HMMT February

Using a little trig, we have $B C=2 \sin 18, A C=2 \sin 36$, and $A B=2 \sin 30$ (see left diagram). Call these $a, b$, and $c$, respectively. By the law of cosines, $b^{2}=a^{2}+c^{2}-2 a c \cos A B C$, therefore $\cos A B C=\frac{\sin ^{2} 18+\sin ^{2} 30-\sin ^{2} 36}{2 \sin 18 \sin 30}$. In the right diagram below we let $x=2 \sin 18$ and see that $x+x^{2}=1$, hence $\sin 18=\frac{-1+\sqrt{5}}{4}$. Using whatever trig identities you prefer you can find that $\sin ^{2} 36=\frac{5-\sqrt{5}}{4}$, and of course $\sin 30=\frac{1}{2}$. Now simplification yields $\sin ^{2} 18+\sin ^{2} 30-\sin ^{2} 36=0$, so $\angle A B C=\mathbf{90}^{\circ}$. Note that this means that if a regular pentagon, hexagon, and decagon are inscribed in a circle, then we can take one side from each and form a right triangle.