Difference between revisions of "2002 AIME I Problems/Problem 10"
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Revision as of 18:55, 4 July 2013
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
