Dexter's Laboratory has $2024$ robots, each with a program setup by Dexter. One day, his naughty sister Dee Dee intrudes and writes an integer in $\{1, 2, \dots, 113\}$ on each of the robot's forehead. Each robot detects the numbers on all other robots' foreheads, and guess its own number base on its program, individually and simultaneously. Find the largest positive integer $k$ such that Dexter can setup the programs so that, no matter how the numbers distribute, there are always at least $k$ robots who guess their numbers right. Proposed by sn6dh

For each positive integer $k$, define $r(k)$ as the number of runs of $k$ in base-$2$, where a run is a collection of consecutive $0$s or consecutive $1$s without a larger one containing it. For example, $(11100100)_2$ has $4$ runs, namely $111-00-1-00$. Also, $r(0) = 0$. Given a positive integer $n$, find all functions $f : \mathbb{Z} \rightarrow\mathbb{Z}$ such that \[\sum_{k=0}^{2^n-1} 2^{r(k)}f(k+(-1)^{k} x)=(-1)^{x+n}\text{ for all integer $x$.}\] Proposed by YaWNeeT

Let $ABC$ be a triangle such that the angular bisector of $\angle BAC$, the $B$-median and the perpendicular bisector of $AB$ intersect at a single point $X$. Let $H$ be the orthocenter of $ABC$. Show that $\angle BXH = 90^{\circ}$. Proposed by usjl

Let $I$ be the incenter of triangle $ABC$, and let $\omega$ be its incircle. Let $E$ and $F$ be the points of tangency of $\omega$ with $CA$ and $AB$, respectively. Let $X$ and $Y$ be the intersections of the circumcircle of $BIC$ and $\omega$. Take a point $T$ on $BC$ such that $\angle AIT$ is a right angle. Let $G$ be the intersection of $EF$ and $BC$, and let $Z$ be the intersection of $XY$ and $AT$. Prove that $AZ$, $ZG$, and $AI$ form an isosceles triangle. Proposed by Li4 and usjl.

Find all positive integers $n$ and sequence of integers $a_0,a_1,\ldots, a_n$ such that the following hold: 1. $a_n\neq 0$; 2. $f(a_{i-1})=a_i$ for all $i=1,\ldots, n$, where $f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$. Proposed by usjl

Let $ABC$ and $A'B'C'$ be two triangles so that the midpoints of $\overline{AA'}, \overline{BB'}, \overline{CC'}$ form a triangle as well. Suppose that for any point $X$ on the circumcircle of $ABC$, there exists exactly one point $X'$ on the circumcircle of $A'B'C'$ so that the midpoints of $\overline{AA'}, \overline{BB'}, \overline{CC'}$ and $\overline{XX'}$ are concyclic. Show that $ABC$ is similar to $A'B'C'$. Proposed by usjl