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

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==Solution 2 (Desperate Bash)==
 
==Solution 2 (Desperate Bash)==
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We want to find the area of <math>DCFG</math>. This is the same as <math>[ADC]</math> - <math>[AGF]</math>. We know <math>[ADC]</math> is <math>\frac{1}{2}absinC
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From the Pythagorean Theorem, </math>BC=35$.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2002|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2002|n=I|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 13:07, 3 July 2015

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 $\boxed{148}$ as the answer.

Solution 2 (Desperate Bash)

We want to find the area of $DCFG$. This is the same as $[ADC]$ - $[AGF]$. We know $[ADC]$ is $\frac{1}{2}absinC From the Pythagorean Theorem,$BC=35$.

See also

2002 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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