Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 11: Line 11:
 
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>.
 
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>\boxed{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 15:00, 22 October 2011

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.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_I_Problems/Problem_10&oldid=42668"