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

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TopNewestLet \( \angle IBC= x^\circ\) 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)

\( \Rightarrow \boxed{\angle QPR=55^\circ}\) – Aneesh Kundu · 2 years, 11 months ago

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– Pranshu Gaba · 2 years, 11 months ago

Thanks :)Log in to reply

55 – Abhishek Bakshi · 2 years, 11 months ago

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– Pranshu Gaba · 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.Log in to reply

is it 55 ??? – Nitish Deshpande · 2 years, 11 months ago

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– Aneesh Kundu · 2 years, 11 months ago

Yes it isLog in to reply

55 – Sahil Nare · 2 years, 4 months ago

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80?? – Lakshya Kumar · 2 years, 11 months ago

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40?? – Harsh Khatri · 2 years, 11 months ago

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