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# 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
2 years, 11 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)

$$\Rightarrow \boxed{\angle QPR=55^\circ}$$ · 2 years, 11 months ago

Thanks :) · 2 years, 11 months ago

55 · 2 years, 11 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. · 2 years, 11 months ago

is it 55 ??? · 2 years, 11 months ago

Yes it is · 2 years, 11 months ago

55 · 2 years, 4 months ago

80?? · 2 years, 11 months ago