1960 IMO SL #6

An isosceles trapezoid with bases $a$ and $c$ and altitude $h$ is given. 

a) On the axis of symmetry of this trapezoid, find all points $P$ such that both legs of the trapezoid subtend right angles at $P$; 

b) Calculate the distance of $p$ from either base; 

c) Determine under what conditions such points $P$ actually exist. Discuss various cases that might arise.

Since the question is related to the axis of symmetry of the trapezoid, we consider the midpoints of both bases. Let the isosceles trapezoid be $ABCD$, and denote $E,F$ as the midpoints of bases $AB, CD$. 

a) Since $\angle BPC = 90^\circ$, $X$ lies on the circle with diameter $BC$. But it's given that $P \in EF$, so it follows that $P$ is the intersection of the circle with diameter $BC$ and $EF$. $\square$

b) Let $x=EF$. Now, because $\triangle BEP \sim PFC$, we have $$x(h-x)=\frac{ab}{4}$$ $$\implies x=\frac{h \pm \sqrt{h^2-ab}}{2}.$$ $\square$

c) If we have $h^2>ab$, then there are two possibilities for $P$. If $h^2=ab$, then there is one possibility for $P$. Lastly, if $h^2<ab$, there are no possibilities for $P$. $\square$