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Geometry Difficulty 4.5 AIME Prove it Saudi Arabia

Let ABCDABCD be a square of center OO. The parallel to ADAD through OO intersects ABAB and CDCD at MM and NN and a parallel to ABAB intersects diagonal ACAC at PP. Prove that
OP4+(MN2)4=MP2⋅NP2 OP^{4} + \left(\frac{MN}{2}\right)^{4} = MP^{2} \cdot NP^{2}

Solution

Let AB=2aAB = 2a and let QQ be the intersection point of the parallel to ABAB with MNMN. Let x=OQx = OQ. Then PQ=xPQ = x. We have
OP2=2x2,MP2=x2+(a−x)2NP2=x2+(a+x)2,MN=2a \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}

Figure 1

The relation is equivalent to
4x4+a4=[x2+(a−x)2][x2+(a+x)2] 4x^{4} + a^{4} = \left[x^{2} + (a - x)^{2}\right]\left[x^{2} + (a + x)^{2}\right]
that is
4x4+a4=(2x2+a2−2ax)(2x2+a2+2ax) 4x^{4} + a^{4} = \left(2x^{2} + a^{2} - 2ax\right)\left(2x^{2} + a^{2} + 2ax\right)
hence
4x4+a4=(2x2+a2)2−4a2x2 4x^{4} + a^{4} = \left(2x^{2} + a^{2}\right)^{2} - 4a^{2}x^{2}
and we are done.

Figure 1

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.