Question

Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $\overline{AB},$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of $\triangle CDM$ can be expressed as $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m + n + p.$

Answer

Solution:

The hypotenuse is $AB = \sqrt{7^2 + 24^2} = 25,$ and the median to the hypotenuse gives $CM = \frac{25}{2}.$ Since $AD = BD,$ point $D$ lies on the perpendicular to $AB$ at $M,$ so $DM \perp AB$ and $DM = \sqrt{15^2 - \left(\tfrac{25}{2}\right)^2} = \sqrt{\tfrac{275}{4}} = \tfrac{5\sqrt{11}}{2}.$

Let $\beta = \angle AMC.$ In triangle $AMC$ with $AM = CM = \frac{25}{2}$ and $AC = 7,$ the law of cosines gives $\cos\beta = \frac{\left(\frac{25}{2}\right)^2 + \left(\frac{25}{2}\right)^2 - 7^2} {2 \cdot \frac{25}{2} \cdot \frac{25}{2}} = \frac{527}{625}.$ Since $C$ and $D$ are on the same side of $AB$ and $MD \perp AB,$ we have $\angle CMD = 90^\circ - \beta,$ so $\sin\angle CMD = \cos\beta.$

Therefore $[CDM] = \frac{1}{2} \cdot \frac{25}{2} \cdot \frac{5\sqrt{11}}{2} \cdot \frac{527}{625} = \frac{527\sqrt{11}}{40},$ and $m + n + p = 527 + 11 + 40 = 578.$

2003 AIME II Problem 11

Solution:

Problem 11 in Other Years