IMO 1961 - Problem 4

Problem: In the interior of $\triangle{P_1P_2P_3}$ a point $P$ is given. Let $Q_1,Q_2$ and $Q_3$ respectively be the intersections of $PP_1,PP_2,PP_3$ with the opposing edges of $\triangle{P_1P_2P_3}$. Prove that among the ratios $\frac{PP_1}{PQ_1},\frac{PP_2}{PQ_2}$ and $\frac{PP_3}{PQ_3}$ there exists at least one not larger than $2$ and at least one not smaller than $2$.
Solution: Let $x_i=\frac{PP_i}{PQ_i}$ for $i=1,2,3$. For all $i$ we have
$\frac{1}{x_i+1}=\frac{PQ_i}{P_iQ_i}=\frac{S_{PP_jP_k}}{S_{P_1P_2P_3}}$
where the indices $j$ and $k$ are distinct and different from $i$. Hence we have $f(x_1,x_2,x_3)=\frac{1}{x_1+1}+\frac{1}{x_2+1}+\frac{1}{x_3+1}=\frac{S_{PP_2P_3}+S_{PP_1P_3}+S_{PP_2P_1}}{S_{P_1P_2P_3}}=1$.
It follows that $\frac{1}{x_{i}+1}\ge \frac{1}{3}$ for some $i$ and $\frac{1}{x_{j}+1}\le \frac{1}{3}$ for some $j$. Consequently, $x_i\le 2$ and $x_j\ge 2$.