2003 AIME II Problems/Problem 11

Problem

Triangle $ABC$ is a right triangle with $AC = 7,$ $BC = 24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD = BD = 15.$ Given that the area of triangle $CDM$ may 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 $m + n + p.$

Solution

Solution 1

We use the Pythagorean Theorem on $ABC$ to determine that $AB=25.$

Let $N$ be the orthogonal projection from $C$ to $AB.$ Thus, $[CDM]=\frac{(DM)(MN)} {2}$ , $MN=AM-AN$ , and $[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.$

From the third equation, we get $CN=\frac{168} {25}.$

By the Pythagorean Theorem in $\Delta ACN,$ we have

$AN=\sqrt{(\frac{24 \cdot 25} {25})^2-(\frac{24 \cdot 7} {25})^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.$

Thus, $MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.$

In $\Delta ADM$ , we use the Pythagorean Theorem to get $DM=\sqrt{15^2-(\frac{25} {2})^2}=\frac{5} {2} \sqrt{11}.$

Thus, $[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.$

Hence, the answer is $527+11+40=\boxed{578}.$

Solution 2

By the Pythagorean Theorem in $\Delta AMD$ , we get $DM=\frac{5\sqrt{11}} {2}$ . Since $ABC$ is a right triangle, $M$ is the circumcenter and thus, $CM=\frac{25} {2}$ . We let $\angle CMD=\theta$ . By the Law of Cosines,

$2 \cdot (12.5)^2-2 \cdot (12.5)^2 * \cos (90+\theta).$

It follows that $\sin \theta = \frac{527} {625}$ . Thus, $[CMD]=\frac{1} {2} (12.5) (\frac{5\sqrt{11}} {2})(\frac{527} {625})=\frac{527\sqrt{11}} {40}$ .

2003 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2003 AIME II Problems/Problem 11

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