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Difference between revisions of "2002 AIME I Problems/Problem 10"
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== Problem ==
 
== Problem ==
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In the diagram below, angle <math>ABC</math> is a right angle. Point <math>D</math> is on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects angle <math>CAB</math>. Points <math>E</math> and <math>F</math> are on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>AE=3</math> and <math>AF=10</math>. Given that <math>EB=9</math> and <math>FC=27</math>, find the integer closest to the area of quadrilateral <math>DCFG</math>. <center>[[Image:AIME_2002I_Problem_10.png]]</center>
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In the diagram below, angle <math>ABC</math> is a right angle. Point <math>D</math> is on <math>\overline{BC}</math>, and <math>\overline{AD}</math> bisects angle <math>CAB</math>. Points <math>E</math> and <math>F</math> are on <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, so that <math>AE=3</math> and <math>AF=10</math>. Given that <math>EB=9</math> and <math>FC=27</math>, find the integer closest to the area of quadrilateral <math>DCFG</math>.
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<center>[[Image:AIME_2002I_Problem_10.png]]</center>
  
 
== Solution ==
 
== Solution ==
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By the Pythagorean Theorem, <math>BC=35</math>. Letting <math>BD=x</math> we can use the angle bisector theorem on triangle ABC to get <math>x/12=(35-x)/37</math>, and solving gives <math>BD=60/7</math> and <math>DC=185/7</math>. Now, the area of triangle AGF is 10/3 that of triangle AEG, since they share a common side an angle, so the area of triangle AGF is 10/13 the area of triangle AEF. Since the area of a triangle is 1/2absinC, the area of AEF is 525/37 and the area of AGF=5250/581. The area of triangle ABD is 360/7. The area of the whole triangle ABC is 210. Subtracting the areas of ABD and AGF from 210 and finding the closest integer gives 148 as the answer.
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By the Pythagorean Theorem, <math>BC=35</math>. Letting <math>BD=x</math> we can use the angle bisector theorem on triangle <math>ABC</math> to get <math>x/12=(35-x)/37</math>, and solving gives <math>BD=60/7</math> and <math>DC=185/7</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>.
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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.
  
 
== 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}}

Revision as of 13:25, 19 April 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 $5250/581$.

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 (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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