Problem 11879
(American Mathematical Monthly, Vol.122, December 2015) Proposed by S. Siboni (Italy).
For positivea, b, and c, prove that there exist positive α, β, and γ with α + β + γ = π such that a
sin α = b
sin β = c sin γ if and only if|b − c| < a < b + c.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
If |b − c| < a < b + c then
a < b + c, b = b − c + c ≤ |b − c| + c < a + c, c = c − b + b ≤ |b − c| + b < a + b which imply that there exists a non-degenerate triangle with a, b, c as its side lengths.
Let α, β, γ the radian measures of the corresponding opposite angles, then they are positive numbers such that α + β + γ = π and, by the law of sines,
a
sin α = b
sin β = c sin γ.
Now assume that there are positive α, β, and γ with α + β + γ = π such that a
sin α = b
sin β = c sin γ. Then b = a sin β/ sin α, c = a sin γ/ sin α and it suffices to show that
| sin β − sin γ| < sin α < sin β + sin γ that is
sin α < sin β + sin γ, sin β < sin α + sin γ, sin γ < sin α + sin β.
We show the first inequality:
sin α = sin(π − β − γ) = sin(β + γ) = sin β cos γ + sin γ cos β < sin β + sin γ
because sin β > 0, sin γ > 0, cos β < 1, and cos γ < 1. The other two can be verified in a similar
way.
