103 Trigonometry Problems

40 103 Trigonometry Problems
new lines are also concurrent, and the point of concurrency is the second Brocard
point. This is the reason we say that the two Brocard points are isogonal conjugates
of each other.
C
P
B
Figure 1.42.
A
Example 1.13. [AIME 1999] Point P is located inside triangle ABC (Figure 1.42)
so that angles PAB, PBC, and PCA are all congruent. The sides of the triangle
have lengths |AB| =13, |BC|=14, and |CA|=15, and the tangent of angle PAB
is m/n, where m and n are relatively prime positive integers. Find m + n.
Solution: Let α = ̸ PAB = ̸ PBC = ̸ PCA and let x, y, and z denote |PA|,
|PB|, and |PC|. Apply the law of cosines to triangles PCA, PAB, and PBC to
obtain
x 2 = z 2 + b 2 − 2bz cos α,
y 2 = x 2 + c 2 − 2cx cos α,
z 2 = y 2 + a 2 − 2ay cos α.
Sum these three equations to obtain 2(cx + ay + bz) cos α = a 2 + b 2 + c 2 . Because
the combined area of triangles PAB, PBC, and PCA is equal to
the preceding equation can be rewritten as
tan α =
4[ABC]
a 2 + b 2 + c 2 .
(cx+ay+bz) sin α
2
,
With a = 14, b = 15, and c = 13, use Heron’s formula to find that [ABC] =84.
It follows that tan α = 168
295
,som + n = 463.