103 Trigonometry Problems

5. Solutions to Advanced Problems 163
Note: Using a similar method, one can show that
1
n − 1 + a 1
+
1
n − 1 + a 2
+···+
1
n − 1 + a n
≤ 1,
where a 1 ,a 2 ,...,a n are positive real numbers such that a 1 a 2 ···a n = 1.
An interesting exercise is to provide a trigonometry interpretation for the last
inequality.
37. [MOSP 2001] Find all triples of real numbers (a,b,c)such that a 2 −2b 2 = 1,
2b 2 − 3c 2 = 1, and ab + bc + ca = 1.
Solution: Since a 2 − 2b 2 = 1, a ̸= 0. Since 2b 2 − 3c 2 = 1, b ̸= 0. If
c = 0, then b = 1/ √ 2 and a = √ 2. It is easy to check that (a,b,c) =
( √2,
1/
√
2, 0
)
is a solution of the system. We claim that there are no other
valid triples.
We approach the problem indirectly by assuming that there exists a triple of
real numbers (a,b,c), with abc ̸= 0, such that the equations hold. Without
loss of generality, we assume that two of the numbers are positive; otherwise,
we can consider the triple (−a,−b, −c). Without loss of generality, we assume
that a and b are positive. (The first two equations are independent of the signs
of a,b,c, and the last equation is symmetric with respect to a,b, and c.) By
Introductory Problem 21, we may assume that a = cot A, b = cot B, and
c = cot C, with 0