2001 AIME I Problems/Problem 4
Contents
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)
First, draw a good diagram.
We realize that , and . Therefore, as well, making an isosceles triangle. and are congruent, so . We now drop an altitude from , and call the foot this altitude point .
By 30-60-90 triangles, and .
We also notice that is an isosceles right triangle. is congruent to , which makes . The base is , and the altitude . We can easily find that the area of triangle is , so .
-youyanli
Solution 3(Speedy and Simple)
After drawing line AT, we see that we have two triangles: (ABT) with 45, 30, and 105 degrees, and (ATC), with 30, 75, 75 degrees. If we can sum these two triangles' areas, we have our answer. Let's take care of (ATC) first. We see that ATC is a iscoceles triangle, with AT = AC = 24. Because the area of a triangle is , we have , which is equal to 144. Now on to triangle (ABT). Draw the altitude from angle T to AB, and call the point of intersection D. This splits (ABT) into 2 triangles, one with 30-60-90 (ADT), and another with 45-45-90 (BDT). Now, because we know that AT is 24, we have by special right triangle ratios the area of ADT and the area of BDT , which gives $72*{\sqrt{3}}+72}$ (Error compiling LaTeX. Unknown error_msg). Adding this to ATC we have , from which we sum to get 291.
~MathCosine
Video Solution by OmegaLearn
https://youtu.be/BIyhEjVp0iM?t=526
~ pi_is_3.14
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 |
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