# Pre-RMO 2014/16

In a triangle $ABC$, let $I$ denote the incenter. Let the line $AI, BI$ and $CI$ intersect the incircle at $P, Q,$ and $R$, respectively. If $\angle BAC = 40^\circ$, what is the value of $\angle QPR$ in degrees.

This note is part of the set Pre-RMO 2014 Note by Pranshu Gaba
5 years, 3 months ago

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Let $\angle IBC= x^\circ$ and $\angle ICB=y^\circ$

$\Rightarrow \angle ABC=2x^\circ \Rightarrow \angle ACB=2y^\circ$ (incenter is the point of concurreny of the angle bisectors)

By apply angle sum, we get

$x^\circ+y^\circ=70^\circ$

$\angle BIC= 180^\circ -(x^\circ +y^\circ)=110^\circ$

$\angle QPR=110^\circ=2\times \angle QPR$ (Angle subtended at the center is double the angle subtended at the circumference)

$\Rightarrow \boxed{\angle QPR=55^\circ}$

- 5 years, 3 months ago

Thanks :)

- 5 years, 3 months ago

is it 55 ???

- 5 years, 3 months ago

Yes it is

- 5 years, 3 months ago

55

- 5 years, 3 months ago

How did you solve this problem? I assumed it to be an isosceles triangle. I got the correct answer, but couldn't figure out using a proper method.

- 5 years, 3 months ago

55

- 4 years, 8 months ago

40??

- 5 years, 3 months ago

80??

- 5 years, 3 months ago

Nooo

- 1 year, 6 months ago