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IMO Team Selection Contest II

Geometry Difficulty 7.9 National Olympiad, round 2 Prove it Estonia

Let $ABC$ be a triangle with $AB = \frac{AC}{2} + BC$ . Consider the two semicircles outside the triangle with diameters $AB$ and $BC$ . Let $X$ be the orthogonal projection of $A$ onto the common tangent line of those semicircles. Find $\angle CAX$ .

Solution

Let $K$ and $L$ be the midpoints of the sides $AB$ and $BC$ , respectively, and $M$ and $N$ the feet of perpendiculars from $K$ and $L$ , respectively, to the common tangent of the semicircles (Fig. 28). Then $\angle CAX = \angle LKM$ . As $KM$ and $LN$ are radii of the semicircles, $KM = \frac{AB}{2}$ and $LN = \frac{BC}{2}$ . Let $Y$ be the intersection point of line $KL$ with the common tangent of the semicircles. As triangles $KYM$ and $LYN$ are similar, $\frac{KY}{LY} = \frac{KM}{LN}$ .

Fig. 28

Thus $\frac{KL+LY}{LY} = \frac{AB}{BC}$ , whence $LY = \frac{KL \cdot BC}{AB-BC} = \frac{AC}{2} \cdot \frac{BC}{AC} = BC$ . Therefore $\sin \angle NYL = \frac{LN}{LY} = \frac{BC}{AB} = \frac{1}{2}$ , implying $\angle NYL = 30^\circ$ . Consequently, $\angle CAX = \angle LKM = 90^\circ - \angle NYL = 60^\circ$ .

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