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Romanian Mathematical Olympiad

Geometry Difficulty 5.9 AIME, harder Prove it Romania

In the right parallelepiped $ABCD'A'B'C'D'$ , with $AB = 12\sqrt{3}$ cm and $AA' = 18$ cm, we consider the points $P \in [AA']$ and $N \in [A'B']$ such that $A'N = 3B'N$ . Determine the length of the line segment $[AP]$ such that for any position of the point $M \in [BC]$ , the triangle $MNP$ is right angled at $N$ .

Damian Marinescu

Solution

We have $BC \perp (ABB')$ , therefore $BC \perp PN$ . Since $PN \perp NM$ , it follows that $PN \perp (NBC)$ , hence $PN \perp NB$ , that is, triangle $NBP$ is right angled at $N$ . Let $AP = x$ ; we obtain $BP^2 = x^2 + 432$ , $PN^2 = (18-x)^2 + 243$ and $BN^2 = 351$ . But $BP^2 = PN^2 + BN^2$ , and hence $x = 13, 5$ cm.

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