Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Find the answer

Let $A B C D$ be an isosceles trapezoid with $A D=B C=255$ and $A B=128$ . Let $M$ be the midpoint of $C D$ and let $N$ be the foot of the perpendicular from $A$ to $C D$ . If $\angle M B C=90^{\circ}$ , compute $\tan \angle N B M$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Construct $P$ , the reflection of $A$ over $C D$ . Note that $P, M$ , and $B$ are collinear. As $\angle P N C=\angle P B C=$ $90^{\circ}, P N B C$ is cyclic. Thus, $\angle N B M=\angle N C P$ , so our desired tangent is $\tan \angle A C N=\frac{A N}{C N}$ . Note that $N M=\frac{1}{2} A B=64$ . Since $\triangle A N D \sim \triangle M A D$ , $\frac{255}{64+N D}=\frac{N D}{255}$ Solving, we find $N D=225$ , which gives $A N=120$ . Then we calculate $\frac{A N}{C N}=\frac{120}{128+225}=\frac{120}{353}$ .

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