2001 AIME I Problems/Problem 4
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
After chasing angles, and , meaning is an isosceles triangle and .
Using law of sines on , we can create the following equation:
and , so .
We can then use the Law of Sines area formula to find the area of the triangle.
can be found through the sin addition formula.
Therefore, the area of the triangle is
Solution 2 (no trig)
Firstly, draw a good diagram.
We realize that the measure of angle C is 75 degrees. The measure of angle CAT is 30 degrees. Therefore, the measure of angle CTA must also be 75 degrees, making CAT an isosceles triangle. AT and AC are congruent, so AC must equal 24.
We now drop an altitude from C, and call the foot of it a point D on side AB. By 30-60-90 triangles, AD equals twelve and CD equals .
We also notice that CDB is an isosceles right triangle. CD is congruent to BD, which makes BD also equal to . The base AB is 12+, and the altitude CD is . We can easily find that the area triangle ABC is 216+, so a+b+c=.
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.