HMMT November, 2019

Geometry Difficulty 4.8 AIME Prove it United States

Problem:

Let $ABCD$ be an isosceles trapezoid with $AD = BC = 255$ and $AB = 128$ . Let $M$ be the midpoint of $CD$ and let $N$ be the foot of the perpendicular from $A$ to $CD$ . If $\angle MBC = 90^{\circ}$ , compute $\tan \angle NBM$ .

Solution

Solution:

Construct $P$ , the reflection of $A$ over $CD$ . Note that $P$ , $M$ , and $B$ are collinear. As $\angle PNC = \angle PBC = 90^{\circ}$ , $PNBC$ is cyclic. Thus, $\angle NBM = \angle NCP$ , so our desired tangent is $\tan \angle ACN = \frac{AN}{CN}$ . Note that $NM = \frac{1}{2} AB = 64$ . Since $\triangle AND \sim \triangle MAD$ ,
$\frac{255}{64 + ND} = \frac{ND}{255}$
Solving, we find $ND = 225$ , which gives $AN = 120$ . Then we calculate $\frac{AN}{CN} = \frac{120}{128 + 225} = \frac{120}{353}$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.