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 to get , and solving gives and .
The area of triangle is that of triangle , since they share a common side and angle, so the area of triangle is the area of triangle .
Since the area of a triangle is , the area of is and the area of is .
The area of triangle is , and the area of the entire triangle is . Subtracting the areas of and from and finding the closest integer gives 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 |