An obtuse angled triangle has integral sides and one acute angle is twice the other

Let $x$ be the smallest acute angle. Then the other acute angle that is twice that is $2x$ , and the third obtuse angle is $180° - 3x$ . Let $a$ be the side opposite to $x$ , $b$ be the side opposite to $2x$ , and $c$ be the side opposite to $180° - 3x$ . Then $a < b < c$ . Also, since $c$ is opposite an obtuse angle, $c^2 > a^2 + b^2$ .

By law of sines, $\frac{\sin x}{a} = \frac{\sin 2x}{b}$ . Since $\sin 2x = 2 \sin x \cos x$ , $\frac{\sin x}{a} = \frac{2 \sin x \cos x}{b}$ , which simplifies to $\cos x = \frac{b}{2a}$ . This means that $b < 2a$ , and $\cos^2 x = \frac{b^2}{4a^2}$ , and $\sin^2 x = 1 - \frac{b^2}{4a^2}$ .

Also by law of sines, $\frac{\sin x}{a} = \frac{\sin (180° - 3x)}{c}$ . Since $\sin (180° - 3x) = \sin 3x = 3 \sin x - 4 \sin^3 x$ , $\frac{\sin x}{a} = \frac{3 \sin x - 4 \sin^3 x}{c}$ , which simplifies to $c = a(3 - 4 \sin^2 x)$ . Also from above $\sin^2 x = 1 - \frac{b^2}{4a^2}$ , so $c = a(3 - 4(1 - \frac{b^2}{4a^2}))$ which simplifies to $c = \frac{b^2}{a} - a$ .

So now we must have $a < b < c$ , $a < b < 2a$ , $c^2 > a^2 + b^2$ , and $c = \frac{b^2}{a} - a$ for integer solutions for $a$ , $b$ , and $c$ . Since $c$ is an integer solution, $a$ must divide into $b^2$ . Starting with $a = 1$ and testing values for $b$ such that $a < b < 2a$ and $a$ divides into $b^2$ , we find that the first $c$ value such that $c > b$ and $c^2 > a^2 + b^2$ occurs when $a = 16$ , $b = 28$ , and $c = 33$ , for a perimeter of $\boxed{77}$ .