PL-OM 2024

Let $X$ be an interior point of a rectangle $ABCD$ . Let the bisectors of $\angle DAX$ and $\angle CBX$ intersect in $P$ . A point $Q$ satisfies $\angle QAP=\angle QBP=90^\circ$ . Show that $PX=QX$ .

Let $n$ be a positive integer. Bolek draws $2n$ points in the plane, no two of them defining a vertical or a horizontal line. Then Lolek draws for each of these $2n$ points two rays emanating from them, one of them vertically and the other one horizontally.

Lolek wants to maximize the number of regions in which these rays divide the plane. Determine the largest number $k$ such that Lolek can obtain at least $k$ regions independent of the points chosen by Bolek.

We are given an integer $n \ge 2024$ and a sequence $a_1,a_2,\dots,a_{n^2}$ of real numbers satisfying $$\vert a_k-a_{k-1}\vert \le \frac{1}{k} \text{and} \vert a_1+a_2+\dots+a_k\vert \le 1$$ for $k=2,3,\dots,n^2$ . Show that $\vert a_{n(n-1)}\vert \le \frac{2}{n}$ . Note: Proving $\vert a_{n(n-1)}\vert \le \frac{75}{n}$ will be rewarded partial points.

Let $ABCD$ be a parallelogram. Let $X \notin AC$ lie inside $ABCD$ so that $\angle AXB = \angle CXD = 90^ {\circ}$ and $\Omega$ denote the circumcircle of $AXC$ . Consider a diameter $EF$ of $\Omega$ and assume neither $E, \ X, \ B$ nor $F, \ X, \ D$ are collinear. Let $K \neq X$ be an intersection point of circumcircles of $BXE$ and $DXF$ and $L \neq X$ be such point on $\Omega$ so that $\angle KXL = 90^{\circ}$ . Prove that $AB = KL$ .