IMO Team Selection Contest II

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, $\angle CAX=\angle LKM=90^{\circ }-\angle NYL=60^{\circ }$.