103 Trigonometry Problems

5. Solutions to Advanced Problems 153
A
F
A
B
O
M
D
I
E
C
E
C
O (F)
I
D M
B
Figure 5.8.
Second Solution: As shown in Figure 5.8, the incircle touches the sides
AB, BC, and CA at D, E, and F , respectively. Let M be the foot of the
perpendicular line segments from O to side BC. Then the line OM is the
perpendicular bisector of BC, and |BM| =|CM|. From equal tangents, we
have |AF |=|AE|, |BD|=|BF|, and |CD|=|CE|. Because c>b, M lies
on segment BD. We find that
√
2|OI|=c − b = (|AF |+|FB|) − (|AE|+|EC|)
=|FB|−|EC| =|BD|−|DC|.
We deduce that |BD|=|BM|+|MD| and |DC| =|CM|−|DM|. Hence
√
2|OI|=2|DM|,or|OI|=
√
2|DM|. Thus lines OI and DM form a 45
◦
angle, which implies that either OI ⊥ AB or OI ‖ AB. We consider these
two cases separately.
• First Case: In this case, we assume that OI ⊥ AB. Then OI is the
perpendicular bisector of side AB; that is, the incenter lies on the perpendicular
bisector of side AB. Thus triangle ABC must be isosceles,
with |AC| =|BC|, and so A = B = 45 ◦ and sin A =
√
2
2 .
• Second Case: In this case, we assume that OI ‖ AB. Let N be the
midpoint of side AB. Then OIFN is a rectangle. Note that ̸ AON =
̸ C, and thus
R cos ̸
AON = R cos C =|ON|=|IF|=r.
By the solution to Introductory Problem 27, we have
cos C = R r
= cos A + cos B + cos C − 1,