INMO 2025

We have $a^{2023} \equiv -b^{2023} \pmod{p}$ $b^{2024} \equiv -c^{2024} \pmod{p}$ $c^{2025} \equiv -a^{2025} \pmod{p}$

Thus, $a^{2023 \cdot 2024 \cdot 2025} \equiv b^{2023 \cdot 2024 \cdot 2025} \pmod{p}$ by the first congruence (raise to appropriate power), $\equiv -c^{2023 \cdot 2024 \cdot 2025} \pmod{p}$ by the second congruence, $\equiv -a^{2023 \cdot 2024 \cdot 2025} \pmod{p}$ by the third congruence.

Thus, $p \mid 2 \cdot a^{2023 \cdot 2024 \cdot 2025}$ and hence $p \mid a$ since $p$ is odd. Now $p \mid a$ implies $p \mid b$ from the first equation, and $p \mid b$ implies $p \mid c$ from the second equation. $\square$

Alternative Solution. Set $k = 2023$. If one of $a, b, c$ is divisible by $p$, then all of them are. So assume for contradiction that none of $a, b, c$ is divisible by $p$. Then

$a^{k(k+2)} \equiv (-b^k)^{k+2} \equiv -b^{k(k+2)} \pmod{p}$

and

$a^{k(k+2)} \equiv (-c^{k+2})^k \equiv -c^{k(k+2)} \pmod{p}.$

So $b^{k(k+2)} \equiv c^{k(k+2)} \pmod{p}$. But then

$c^{k(k+2)} \cdot c \equiv c^{(k+1)^2} \equiv (-b^{k+1})^{k+1} \equiv b^{(k+1)^2} \equiv b^{k(k+2)} \cdot b \equiv c^{k(k+2)} \cdot b \pmod{p}$

which forces $b \equiv c \pmod{p}$. Thus

$0 \equiv b^{k+1} + c^{k+1} \equiv 2b^{k+1} \pmod{p}$

implying $p \mid b$, a contradiction. $\square$

The possible values are 1, 2, and 2024.

Construction. The function $f(x) = 1$ for all $x \in \mathbb{N}$ works. Also, $f(x) = 1$ for all $x \neq 2024$ and $f(2024) = 2$ works. Finally, $f(x) = x$ for all $x$ works.

Proof these are the only values. Denote $\text{Im}(f) = \{f(x) \mid x \in \mathbb{N}\}$. The cardinal set $\{1\}$ gives $f(1) = 1$.

Case 1: $\text{Im}(f)$ is unbounded. Fix any $n > 1$. Pick $n-1$ distinct integers $k_1, \dots, k_{n-1}$ such that $f(k_i) \notin \{n, f(n)\}$ and $f(k_i)$ are all pairwise distinct. Then $\{n, k_1, \dots, k_{n-1}\}$ is cardinal with $n$ elements. So $\{f(n), f(k_1), \dots, f(k_{n-1})\}$ is cardinal with $n$ distinct elements, meaning $n$ lies in this set, hence $f(n) = n$. This gives the identity function.

Case 2: $\text{Im}(f)$ is bounded by some $M$.

Claim: For any $a$ with $1 \le a \le M$, if infinitely many $n$ satisfy $f(n) = a$, then $a = 1$.

Proof. Let $b > 1$ with $f(b) = a$. Choose $b - 1$ other integers $c_1, \dots, c_{b-1}$ with $f(c_i) = a$. Then $\{b, c_1, \dots, c_{b-1}\}$ is cardinal, so its image $\{a\}$ must be cardinal, forcing $a = 1$. $\square$

So for every $2 \le m \le M$, only finitely many $x$ satisfy $f(x) = m$. Thus there exists $N > 1$ such that $f(n) = 1$ for all $n \ge N$.

For every $1 < l < N$, the cardinal set $\{l, N+1, N+2, \dots, N+l-1\}$ maps to $\{1, f(l)\}$, which is cardinal only when $f(l) = 1$ or $f(l) = 2$.

By the above, $f(2024)$ can only be $\boxed{1, 2, \text{ or } 2024}$. $\square$

Fix an index $i \in \{1, 2, 3\}$. Let $D_1, D_2, D_3$ be the points of tangency of the incircle of $\triangle A_1A_2A_3$ with sides $A_2A_3$, $A_3A_1$, $A_1A_2$ respectively.

The key observation is that given a line $\ell$ in the plane, the image of $\ell$ under $\tau_{i+2} \circ \tau_i \circ \tau_{i+1}$ is a line parallel to $\ell$. Indeed, $\ell$ is rotated thrice by angles equal to the angles of $\triangle A_1A_2A_3$, and the composition induces a half-turn and translation since the angles sum to $180°$.

Since $D_i$ is a fixed point of this transformation (by the chain $D_i \xrightarrow{\tau_{i+1}} D_{i+2} \xrightarrow{\tau_i} D_{i+1} \xrightarrow{\tau_{i+2}} D_i$), the line $\overline{PD_i}$ maps to $\overline{P_iD_i}$. But these lines are parallel and both pass through $D_i$, so they coincide. Since each rotation preserves distances, $P_i$ is the reflection of $P$ in $D_i$.

Therefore, $\triangle P_1P_2P_3$ is obtained by applying a homothety with ratio 2 and center $P$ to $\triangle D_1D_2D_3$. The circumradius of $\triangle P_1P_2P_3$ is twice the circumradius of $\triangle D_1D_2D_3$, i.e., twice the inradius of $\triangle A_1A_2A_3$.

Since the inradius $r$ satisfies $r \le \frac{R}{2}$ where $R$ is the circumradius (with equality iff the triangle is equilateral), we get that the circumradius of $\triangle P_1P_2P_3$ is $2r \le R$, the radius of $\Gamma$. $\square$

Let $M = n^2 + \frac{n}{2}$.

Lemma 1. $B_n - A_n < \frac{1}{2}$.

Indeed, $B_n - A_n = \sum_{k=1}^{n} \left(\sqrt{n^2 + 2k} - \sqrt{n^2 + 2k - 1}\right) = \sum_{k=1}^{n} \frac{1}{\sqrt{n^2 + 2k} + \sqrt{n^2 + 2k - 1}} < \sum_{k=1}^{n} \frac{1}{2n} = \frac{1}{2}.$

Lemma 2. $A_n < M < B_n$.

Proof. We have $A_n - n^2 = \sum_{k=1}^{n} \left(\sqrt{n^2 + 2k - 1} - n\right) = \sum_{k=1}^{n} \frac{2k - 1}{\sqrt{n^2 + 2k - 1} + n} < \sum_{k=1}^{n} \frac{2k-1}{2n} = \frac{n^2}{2n} = \frac{n}{2},$ so $A_n < M$. Similarly, $B_n - n^2 = \sum_{k=1}^{n} \frac{2k}{\sqrt{n^2 + 2k} + n} > \sum_{k=1}^{n} \frac{2k}{2n + 1} = \frac{n(n+1)}{2n+1} > \frac{n}{2},$ so $B_n > M$. $\square$

When $n$ is even, $M$ is an integer. Then $\lfloor A_n \rfloor < M$ but $\lfloor B_n \rfloor \ge M$, so $\lfloor A_n \rfloor \neq \lfloor B_n \rfloor$.

When $n$ is odd, $M$ is a half-integer, so $M - \frac{1}{2}$ and $M + \frac{1}{2}$ are consecutive integers. By both lemmas: $M - \frac{1}{2} < A_n < M < B_n < M + \frac{1}{2},$ which gives $\lfloor A_n \rfloor = \lfloor B_n \rfloor = M - \frac{1}{2}$.

The answer is all odd integers $n \ge 3$. $\square$

We have $FG = FA$ since $FG$ is parallel to $BC$ and the triangle is isosceles. Also $\triangle GAE$ is a right-angled triangle (since $\angle AEG$ subtends a diameter-related arc). Thus, if $F'$ is the midpoint of $GE$, then $\angle GAF = \angle FGA = \angle F'GA = \angle GAF'$, which implies $F \equiv F'$. So $F$ is the midpoint of $GE$.

Let $O$ be the circumcenter of $\triangle EBG$. Then $\angle FOE = \angle GBE = \angle ABE = \angle ACE = \angle FCE.$

Thus $\angle FOE = \angle FCE$, which means $F$, $O$, $C$, $E$ are concyclic. That is, $O$ lies on the circumcircle of $\triangle ECF$. $\square$

Let $n = 2024$ and $k = 1000$. We claim the maximum is $k(2n - k) = 1000 \times 3048 = \mathbf{3{,}048{,}000}$.

Upper bound. Call a row/column bad if it has at least one red square. After the first move, there are exactly $k + 1$ bad rows and columns: the chosen row/column plus the $k$ perpendicular lines through the colored squares. Any subsequent move increases the number of bad rows/columns by at least 1. Since there are only $2n$ rows and columns total, at most $2n - k$ moves can be made. Each move colors $k$ squares, giving at most $k(2n - k)$ red squares.

Construction. Choose each of the $n$ columns in the first $n$ moves, coloring the top $k$ cells in each. After this, the bottom $n - k$ rows are entirely white, so we can make $n - k$ more moves. Total moves: $n + (n - k) = 2n - k$. Total red squares: $k(2n - k)$. $\square$