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Romanian Mathematical Olympiad

Geometry Difficulty 5.3 AIME, harder Prove it Romania

Let $ABCD A'B'C'D'$ be a cube. On the segments $BC$ and $DD'$ we take the points $M$ and $N$ respectively, such that $BM = DN$ . Prove that line $A'M$ is perpendicular to plane $(AB'N)$ .

Cătălin Barbu

Solution

$BC \perp (ABB')$ and $AB' \subset (ABB')$ yields $AB' \perp BC$ . Since $AB' \perp A'B$ (diagonals of the square $ABB'A'$ ), we get $AB' \perp (A'BC)$ . Because $A'M \subset (A'BC)$ , it follows that $A'M \perp AB'$ . (1)

Let $E \in (AD)$ be such that $AE = BM$ . Then $ABME$ is a rectangle, so $AB \parallel ME$ . Because $AB \perp (ADA')$ and $AN \subset (ADA')$ , it follows that $AN \perp ME$ . (2)

From $\triangle A'AE \equiv \triangle ADN$ (L.L.) one gets $\angle DAN = \angle AA'E$ , whence $\angle AA'E + \angle A'AN = \angle DAN + \angle A'AN = 90^\circ$ . Thus, $AN \perp A'E$ . Using now relation (2) we get $AN \perp (A'EM)$ , so $AN \perp A'M$ . Taking into account this last relation and (1), we conclude that $A'M \perp (AB'N)$ .

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