Let $ABC$ be an acute angled triangle and let $P, Q$ be points on $AB, AC$ respectively, such that $PQ$ is parallel to $BC$. Points $X, Y$ are given on line segments $BQ, CP$ respectively, such that $\angle AXP = \angle XCB$ and $\angle AYQ = \angle YBC$. Prove that $AX = AY$. Proposed by Ervin Maci$\acute{c},$ Bosnia and Herzegovina
Problem
Source: IMSC 2024 Day 1 Problem 2
Tags: geometry, imsc
Randomization
28.06.2024 17:21
Consider the homothety centred at $A$ which maps line $PQ \mapsto BC$. Let $X'$ denote the image of $X$ in this map. From the angles, we have \[\angle XCB = \angle AXP = \angle AX'B = \angle XX'B\]and thus points $X, X', B, C$ are concylic. Since $BX \parallel CX'$, we obtain that $BXCX'$ is an isosceles trapezoid, whence $XX' = BC$. Similarily, $YY' = BC$. Now it is clear that $AX = AY$ from the homothety, so the problem is solved.
wassupevery1
28.06.2024 22:18
Let $X_1$ be intersection of the line through $A$, parallel to $BC$ and line $BQ$. I claim that $AX=AX_1$.
Let $X_2$ be the intersection of $AX$ and the circumcircle of triangle $BXC$, different from $X$.
Then $\angle AXP = \angle XCB = \angle AX_2B$, hence $PX \parallel BX_2$. This means the homothety sending $PQ$ to $BC$ also sends $X$ to $X_2$, hence $CX_2 \parallel QX$.
Consequently $\angle AXX_1 = \angle AX_2C = \angle QBC = \angle AX_1X$, so $AX=AX_1$.
Similarly, let $Y_1$ be intersection of the line through $A$, parallel to $BC$ and line $CP$. Then $AY=AY_1$.
Now, $\frac{AX_1}{PQ} = \frac{BP}{BA} = \frac{CQ}{CA} = \frac{AY_1}{PQ}$, hence $AX=AX_1=AY_1=AY$, as desired.
Z4ADies
30.06.2024 16:11
This problem has a lot of configurations. Let $BQ \cap CP$ at $M$. It is known that arbitrary $X,Y$ s.t $PQYX$ cyclic is equivalent to $BXYC$ cyclic. Thus,let's take $X,Y$ s.t $AX,AY$ are tangent to $(MXY)$ and $PXYC$ cyclic.We will prove $\angle AYQ=\angle YBC$ and $\angle AXP=\angle XBC$.After some angle chasing we need to found that $YM$ is angle bisector of $\angle QYB$. In other words,$\frac{YQ}{YB}=\frac{QM}{MB}(?)$ $\implies$ $QM*YB=YQ*MB(?)$. Also,from angle chasing $\triangle PXC$ and $\triangle QYB$ are similar. From sine thrm to $\triangle QMY,\triangle QYB,\triangle MYB$ we get equation is true. Ps:The uniqueness of $X,Y$ in this condition can be proved by taking another $X',Y'$ and angle chasing like that after some same results it will be clear that $X' \equiv X$ and $Y' \equiv Y$.
math_comb01
01.07.2024 10:40
Cute Problem! Let $X^{*} \in {AX}$ so that $AX/AX^{*}=AP/AB=AQ/AC$, by angle condition $BXCX^{*}$ is cyclic trapezium and therefore $XX^{*}=YY^{*}=BC$ which implies $AX=AY$ by homothety. Remark: Notice that the condition yields $\measuredangle BXP+ \measuredangle CXQ = 180$, I suspect there is a solution using isogonal conjugates as well.