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

#Geometry #Pre-RMO

Note by Pranshu Gaba
6 years, 8 months ago

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Comments

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

ABC=2xACB=2y \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+y=70 x^\circ+y^\circ=70^\circ

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

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

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

Aneesh Kundu - 6 years, 8 months ago

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Thanks :)

Pranshu Gaba - 6 years, 8 months ago

is it 55 ???

Nitish Deshpande - 6 years, 8 months ago

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Yes it is

Aneesh Kundu - 6 years, 8 months ago

55

Abhishek Bakshi - 6 years, 8 months ago

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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.

Pranshu Gaba - 6 years, 8 months ago

55

Sahil Nare - 6 years, 1 month ago

40??

Harsh Khatri - 6 years, 8 months ago

80??

Lakshya Kumar - 6 years, 8 months ago

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Nooo

Anish Manna - 2 years, 10 months ago
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