Saudi Arabia Mathematical Competitions

Let $AB=2a$ and let $Q$ be the intersection point of the parallel to $AB$ with $MN$. Let $x=OQ$. Then $PQ=x$. We have $$\begin{array}{c}O{P}^2=2x^2,M{P}^2=x^2+(a-x{)}^2\\ N{P}^2=x^2+(a+x{)}^2,MN=2a\end{array}$$



The relation is equivalent to $$4x^4+a^4=\left[x^2+(a-x{)}^2\right]\left[x^2+(a+x{)}^2\right]$$ that is $$4x^4+a^4=\left(2x^2+a^2-2ax\right)\left(2x^2+a^2+2ax\right)$$ hence $$4x^4+a^4={\left(2x^2+a^2\right)}^2-4a^2{x}^2$$ and we are done.