2002 AIME I Problems/Problem 10
Problem
In the diagram below, angle is a right angle. Point is on , and bisects angle . Points and are on and , respectively, so that and . Given that and , find the integer closest to the area of quadrilateral .
Solution
By the Pythagorean Theorem, . Letting we can use the angle bisector theorem on triangle ABC to get , and solving gives and . 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.
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 |