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Equilateral Triangle (Regional Mathematics Olympiad 2015 solution to Problem 5)

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: 

Screen Shot 2015-12-06 at 10.41.28 PM

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 \(\angle A = \angle C \)

Screen Shot 2015-12-06 at 11.18.07 PM

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 I.S.I. & C.M.I. course, Cheenta Math Olympiad Program, discuss these topics in the ‘Geometry’ module.
  • Book Suggestions: Challenges and Thrills of Pre-College Mathematics
December 7, 2015

5 comments

  1. all we need to see is that A,P,R,B’ are concyclic

  2. what if we do not use the information on kite

  3. but i could solve it without the information on kite just as i is the incentre so point R is the point where the incircle meets and its right angle similarly on the left hand side also and just consider the two right triangles RIC and just vertically opposite to it and provet the rest same way……

    • R is not necessarily the point where incircle meets. It is the point where angle bisector meets the opposite side. You must drop perpendicular from I on the sides to get the points through which incircle pass

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