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 1
![[asy] size(180); pointpen = black; pathpen = black+linewidth(0.7); pair A=(0,0),B=(12+12*3^.5,0),C=(12,12*3^.5),D=foot(C,A,B),T=IP(CR(A,24),B--C); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(D(MP("T",T,NE))--A); D(MP("D",D)--C,linetype("6 6") + linewidth(0.7)); MP("24",(A+3*T)/4,SE); D(anglemark(C,B,A,65)); D(anglemark(B,A,C,65)); D(rightanglemark(C,D,B,50)); MP("30^{\circ}",A,(4,1)); MP("45^{\circ}",B,(-3,1)); [/asy]](http://latex.artofproblemsolving.com/f/d/8/fd80ef0be57cda8d5c014431604ac07fc2c9a044.png)
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
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 |