103 Trigonometry Problems

102 103 Trigonometry Problems
Euler’s formula states that |OI| 2 = R 2 − 2Rr, where O and I are the
circumcenter and incenter of triangle ABC. Because |OI| 2 ≥ 0, we have
R ≥ 2r, or
R r ≤ 2 1 , from which (b) follows.
Note: Relation (∗) also has a geometric interpretation.
A
C1
O
B1
B
A1
C
Figure 4.3.
As shown in the Figure 4.3, let O be the circumcenter, and let A 1 ,B 1 ,C 1 be the
feet of the perpendiculars from O to sides BC,CA,AB, respectively. (Thus
A 1 ,B 1 ,C 1 are the midpoints of sides BC,CA,AB, respectively.) Because
̸ AOB = 2C and triangle AOB is isosceles with |OA| =|OB| =R, we
have |OC 1 |=R cos C. Likewise, |OB 1 |=R cos B and |OA 1 |=R cos A.
It suffices to show that
|OA 1 |+|OB 1 |+|OC 1 |=R + r.
Note that |OA|=|OB|=|OC|=R and |BA 1 |=|A 1 C|, |CB 1 |=|B 1 A|,
|AC 1 |=|C 1 B|. Hence |AB| =2|A 1 B 1 |, |BC|=2|B 1 C 1 |, |CA|=2|C 1 A 1 |.
Let s denote the semiperimeter of triangle ABC.Applying Ptolemy’s theorem
to cyclic quadrilaterals OA 1 CB 1 , OB 1 AC 1 , OC 1 BA 1 yields
Adding the above gives
|A 1 B 1 |·|OC|=|A 1 C|·|OB 1 |+|CB 1 |·|OA 1 |,
|B 1 C 1 |·|OA|=|B 1 A|·|OC 1 |+|AC 1 |·|OB 1 |,
|C 1 A 1 |·|OB|=|C 1 B|·|OA 1 |+|BA 1 |·|OC 1 |.
Rs =|OA 1 |(s −|A 1 B|) +|OB 1 |(s −|B 1 C|) +|OC 1 |(s −|C 1 A|)
= s(|OA 1 |+|OB 1 |+|OC 1 |) −[ABC]
= s(|OA 1 |+|OB 1 |+|OC 1 |) − rs,