# 2007 Irish Mathematical Olympiad, Problem 2

Prove that a triangle ABC is right-angled if and only if

sin^{2}A + sin^{2}B + sin^{2}C = 2.

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Let the three side lengths of triangle ABC be a, b and c.

Apply the law of sines, we obtain

\frac{a^{2}}{sin^{2}A} = \frac{b^{2}}{sin^{2}B} = \frac{c^{2}}{sin^{2}C} = \frac{a^{2} + b^{2} + c^{2}}{sin^{2}A + sin^{2}B + sin^{2}C} = \frac{a^{2} + b^{2} + c^{2}}{2}.

Now applying the law of cosines, we obtain

a^2=b^{2}+c^{2} - 2bc\cdot cosA

and substitute a^{2} into the above equation

\frac{a^{2}}{sin^{2}A}= \frac {2(b^{2} + c^{2}- bc\cdot cosA)}{2}= b^{2} + c^{2}- bc\cdot cosA or
b^{2}+c^{2}- 2bccosA = (b^{2}+c^{2}-bc\cdot cosA)sin^{2}A or
(b^{2}+c^{2})(1 - sin^{2}A) = bc\cdot cosA (2 - sin^{2}A)
(b^{2}+c^{2})cos^{2}A = bc\cdot cosA(1 + cos^{2}A)
(b^{2}+c^{2})cosA = bc\cdot (1 + cos^{2}A),

(if \angle A is not right.)

bc\cdot cos^{2}A –(b^{2}+c^{2})cosA + bc = 0.

Solve for cosA, we have cosA = b/c and c/b, which implies that either angle B or angle C is right.

Conversely, assume that one of the angles, say angle A, is right.

Then sin^{2}A = 1, sin^{2}B = b^{2}/a^{2} and sin^{2}C = c^{2}/a^{2}, and sin^{2}B + sin^{2}C= (b^{2}+c^{2})/a^{2} = 1.

Therefore, sin^{2}A + sin^{2}B + sin^{2}C = 2.

Note: another proof is available elsewhere. A slight modification of the problem was also suggested (but did not get it) by Poland for the 1967 IMO.