Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.7 AIME, harder Prove it Estonia

The sides $AB$ and $AC$ of the triangle $ABC$ touch the circle $c$ respectively at points $B'$ and $C'$ . The center $L$ of the circle $c$ lies on the side $BC$ . The circumcenter $O$ of triangle $ABC$ lies on the shorter arc $B'C'$ of the circle $c$ . Prove that the circumcircle of $ABC$ and the circle $c$ meet at two points.

Solution

Let $r$ be the circumradius of $ABC$ , let $s$ be the radius of $c$ and $\alpha = \angle BAC$ (Fig. 15).

By tangency, $|AB'| = |AC'|$ . Thus $\angle C'B'A = \angle B'C'A = \frac{\pi}{2} - \frac{\alpha}{2}$ whence, by property of inscribed angle, $\angle B'OC' = \pi - (\frac{\pi}{2} - \frac{\alpha}{2}) = \frac{\pi}{2} + \frac{\alpha}{2}$ .

Clearly $\angle B'OC' > \angle BOC = 2\alpha$ , leading to $\frac{\pi}{2} + \frac{\alpha}{2} > 2\alpha$ . Hence $\alpha < \frac{\pi}{3}$ .

Now let $K$ be the midpoint of side $BC$ . From the right triangle $KOC$ , one gets $|KO| = |OC|\cos\angle KOC = r\cos\alpha$ .

By the inequality obtained above, $\cos\alpha > \cos\frac{\pi}{3} = \frac{1}{2}$ .

On the other hand, $|KO| \le |LO| = s$ , leading to $\frac{1}{2}r < r\cos\alpha = |KO| \le s$ or $r < 2s$ .

As $c$ passes through the circumcenter of $ABC$ , this inequality shows that these circles must intersect.

Fig. 15

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