2002 AIME I Problems/Problem 10

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 $DCFG$ .

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 $[ADC]$ - $[AGF]$ .

Let us calculate $[ADC]$ first. We know $[ADC]$ is $\frac{AD*AC\sin{A}}{2}$ . Firstly, we need AD. From the angle bisector theorem From the Pythagorean Theorem, $BC=35$ .

2002 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2002 AIME I Problems/Problem 10

Contents

Problem

Solution

Solution 2 (Desperate Bash)

See also