## Problem 1

An obtuse angled triangle has integral sides and one acute angle is twice the other. Find the smallest possible perimeter.


## Solution

Answer: $77=16+28+33$.

Let the triangle be $\mathrm{ABC}$ with angle $\mathrm{A}$ obtuse and angle $\mathrm{B}=2$ angle $\mathrm{C}$. Let the sides be $\mathrm{a}, \mathrm{b}, \mathrm{c}$ as usual. Note that $\mathrm{a}>\mathrm{b}>\mathrm{c}$. We have $\mathrm{b} \sin \mathrm{C}=\mathrm{c} \sin 2 \mathrm{C}$ and $\mathrm{c}^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab} \sin \mathrm{C}$. Hence, $\mathrm{b} / 2 \mathrm{c}=\sin 2 \mathrm{C} / \sin \mathrm{C}=\cos \mathrm{C}=$ $\left(a^{2}+b^{2}-c^{2}\right) / 2 a b$. So $a b^{2}=a^{2} c+b^{2} c-c^{3}$. Hence $b^{2}(a-c)=c\left(a^{2}-c^{2}\right)$. Dividing by $a-c$ we get $b^{2}=c(a+c)$.

Now the triangle with smallest perimeter will have $\mathrm{a}, \mathrm{b}, \mathrm{c}$ relatively prime (otherwise we could divide by the common factor). Hence $\mathrm{c}$ must be a square. For if $\mathrm{c}$ and $\mathrm{a}+\mathrm{c}$ have a common factor, then so do a and $\mathrm{c}$ and hence $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$, which means they cannot be the minimal set. Clearly $\mathrm{c}$ is not 1 (or the triangle would have nil area). $\mathrm{c}=4$ gives $\mathrm{a}=5$, which is too short, or $\mathrm{a} \geq 21$, which is too long. $\mathrm{c}=9$ gives $\mathrm{a}=7$ (too short), 18 (3 divides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ ), or $\geq 40$ (too long). $\mathrm{c}=16$ gives $\mathrm{a}=20$ (too short) or $\mathrm{a}=33$ which works. Larger $\mathrm{c}$ gives $\mathrm{a}$ larger perimeter. $\operatorname{Eg} \mathrm{c}=25$ gives $\mathrm{a}=56, \mathrm{~b}=45$ (perimeter 126).

