Pre-RMO 2014/16

In a triangle ABCABC, let II denote the incenter. Let the line AI,BIAI, BI and CICI intersect the incircle at P,Q,P, Q, and RR, respectively. If BAC=40\angle BAC = 40^\circ, what is the value of QPR\angle QPR in degrees.


This note is part of the set Pre-RMO 2014

Note by Pranshu Gaba
4 years, 8 months ago

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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 - 4 years, 8 months ago

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Thanks :)

Pranshu Gaba - 4 years, 8 months ago

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is it 55 ???

Nitish Deshpande - 4 years, 8 months ago

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Yes it is

Aneesh Kundu - 4 years, 8 months ago

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55

Abhishek Bakshi - 4 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 - 4 years, 8 months ago

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55

Sahil Nare - 4 years, 1 month ago

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40??

Harsh Khatri - 4 years, 8 months ago

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80??

Lakshya Kumar - 4 years, 8 months ago

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Nooo

Anish Manna - 10 months, 3 weeks ago

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