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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.0 AIME Prove it United States

Problem:

$ABCD$ is a cyclic quadrilateral in which $AB = 3$ , $BC = 5$ , $CD = 6$ , and $AD = 10$ . $M$ , $I$ , and $T$ are the feet of the perpendiculars from $D$ to lines $AB$ , $AC$ , and $BC$ respectively. Determine the value of $MI / IT$ .

Solution

Solution:

Answer: $\frac{25}{9}$ . Quadrilaterals $AMID$ and $DICT$ are cyclic, having right angles $\angle AMD$ , $\angle AID$ , and $\angle CID$ , $\angle CTD$ respectively. We see that $M$ , $I$ , and $T$ are collinear. For, $m \angle MID = \pi - m \angle DAM = \pi - m \angle DAB = m \angle BCD = \pi - m \angle DCT = \pi - m \angle DIT$ . Therefore, Menelaus' theorem applied to triangle $MTB$ and line $ICA$ gives
$\frac{MI}{IT} \cdot \frac{TC}{CB} \cdot \frac{BA}{AM} = 1$
On the other hand, triangle $ADM$ is similar to triangle $CDT$ since $\angle AMD \cong \angle CTD$ and $\angle DAM \cong \angle DCT$ and thus $AM / CT = AD / CD$ . It follows that
$\frac{MI}{IT} = \frac{BC \cdot AM}{AB \cdot CT} = \frac{BC \cdot AD}{AB \cdot CD} = \frac{5 \cdot 10}{3 \cdot 6} = \frac{25}{9}$

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