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Difference between revisions of "2003 AIME II Problems/Problem 11"

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== Solution ==
 
== Solution ==
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{{solution}}
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We use the [[Pythagorean Theorem]] on <math>ABC</math> to determine that <math>AB=25.</math>
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Let <math>N</math> be the orthogonal projection from <math>C</math> to <math>AB.</math> Thus, <math>[CDM]=\frac{(DM)(MN)} {2}</math>, <math>MN=AM-AN</math>, and <math>[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.</math>
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 +
From the third equation, we get
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<math>CN=\frac{168} {25}.</math>
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 +
By the [[Pythagorean Theorem]] in <math>\Delta ACN,</math> we have
 +
 
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<math>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}.</math>
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Thus,
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<math>MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.</math>
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 +
In <math>\Delta ADM</math>, we use the [[Pythagorean Theorem]] to get
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<math>DM=\sqrt{15^2-(\frac{25} {2})^2}=\frac{5} {2} \sqrt{11}.</math>
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 +
Thus,
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<math>[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}=\frac{527\sqrt{11}} {40}.</math>
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Hence, the answer is <math>527+11+40=\boxed{578}.</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2003|n=II|num-b=10|num-a=12}}
 
{{AIME box|year=2003|n=II|num-b=10|num-a=12}}

Revision as of 10:35, 27 August 2008

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

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 $527+11+40=\boxed{578}.$

See also

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