Difference between revisions of "2001 AIME I Problems/Problem 4"
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Thus the answer is <math>216 + 72 + 3 = \boxed{291}</math> | Thus the answer is <math>216 + 72 + 3 = \boxed{291}</math> | ||
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+ | Note: We could also get the lengths (and area) of the triangle by drawing a perpendicular from <math>T</math> to <math>AB</math>, forming a <math>30-60-90</math> and a <math>45-45-90</math> triangle. | ||
== See also == | == See also == |
Revision as of 11:46, 18 August 2017
Contents
[hide]Problem
In triangle , angles and measure degrees and degrees, respectively. The bisector of angle intersects at , and . The area of triangle can be written in the form , where , , and are positive integers, and is not divisible by the square of any prime. Find .
Solution 1
Let be the foot of the altitude from to . By simple angle-chasing, we find that , and thus . Now is a right triangle and is a right triangle, so . The area of
and the answer is .
Solution 2
Since has a measure of , and thus has sines and cosines that are easy to compute, we attempt to find and , and use the formula that
By angle chasing, we find that is a triangle with and . Thus .
Switching to the lower triangle , , and , with .
Using the Law of Sines on :
We now plug in , and into the formula for the area:
Thus the answer is
Note: We could also get the lengths (and area) of the triangle by drawing a perpendicular from to , forming a and a triangle.
See also
2001 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.