Problem 4

In a triangle

ABC

we have

AB=AC

. A circle internally tangent to the circumcircle is also tangent to

AB,AC

P,Q

. Prove that the midpoint of

PQ

is the incenter of triangle ABC.

Step 3 of 5: Apply internal tangency

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In plain words

Tangency fixes the circle's position.

s-R=R-s\sin\alpha,\quad s=\frac{L}{\cos\alpha(1+\sin\alpha)}

Detailed analysis

The small circle lies beyond $O$ on the axis. Internal tangency gives $s-R=R-s\sin\alpha$ . Substituting $R=L/(2\cos\alpha)$ yields the displayed $s$ .