IMO 1960 - Problem 6

Problem: An isosceles trapezoid with bases $a$ and $b$ and height $h$ is given.
(a) On the line of symmetry construct the point $P$ such that both (non-base) sides are seen form $P$ with an angle of $90^0$.
(b) Find the distance of $P$ from one of the bases of the trapezoid .
(c) Under what conditions for $a,b$ and $h$ can be the point $P$ be constructed (analyze all possible cases) ?
Solution: Let $E,F$ respectively be the midpoints of the bases $AB,CD$ of the isosceles trapezoid $ABCD$.
(a) The point $P$ is on the intersection of $EF$ and the circle with diameter $BC$.
(b) Let $x=EP$. Since $\triangle{BEP}\sim \triangle{PFC}$, we have $x(h-x)=\frac{ab}{4}\implies x_{1,2}=\frac{h\pm \sqrt{h^2-ab}}{2}$.
(c) If $h^2>ab$ there are two solutions, if $h^2=ab$ there is only one solution, and if $h^2<ab$ there are no solutions.