Let $ABC$ be a triangle with circumcircle $\Gamma$ and incenter $I.$ Let the internal angle bisectors of $\angle A,\angle B,\angle C$ meet $\Gamma$ in $A',B',C'$ respectively. Let $B'C'$ intersect $AA'$ at $P,$ and $AC$ in $Q.$ Let $BB'$ intersect $AC$ in $R.$ Suppose the quadrilateral $PIRQ$ is a kite; that is, $IP=IR$ and $QP=QR.$ Prove that $ABC$ is an equilateral triangle.
Problem
Source: CRMO 2015 Region 3 (West Bengal) p5
Tags: geometry, circumcircle, incenter, Equilateral Triangle
USER378398
30.09.2018 08:59
Use the fact that the incentre of $ABC$ is the orthocentre of $A'B'C'$. Then some angle chasing allows us to prove $AB'CA'BC'$ as a regular hexagon.
aops29
30.05.2019 13:48
First show that $BA=BC$. Then we have $IA=IB'$. Now, from Fact 5 (incenter-excenter lemma), we see that $AIB'$ is equilateral, and we are done.
guptaamitu1
12.10.2020 19:37
I have a solution which uses only angle chasing and nothing more. So here goes my solution to the problem:(please refer to the figure of official solution) As shown in the official solution, ∠A = ∠C and quadrilaterals PIRQ, APRB' are cyclic which gives us ∠RIQ = ∠RPQ = ∠RPB' = ∠RAB' = ∠CAB' = ∠CBB' = ∠CBI = ∠B/2 ; ∠QIP = ∠QRP = ∠ARP = ∠AB'P = ∠AB'C' = ∠ACC' = ∠C/2. But since PIRQ is a kite, so IQ bisects ∠RIP which implies that ∠B/2 = ∠C/2. Thus, ∠A = ∠B = ∠C and hence, triangle ABC is equilateral.
Anulick
09.10.2023 00:25
Using the fact that the incenter of $ABC$ is the orthocenter of $A'B'C'$. We can say, $\angle IPQ=\angle PRQ=90^o$. We can create such $90^o$ all across the shape. Let $\angle A= \angle C=2 \theta$. This means, $\angle AIB'= \angle B'IC =\angle CIA'=\dots=90-\theta=\phi$. Hence, $\phi=60^o \implies \theta = 30^o$. This means $\angle A = \angle C = 60^o\implies \angle B = 60^0$ Hence, $ABC$ is equilateral.