Let $ABC$ be a triangle with $\angle A = 60^\circ$. Points $E$ and $F$ are the foot of angle bisectors of vertices $B$ and $C$ respectively. Points $P$ and $Q$ are considered such that quadrilaterals $BFPE$ and $CEQF$ are parallelograms. Prove that $\angle PAQ > 150^\circ$. (Consider the angle $PAQ$ that does not contain side $AB$ of the triangle.) Proposed by Alireza Dadgarnia
Problem
Source: 6th Iranian Geometry Olympiad (Intermediate) P5
Tags: IGO, Iran, geometry
Pathological
21.09.2019 06:12
Funny problem! First we have to unmask the problem a little bit. WLOG assume $AB < AC.$ Let $I$ be the incenter of $\triangle ABC.$ Let $T$ be the point so that $AETF$ is a parallelogram. By reflecting the entire problem over the midpoint of $EF$, we need to show that $\angle BTC > 150,$ where we consider the $\angle BTC$ which contains only a set of points of finite area which are on the same side of $BC$ as $A.$ Let $D$ be the reflection of $F$ over $BI$ (also the reflection of $E$ over $CI$). Let $D'$ be the point on $BC$ such that $ID = ID'$ and $D \neq D'.$ It's easy to check by angle-chasing that $D'$ is in the interior of segment $BC.$ We claim that $T$ is the reflection of $D'$ over $ID.$ Before doing so, we make some observations. By some easy angle-chasing, we can get that $AEIF$ is cyclic, $\triangle DEF$ is equilateral, and that $I$ is the center of $\triangle DEF.$ Since $\angle ETF = \angle EAF = 60 = \angle EDF$, we've that $ETDF$ is cyclic. This implies that $IT = ID = ID'$. Furthermore, we have that $\angle DIT = 2 \angle DFT.$ Because $AE || FT$ and $BI \perp DF$, this is equal to $2(90 - \angle BEA) = 2 \left(\frac{\angle B}{2} - 30\right) = \frac{\angle B - \angle C}{2}.$ It suffices only to show that $\angle DID' = \angle DIT$ to prove our claim. We have that $\angle DID' = 180 - 2 \angle IDD' = 2 (90 - \angle BDI) = 2(\angle DIB + \angle IBD - 90) = 2\left(\frac{\angle B}{2} - 30\right) = \frac{\angle B - \angle C}{2}.$ The claim is proven. Now, let $X$ be the point so that $\triangle BCX$ is an equilateral triangle, and $A, X$ are on opposite sides of $BC.$ Observe that $BICX$ is cyclic, and so since $BX = XC$, we've that $D \in IX.$ Consider the circle $\omega$ centered at $X$ with radius $XB.$ Since segment $BC$ is entirely within it, we have that $D'$ is in the interior of $\omega.$ Since $D', T$ are reflections over $IX$, we've that $XT = XD' < XB.$ In other words, $T$ is also in the interior of $\omega.$ This implies that $\angle BTC > 150,$ and so we're done. $\square$
Delta0001
21.09.2019 17:13
This was a definite trig-bash