103 Trigonometry Problems

190 103 Trigonometry Problems
X
X1
O1
Y
O
Figure 5.14.
Z
Then R(Y ) = X, and so triangle ZO 1 O is equilateral. Consequently, triangles
ZO 1 X and ZOY are congruent, and so |O 1 X|=|OY|. Note that x + y + z =
360 ◦ .Wehave
̸ O 1 OX = ̸ ZOX − ̸ ZOO 1 = ̸ ZOX − 60 ◦ = y − 60 ◦ ,
̸ XO 1 O = ̸ XO 1 Z − ̸ OO 1 Z = ̸ YOZ− 60 ◦ = x − 60 ◦ ,
̸ OXO 1 = 180 ◦ − ̸ O 1 OX − ̸ XO 1 O = z − 60 ◦ .
Applying the law of sines to triangle XOO 1 establishes the desired result.
Now we prove our main result. Without loss of generality, suppose that P 4 ,P m ,
and P n are on sides AB, BC, and CA, respectively (Figure 5.15). Let O be
the circumcenter of triangle ABC. Without loss of generality, we assume that
the circumradius of triangle ABC is 1, so |OA|=|OB|=|OC|=1. Let X,
Y , and Z be the centers of P 4 ,P m , and P n , respectively.
Z
C
Y
A
O
B
X
Figure 5.15.
Because |OB| =|OC| and |YB|=|YC|, BOCY is a kite with OY as its
axis of symmetry. Thus, ̸ BOY = ̸ BOC
2
= ̸ A, and ̸ OYB = 180 ◦ /m. Let