Baltic Way 1992 · Problem 17

Figure 1

Let $\angle A C D=2 \alpha$ (see Figure 1). Then $\angle C A D=\frac{\pi}{2}-2 \alpha, \angle A B D=2 \alpha, \angle A D B=\frac{\pi}{2}-\alpha$ and $\angle C D B=\alpha$. The sine theorem applied to triangles $D C P$ and $D A P$ yields

$$\frac{|D P|}{\sin 2 \alpha}=\frac{2}{5 \sin \alpha}$$

and

$$\frac{|D P|}{\sin \left(\frac{\pi}{2}-2 \alpha\right)}=\frac{8}{5 \sin \left(\frac{\pi}{2}-\alpha\right)}$$

Combining these equalities we have

$$\frac{2 \sin 2 \alpha}{5 \sin \alpha}=\frac{8 \cos 2 \alpha}{5 \cos \alpha}$$

which gives $4 \sin \alpha \cos ^{2} \alpha=8 \cos 2 \alpha \sin \alpha$ and $\cos 2 \alpha+1=4 \cos 2 \alpha$. So we get $\cos 2 \alpha=\frac{1}{3}$ and $|C D|=$ $2 \cos 2 \alpha=\frac{2}{3}$.