66th Czech and Slovak Mathematical Olympiad

The circle containing semicircle $q$ is the image of $p$ in homothety with center $M$ and factor $1/2$, hence $Q$ is the midpoint of $RM$. Since triangles $M{P}_{1}Q$, $P_{2}RQ$ share the angle by $Q$, we have $$\frac{S_1}{S_2}=\frac{\frac{1}{2}{P}_{1}Q\cdot MQ\cdot \sin \angle P_{1}QM}{\frac{1}{2}{P}_{2}Q\cdot RQ\cdot \sin \angle P_{2}QR}=\frac{P_{1}Q}{P_{2}Q}.$$ Let $KM=r$, $ML=x$, $P_{1}L=d_1$, $QL=d_2$ (Fig. 1). Points $P_1$, $P_2$ are symmetric about $KM$, therefore $P_{1}L=P_{2}L$ and $P_{2}Q=d_1-d_2$. Denote by $M^{\prime }$ the point such that $M{M}^{\prime }$ is the diameter of $p$. Then triangle $M^{\prime }M{P}_1$ is right and by Geometric Mean Theorem (an altitude splits a right triangle into two similar triangles) we get $d_1^2=x(2r-x)$. Similarly in right triangle $KQM$ we get $d_2^2=x(r-x)$ and thus $$\begin{array}{rl}\frac{S_1}{S_2}& =\frac{P_{1}Q}{P_{2}Q}=\frac{d_1+d_2}{d_1-d_2}=\frac{(d_1+d_2{)}^2}{d_1^2-d_2^2}\\ & =\frac{x(2r-x)+x(r-x)+2\sqrt{x(2r-x)\cdot x(r-x)}}{rx}\\ & =\frac{3r-2x+2\sqrt{(2r-x)(r-x)}}{r}.\end{array}$$ Fig. 1 We view the expression as a function of variable $x$ with parameter $r$. The function is decreasing on $(0,r)$ (both functions $3r-2x$ and $(2r-x)(r-x)$ are decreasing), therefore it attains its maximum $3+2\sqrt{2}$ at $x=0$ and minimum $1$ at $x=r$. By the problem statement, $x\in (0,r)$, thus $1<S_1/S_2<3+2\sqrt{2}=3+\sqrt{8}$.