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.

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## Comments

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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)

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

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Thanks :)

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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.

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is it 55 ???

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Yes it is

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55

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80??

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Nooo

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40??

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