Difference between revisions of "2001 AIME I Problems/Problem 4"
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-youyanli | -youyanli | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/BIyhEjVp0iM?t=526 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See also== | ==See also== |
Revision as of 03:22, 23 January 2023
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
.
![[asy] size(200); defaultpen(linewidth(0.4)+fontsize(8)); pair A,B,C,D,T,F; A = origin; T = scale(24)*dir(30); C = scale(24)*dir(60); B = extension(C,T,A,(1,0)); F = foot(T,A,B); D = foot(C,A,B); draw(A--B--C--A--T, black+0.8); draw(C--D, dashed); label(rotate(degrees(T-A))*"$24$", A--T, N); label(rotate(degrees(C-A))*"$24$", A--C, 2*NW); label("$12\sqrt 3$", C--D, E); label("$12\sqrt 3$", D--B, S); label("$12$", A--D, S); pen p = fontsize(8)+red; MA("45^\circ", C,B,A,2); MA("30^\circ", B,A,T,2.5); MA("30^\circ", T,A,C,3.5); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, N); dot("$T$", T, NE); dot("$D$", D, S); [/asy]](http://latex.artofproblemsolving.com/4/d/f/4df524bae2eda84cea879e883fc1758de646f8c7.png)
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.