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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Let ∠IBC=x∘ and ∠ICB=y∘
⇒∠ABC=2x∘⇒∠ACB=2y∘ (incenter is the point of concurreny of the angle bisectors)
By apply angle sum, we get
x∘+y∘=70∘
∠BIC=180∘−(x∘+y∘)=110∘
∠QPR=110∘=2×∠QPR (Angle subtended at the center is double the angle subtended at the circumference)
⇒∠QPR=55∘
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Thanks :)
is it 55 ???
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Yes it is
55
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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.
55
40??
80??
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Nooo