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Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.5 AIME Prove it Saudi Arabia

Let $ABCD$ be a square of center $O$ . The parallel to $AD$ through $O$ intersects $AB$ and $CD$ at $M$ and $N$ and a parallel to $AB$ intersects diagonal $AC$ at $P$ . Prove that
$OP^{4} + \left(\frac{MN}{2}\right)^{4} = MP^{2} \cdot NP^{2}$

Solution

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
$\begin{gathered} OP^{2} = 2x^{2}, \quad MP^{2} = x^{2} + (a - x)^{2} \\ NP^{2} = x^{2} + (a + x)^{2}, \quad MN = 2a \end{gathered}$

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.

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