AoPSWiki

Our Precalculus course starts on Dec. 4. Master trig, complex numbers, and vectors and matrices in 2 and 3 dimensions. Click here to enroll today!

Personal tools

Search

Toolbox

2003 AIME II Problems/Problem 11

From AoPSWiki

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, $[CMD]=\frac{1} {2} (12.5) (\frac{5\sqrt{11}} {2})(\frac{527} {625})=\frac{527\sqrt{11}} {40}$ .

Solution 3

Suppose $ABC$ is plotted on the cartesian plane with $C$ at $(0,0)$ , $A$ at $(0,7)$ , and $B$ at $(24,0)$ . Then $M$ is at $(12,3.5)$ . Since $\Delta AMD$ is isosceles, $MD$ is perpendicular to $AM$ , and since $AM=12.5$ and $AD=15, MD=2.5\sqrt{11}$ . The slope of $AM$ is $-\frac{7}{24}$ so the slope of $MD$ is $\frac{24}{7}$ . Draw a vertical line through $M$ and a horizontal line through $D$ . Suppose these two lines meet at $X$ . then $MX=\frac{24}{7}DX$ so $MD=\frac{25}{7}DX=\frac{25}{24}MD$ by the pythagorean theorem. So $MX=2.4\sqrt{11}$ and $DX=.7\sqrt{11}$ so the coordinates of D are $(12-.7\sqrt{11},2.5-2.4\sqrt{11})$ . Since we know the coordinates of each of the vertices of $\Delta CMD$ , we can apply the Shoelace Theorem to find the area of $\Delta CMD, \frac{527 \sqrt{11}}{40}$ .

See also

2003 AIME II (Problems • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2003_AIME_II_Problems/Problem_11"

Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.

© Copyright 2008 AoPS Incorporated. All Rights Reserved. • Foundation • Privacy • Contact Us