All-Soviet-Union Mathematical Olympiad

Solution:

$DA = DE$, so $DAE$ is isosceles. $DE$ is parallel to $BC$, so $ABC$ is isosceles, so $BA = AC / (2 \cos A)$. Hence $BD = AC / (2 \cos A) - AC$. But $AE = 2 AC \cos A$, so we have an equation for $c = \cos A$: $4c^2 + 2c - 1 = 0$.

$2\pi/5, 4\pi/5, 6\pi/5, 8\pi/5$ and $10\pi/5$ are the roots of: real part of $(\cos \theta + i \sin \theta)^5 = 1$. Expanding this gives that $\cos 2\pi/5$, $\cos 4\pi/5$, $\cos 6\pi/5$, $\cos 8\pi/5$ and $1$ are the roots of $16c^5 - 20c^3 + 5c - 1 = 0$. Dividing by $(c - 1)$ gives $16c^4 + 16c^3 - 4c^2 - 4c + 1 = (4c^2 + 2c - 1)^2$. So $\cos 2\pi/5$ ($= \cos 8\pi/5$) and $\cos 4\pi/5$ ($= \cos 6\pi/5$) are the roots of $4c^2 + 2c - 1 = 0$.

We know that $A < 90^\circ$ (since $A = C$ and their sum is less than $180^\circ$). Hence $A = 2\pi/5$. So $BD = 2 AC \cos 2\pi/5 = 2 AC \sin \pi/10$, which is the side length for a regular 10-gon inscribed in a circle radius $AC$.