Let $BC$ be a chord of a circle $\Gamma$ and $A$ be a point inside $\Gamma$ such that $\angle BAC$ is acute. Outside $\triangle ABC$, construct two isosceles triangles $\triangle ACP$ and $\triangle ABR$ such that $\angle ACP$ and $\angle ABR$ are right angles. Let lines $BA$ and $CA$ meet $\Gamma$ again at points $E$ and $F$, respectively. Let lines $EP$ and $FR$ meet $\Gamma$ again at points $X$ and $Y$, respectively. Prove that $BX=CY$.
Problem
Source: 2022 Thailand MO Day 1 P1
Tags: geometry
WLOGQED1729
15.06.2022 07:48
Notice that $\triangle ABR \sim \triangle ACP \Rightarrow \frac{RA}{AB}=\frac{PA}{AC} \quad(\star)$ Notice that $\triangle BAC \sim \triangle FAE \Rightarrow \frac{AB}{AF}=\frac{AC}{AE} \quad(\star\star)$ Multiply $(\star),(\star\star) \Rightarrow \frac{RA}{AF}=\frac{PA}{AE}$ Consider $\angle RAF=\angle PAE= 135^\circ-\angle BAC$ So,$\triangle RAF \sim \triangle PAE \Rightarrow \angle RFA=\angle PAE$ $\Rightarrow \angle YFC=\angle BEX$ $\therefore BX=CY \quad\blacksquare$
Tintarn
28.06.2022 12:13
I think there is a typo in the problem statement: $X$ should be the intersection of $EP$ with $\Gamma$ (not of $BP$). At least, this way the problem makes sense and it is also what seems to be implicitly assumed in the solution in #2.
Ld_minh4354
30.03.2023 04:15
The central claim is the following: Claim: $\triangle AFR$ and $\triangle AEP$ are similar. Proof: We have: $$\frac{AR}{AP}=\frac{AB\sqrt{2}}{AC\sqrt{2}}=\frac{AB}{AC}=\frac{AF}{AE}$$and $$\angle FAR=\angle FAB-45^\circ=\angle EAC-45^\circ=\angle EAP$$which prove similarity. Back to the question: It follows from the claim that $\angle YFC=\angle XEB$. Thus arc $BX$ is equal to arc $CY$, and therefore $BX=CY$.