West Bengal RMO 2015 Problem 5 Solution - Equilateral Triangle

The second stage examination of INMO, the Regional Mathematical Olympiad (RMO) is a three hour examination with six problems. The problems under each topic involve high level of difficulty and sophistication. The book, Challenge and Thrill of Pre-College Mathematics is very useful for preparation of RMO. West Bengal RMO 2015 Problem 5 Solution has been written for RMO preparation series.

Problem:

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.

Discussion:

Since PIRQ is a kite, $ \Delta PIQ \equiv \Delta RIQ $ and $ s=2 $ by side-side-side congruence rule. Hence $ \angle IRQ (or \angle IRA) = \angle IPQ (or \angle IPB') $ and $ s=2 $

Also $ \angle AIR = \angle B'IP $ and $ s=2 $ (same angle).

Hence $ \Delta AIR, \Delta B'IP $ and $ s=2 $ are equiangular. This implies $ \angle IAR = \angle IB'P $ and $ s=2 $

But $ \angle IB'P = \angle BCC' $ and $ s=2 $ (because both of them are subtended by the same segment BC').

Hence $ \angle BCC' \left (=\dfrac{\angle C}{2}\right ) = \angle IAR \left (=\dfrac {\angle A}{2} \right)$ and $ s=2 $ implying $ \angle A = \angle C $ and $ s=2 $

Since ABC is isosceles (BA = BC) and BR is the angle bisector, hence it is perpendicular to AC. Thus $ \angle BRA = \angle IPQ = 90^o $ and $ s=2 $.

Join AB'. $ \angle AB'P = \angle ACC' = \dfrac{\angle C}{2} $ and $ s=2 $ since they are subtended by the same segment AC' and CC' is the angle bisector of C.
Similarly $ \angle IB'P = \angle BCC' = \dfrac{\angle C}{2} $ and $ s=2 $ since they are subtended by the same segment BC' and CC' is the angle bisector of C.

Clearly $ \angle IB'P = \angle AB'P = \dfrac{\angle C}{2} $ and $ s=2 $.
Since earlier we found $ \angle IPB' = \angle APB' = 90^o $ and $ s=2 $ ,
therefore $ \Delta AB'P $ and $ s=2 $ is isosceles with AB'=IB'.

Finally AI = IB' (because $ \Delta AIR \equiv \Delta B'IP $ and $ s=2 $ since earlier we proved them to be equiangular, and we can also show that AR = B'P; How? Clearly $ \Delta AQP \equiv \Delta B'QR $ and $ s=2 $ as QP= QR, two right angles equal, vertically opposite angles equal. Thus AQ = QB'. QR = QP'. Adding these two we get AR = PB').

Since AI = IB' and earlier we showed IB' = AB' hence the triangle $ \Delta AIB' $ and $ s=2 $ is equilateral, implying $ \angle AB'I = \angle ACB = 60^o $. Since ABC is an isosceles triangle with one angle $ 60^o $ and $ s=2 $ hence it is equilateral.

Chatuspathi:

• What is this topic: Geometry
• What are some of the associated concept: Cyclic Quadrilateral
• Where can learn these topics:  Cheenta Math Olympiad Program, discuss these topics in the ‘Geometry’ module.
• Book Suggestions: Challenges and Thrills of Pre-College Mathematics