103 Trigonometry Problems

5. Solutions to Advanced Problems 189
̸ ̸ ̸ Because both ABDE and ACDF are cyclic, BDF = CDE = CAB.
Thus, by the Lemma, we have
2([BFC]+[BEC])
=|DF|·|BC|·sin ̸ BDF +|DE|·|BC|·sin ̸ EDC
=|BC|(|DE|+|DF|) sin A ≥ (|DE|+|DF|) 2 sin A.
By equation (†), the last inequality is equivalent to
Likewise, we have
R 2 (sin 2B + sin 2C) 2 sin A ≤ 2[BFC]+2[BEC].
R 2 (sin 2C + sin 2A) 2 sin B ≤ 2[CDA]+2[CFA]
and
R 2 (sin 2A + sin 2B) 2 sin C ≤ 2[AEB]+2[ADB].
Adding the last three inequalities yields the desired result. In view of the
Lemma, it is also clear that equality holds if and only if triangle ABC is
equilateral.
49. [Bulgaria 1998] On the sides of a nonobtuse triangle ABC are constructed
externally a square P 4 , a regular m-sided polygon P m , and a regular n-sided
polygon P n . The centers of the square and the two polygons form an equilateral
triangle. Prove that m = n = 6, and find the angles of triangle ABC.
Solution: The angles are 90 ◦ , 45 ◦ , and 45 ◦ . We prove the following lemma.
Lemma
Let O be a point inside equilateral triangle XYZ. If
̸ YOZ = x, ̸ ZOX = y, ̸ XOY = z.
then
|OX|
sin(x − 60 ◦ ) =
|OY|
sin(y − 60 ◦ ) =
|OZ|
sin(z − 60 ◦ ) .
Proof: As shown in Figure 5.14, let R denote clockwise rotation of 60 ◦
around the point Z, and let R(X) = X 1 and R(O) = O 1 .