equilateral triangle from Indian RMO

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

equilateral triangle from Indian RMO

G H J

Reveal topic tags

Equilateral Triangle

L

G H BBookmark kLocked kLocked NReply

Source: CRMO 2015 Region 3 (West Bengal) p5

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

30629 posts

#1 h Sep 30, 2018, 12:48 AM • 3 Y

Y by Nathanisme, Adventure10, Mango247

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

66 posts

#2 h Sep 30, 2018, 5:59 AM • 2 Y

Y by Adventure10, Mango247

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.

This post has been edited 1 time. Last edited by USER378398, Sep 30, 2018, 6:01 AM
Reason: ......

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

452 posts

aops29

#3 h May 30, 2019, 10:48 AM • 3 Y

Y by Nathanisme, Adventure10, Mango247

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

656 posts

#4 h Oct 12, 2020, 4:37 PM

Y by

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.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

168 posts

#5 h Oct 8, 2023, 9:25 PM

Y by

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.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a