(given to Rémi Varloot). In a triangle $ABC$, angle $\hat{A}$ is twice the angle $\hat{B}$, angle $\hat{C}$ is obtuse, and the lengths of the sides are integers. What is the smallest possible perimeter of this triangle?
. Let $a=BC, b=CA, c=AB$ be the lengths of the sides, and $\beta$ the angle $\widehat{ABC}$ (so $2\beta$ is the angle $\widehat{BAC}$). Let $B'$ be the point on $(AB)$ outside the segment $[AB]$ such that $AB' = b$.

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The triangle $AB'C$ is isosceles, and since the angle $\widehat{BAC}$ is $2\beta$, $\widehat{AB'C} = \beta = \widehat{ABC}$. It follows that the triangle $B'CB$ is also isosceles, and that the two triangles $AB'C$ and $B'CB$ are similar, hence $\frac{CB'}{AB'} = \frac{BB'}{CB}$, or $\frac{a}{b} = \frac{b+c}{a}$, which simplifies to $a^2 = b(b+c)$. If $a, b, c$ are integers, we can assume $b$ and $c$ are coprime, otherwise their common divisor $d$ would also divide $a$ (from this relation), and by dividing $a, b, c$ by $d$, we would have a triangle satisfying the same hypotheses, with a perimeter $d$ times smaller. If $b$ and $c$ (and thus $b$ and $b+c$) are coprime, and if the product $b(b+c)$ is a perfect square, then both $b$ and $b+c$ are perfect squares: $b = m^2, b+c = n^2$, so $a = mn$, and $\frac{a}{b} = \frac{n}{m}$. According to the law of sines, $\frac{a}{b} = \frac{\sin(2\beta)}{\sin(\beta)} = 2\cos(\beta)$. Since the angle $\widehat{ACB} = \pi - 3\beta$ is assumed to be obtuse, $\beta < \frac{\pi}{6}$, so $\sqrt{3} < 2\cos(\beta) = \frac{a}{b} = \frac{m}{n} < 2$. To minimize $a + b + c = mn + n^2$, $m$ and $n$ must be as small as possible, which gives $n = 4$ (since the interval $(\sqrt{3}, 2)$ does not contain any half-integers or third-integers), $m = 7$ (since $\sqrt{3} < \frac{7}{4} < 2$), so $a + b + c = 28 + 49 = 77$.