, 2008

Solution:
We denote $OP = OD = OB = R$, $AC = BC = b$ and $AB = 2a$. Because $OP \perp AC$ and $DQ \perp BC$, then the right triangles $APO$ and $BQD$ are similar and $\angle BDQ = \angle AOP$. So, the triangle $DEO$ is isosceles with $DE = OE$. It follows that
$$\frac{PE}{DE} = \frac{PE}{OE} = \frac{3}{5}$$
Let $F$ and $G$ be the orthogonal projections of the points $E$ and $P$ respectively on the side $AB$ and $M$ is the midpoint of the side $[AB]$. The triangles $OFE$, $OGP$, $OPA$ and $CMA$ are similar. We obtain the following relations
$$\frac{OF}{OE} = \frac{OG}{OP} = \frac{CM}{AC} = \frac{OP}{OA}$$
But $CM = \sqrt{b^2 - a^2}$ and we have $OG = \frac{R}{b} \cdot \sqrt{b^2 - a^2}$. In isosceles triangle $DEO$ the point $F$ is the midpoint of the radius $DO$. So, $OF = R/2$. By using Thales' theorem we obtain
$$\frac{3}{5} = \frac{PE}{OE} = \frac{GF}{OF} = \frac{OG - OF}{OF} = \frac{OG}{OF} - 1 = 2 \cdot \sqrt{1 - \left(\frac{a}{b}\right)^2} - 1$$
From the last relations it is easy to obtain that $\frac{a}{b} = \frac{3}{5}$ and $\frac{AB}{BC} = \frac{6}{5}$.

The problem is solved.