Mongolian Mathematical Olympiad

Let $AB = z$, $PQ = x$, $QC = y$. By Pythagorean theorem:

$$OE^2 = OB^2 + BE^2 \Rightarrow BE^2 = R^2 - (AB - AO)^2 = R^2 - (z - R)^2 = 2Rz - z^2. \ (*)$$

$$AE^2 = BE^2 + AB^2 = 2zR - z^2 + z^2 = 2zR \Rightarrow AE = \sqrt{2zR}. \quad (*)$$
$$\text{Since } AD \text{ is diameter} \Rightarrow \angle AKD = 90^\circ, \angle KBC = 90^\circ \Rightarrow KB^2 = AB \cdot BD = z \cdot ZR = 2ZR \Rightarrow KB = AE = \sqrt{2zR}. \quad (***)$$
By tangent theorem: $KM^2 = KE \cdot KF = (KB - KE)(KB + KF) = KB^2 - BE^2 = 2zR - 2zR - z^2 = z^2$. (by (), ())
$KM = z = AB$. Now let's find area of triangle $\triangle KPQ$.
$$\begin{aligned} PM = PC \Rightarrow KP + KM = PQ + QC \Rightarrow KP = x + yz, \quad KQ = KN + NQ = KM + QC = z + y. \\ \frac{KP \cdot OM}{2} + \frac{PQ \cdot OC}{2} - \frac{KQ \cdot ON}{2} = \left(\frac{x+y-z}{2} - \frac{z+y}{2} + \frac{x}{2}\right) \cdot R = \frac{R}{2}(2x-2z). \\ S_{\triangle KPQ} = \frac{PQ \cdot BC}{2} = \frac{x(AC-AB)}{2} = \frac{x(2R-z)}{2} = \frac{R}{2}(2x-2z) \end{aligned}$$
$$$$
$\text{Hence}$ } $\Rightarrow$ 2Rx - xz = 2Rx - 2zR $\Rightarrow$ x = 2R.