103 Trigonometry Problems

1. Trigonometric Fundamentals 45
[
4 √ 5 − 4 √ 2, 4 √ 5 − 8 √ ]
2 , the slope of line l is 4√ 5−8 √ √ √
2
4 √ 5−4 √ = √ 5−2
√ 2
, and so the
2 5− 2
slope of line PQis
√ √
5 − 2
2 √ 2 − √ 5 = (√ 5 − √ 2)(2 √ 2 + √ √
5) 10 + 1
(2 √ 2 − √ 5)(2 √ 2 + √ 5) = .
3
Second Solution: Let l 1 and l 2 denote two lines, and for i = 1 and 2, let m i and θ i
(with 0 ◦ ≤ θ i < 180 ◦ ) denote the slope and the polar angle of line l i , respectively.
Without loss of generality, we assume that θ 1 >θ 2 .Ifθ is the polar angle formed by
the lines, then θ = θ 1 − θ 2 and
tan θ = tan θ 1 − tan θ 2
1 + tan θ 1 θ 2
= m 1 − m 2
1 − m 1 m 2
by the addition and subtraction formulas.
Let m be the slope of line PQ. Because lines OA and AB have slopes 2 and 1,
respectively, by our earlier discussion we have
m − 1
1 + m = tan ̸ PAB = tan ̸ QAO = 2 − m
1 + 2m ,
or (m − 1)(1 + 2m) = (2 − m)(1 + m). It follows that 3m 2 − 2m − 1 = 0, or
m = 1±√ 10
3
. It is not difficult to see that m = 1+√ 10
3
is the answer to the problem.
(The other value is the slope of line l, the interior bisector of ̸ OAB.)
P
B
A
B
Q
M
O
O
Figure 1.46.
A
Example 1.17. [AIME 1994] The points (0, 0), (a, 11), and (b, 37) are the vertices
of an equilateral triangle (Figure 1.46, right). Find ab.
In both solutions, we set O = (0, 0), A = (a, 11), and B = (b, 37).
First Solution: Let M be the midpoint of segment AB. Then M = ( a+b
2 , 24) ,
OM ⊥ MA, and |OM| = √ 3|MA|. Because −→ (
AM =
a−b
2 , −13) , it follows that