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

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== Problem ==
 
== Problem ==
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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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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== Solution ==
 
== Solution ==

Revision as of 20:36, 3 November 2007

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

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See also