 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, $\Delta PIQ \equiv \Delta RIQ$ by side-side-side congruence rule. Hence $\angle IRQ (or \angle IRA) = \angle IPQ (or \angle IPB')$

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

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

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

Hence $\angle BCC' \left (=\dfrac{\angle C}{2}\right ) = \angle IAR \left (=\dfrac {\angle A}{2} \right)$ implying $\angle A = \angle C$ Since ABC is isosceles (BA = BC) and BR is the angle bisector, hence it is perpendicular to AC. Thus $\angle BRA = \angle IPQ = 90^o$.

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

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

Finally AI = IB’ (because $\Delta AIR \equiv \Delta B'IP$ since earlier we proved them to be equiangular, and we can also show that AR = B’P; How? Clearly $\Delta AQP \equiv \Delta B'QR$ 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 $\Delta AIB'$ is equilateral, implying $\angle AB'I = \angle ACB = 60^o$. Since ABC is an isosceles triangle with one angle $60^o$ 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