AoPS - Problem

Processing math: 90%

Problem

Source: Olimphíada 2022- Problem 2/Level 2

Tags: geometry, Triangle

rcorreaa

26.07.2022 05:57

Let $ABC$ be a triangle and $\omega$ its incircle. $\omega$ touches $AC,AB$ at $E,F$ , respectively. Let $P$ be a point on $EF$ . Let $\omega_1=(BFP), \omega_2=(CEP)$ . The parallel line through $P$ to $BC$ intersects $\omega_1,\omega_2$ at $X,Y$ , respectively. Show that $BX=CY$ .

hectorraul

26.07.2022 09:51

Hint: Consider $X = \omega_1 \cap BC$ and $Y = \omega_2 \cap BC$ .

paccongnong19hy

28.07.2022 17:20

We have $\triangle AEF$ is isosceles, $BFXP$ and $PEYC$ are cyclic quadrilaterals so there is $(XB,XP)\equiv(FB,FP)\equiv (EP,EC) \equiv (YP,YC) \mod \pi \Rightarrow BXCY$ is an isosceles trapezoid $\Rightarrow BX=CY$