item Repeat until some perfect matching $\mathcal{M}$ is possible in the
 remaining graph, i.e. there are no more bad sets (and then terminate
 once that occurs).
 \end{itemize}
 Since Pip is a universal vertex, it's impossible to delete Pip, so
 the algorithm does indeed terminate with nonempty $\mathcal{M}$.

 We commit to assigning each of person in $\mathcal{M}$ their matched
 arc (in particular if there are no bad sets at all, the problem is
 already solved). Now we finish the problem by induction on $n$ (for
 the remaining people) by simply deleting the arcs used up by $\mathcal{M}$.

 To see why this deletion-induction works, consider any particular
 person Quinn not in $\mathcal{M}$. By definition, Quinn is not happy
 with any of the $\operatorname{arcs}$ in $\mathcal{M}$. So when
 an $\operatorname{arc}\mathcal{A}$ of $\mathcal{M}$ is deleted,
 it had value less than 1 for Quinn so in particular it couldn't contain
 entirely any of Quinn's arcs. Hence at most one endpoint among Quinn's
 arcs was in the deleted $\operatorname{arc}\mathcal{A}$. When this
 happens, this causes two arcs of Quinn to merge, and the merged value
 is

 \[
 (\geq1)+(\geq1)-(\leq1)\quad\geq1
 \]

 meaning the induction is OK.","(Partial)
 1. Observed that when an arc with scores smaller than $1$ to some person $X$ is deleted, the problem condition still holds for $X$.
 2. Applied Hall's theorem to the arcs of a fixed person and the $n$ people.
(Almost)
 1. Found a perfect matching and uses induction, but didn't explain why the induction works.
 2. Full proof, but didn't mention that there exists a nonempty set that satisfies Hall's condition.",Combinatorics,IMO-hard, ,USAMO 2025,"import Mathlib

/-- A non-empty finset of cupcakes is consecutive if it consists of indices `start, start+1, ..., start+len-1`
in the cyclic group structure of `Fin m`. -/
def IsConsecutive {m : ℕ} (A : Finset (Fin m)) : Prop :=
  A.Nonempty ∧ ∃ (start : Fin m) (len : ℕ),
    A.card = len ∧ len > 0 ∧
    -- k = start + i (mod m). Fin m addition is cyclic.
    ∀ k : Fin m, k ∈ A ↔ ∃ i < len, k = start + i

/-- The partitioning hypothesis: For each person `p`, the circle can be partitioned into `n`
consecutive groups, each giving `p` a total score of at least 1. -/
def satisfies_partition_condition {m n : ℕ} (scores : Fin n → Fin m → ℝ) : Prop :=
  ∀ (p : Fin n),
    ∃ (groups : Fin n → Finset (Fin m)),
      -- 1. Groups are consecutive
      (∀ i : Fin n, IsConsecutive (groups i)) ∧
      -- 2. Groups form a partition (unique containment)
      (∀ c : Fin m, ∃! i : Fin n, c ∈ groups i) ∧
      -- 3. Score condition
      (∀ i : Fin n, Finset.sum (groups i) (fun c => scores p c) ≥ 1)

/-- The conclusion: It is possible to distribute the cupcakes such that each person `p` receives a total score of at least 1. -/
def exists_good_distribution {m n : ℕ} (scores : Fin n → Fin m → ℝ) : Prop :=
  ∃ (D : Fin m → Fin n),
    ∀ (p : Fin n),
      let received_cupcakes : Finset (Fin m) := Finset.univ.filter (fun c => D c = p)
      Finset.sum received_cupcakes (fun c => scores p c) ≥ 1

/-- The main theorem, stating that the existence of a good score-partition for every person
implies the existence of a distribution of total score >= 1 for every person. -/
theorem PBAdvanced030 {m n : ℕ} (scores : Fin n → Fin m → ℝ) (hm : 1 ≤ m) (hn : 1 ≤ n)
    (hmn : n ≤ m) (h_nonneg : ∀ (p : Fin n) (c : Fin m), 0 ≤ scores p c) :
  satisfies_partition_condition scores →
  exists_good_distribution scores := by sorry"