2002 AIME I Problems/Problem 10: Difference between revisions
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== Solution == | == Solution == | ||
By the Pythagorean Theorem, <math>BC=35</math>. Letting <math>BD=x</math> we can use the angle bisector theorem on triangle ABC to get <math>x/12=(35-x)/37</math>, and solving gives <math>BD=60/7</math> and <math>DC=185/7</math>. 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 == | == 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 22:44, 19 February 2008
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 | ||
