Difference between revisions of "2002 AIME I Problems/Problem 10"

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The area of triangle <math>AGF</math> is <math>10/3</math> that of triangle <math>AEG</math>, since they share a common side and angle, so the area of triangle <math>AGF</math> is <math>10/13</math> the area of triangle <math>AEF</math>.
 
The area of triangle <math>AGF</math> is <math>10/3</math> that of triangle <math>AEG</math>, since they share a common side and angle, so the area of triangle <math>AGF</math> is <math>10/13</math> the area of triangle <math>AEF</math>.
  
Since the area of a triangle is <math>\frac{ab\sin{C}}2</math>, the area of <math>AEF</math> is <math>525/37</math> and the area of <math>AGF</math> is <math>5250/581</math>.
+
Since the area of a triangle is <math>\frac{ab\sin{C}}2</math>, the area of <math>AEF</math> is <math>525/37</math> and the area of <math>AGF</math> is <math>5250/481</math>.
  
 
The area of triangle <math>ABD</math> is <math>360/7</math>, and the area of the entire triangle <math>ABC</math> is <math>210</math>. Subtracting the areas of <math>ABD</math> and <math>AGF</math> from <math>210</math> and finding the closest integer gives <math>148</math> as the answer.
 
The area of triangle <math>ABD</math> is <math>360/7</math>, and the area of the entire triangle <math>ABC</math> is <math>210</math>. Subtracting the areas of <math>ABD</math> and <math>AGF</math> from <math>210</math> and finding the closest integer gives <math>148</math> as the answer.

Revision as of 02:11, 8 August 2008

Problem

In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$. Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10$. Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG$.

AIME 2002I Problem 10.png

Solution

By the Pythagorean Theorem, $BC=35$. Letting $BD=x$ we can use the angle bisector theorem on triangle $ABC$ to get $x/12=(35-x)/37$, and solving gives $BD=60/7$ and $DC=185/7$.

The area of triangle $AGF$ is $10/3$ that of triangle $AEG$, since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$.

Since the area of a triangle is $\frac{ab\sin{C}}2$, the area of $AEF$ is $525/37$ and the area of $AGF$ is $5250/481$.

The area of triangle $ABD$ is $360/7$, and the area of the entire triangle $ABC$ is $210$. Subtracting the areas of $ABD$ and $AGF$ from $210$ and finding the closest integer gives $148$ as the answer.

See also

2002 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AIME Problems and Solutions
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