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:


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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

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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