Indian Mathematical Olympiad 2018

Suppose that the points $A,B,C$ are represented by complex numbers $a,b,0$ respectively. Without loss of generality, we can assume that $a$ is a positive real. We therefore also require that $u = |a| = a$ , $v =|b|$ and $w = |a-b|$ are positive integers. We can also assume that $u \ge v$ .

The fact that $C,D,E,G$ are concyclic means that the four complex numbers $0,\tfrac12a,\tfrac12b,\tfrac13(a+b)$ are concyclic. This means that their cross-ratio is real, and hence $\frac{b(2b-a)}{2(b^2 - a^2)} \; = \; \frac{\frac12b (\frac13b - \frac16a)}{\frac12(b-a)\frac13(b+a)} \; = \; \tfrac12\lambda$ for some real $\lambda$ . This implies that $(2-\lambda)b^2 - ab + \lambda a^2 \; = \; 0$ Since we want $ABC$ to be a nontrivial triangle, $b$ cannot be a real multiple of $a$ , and so we deduce that $4\lambda(2-\lambda) > 1$ , with $b \; = \; \frac{a}{2(2-\lambda)}\left\{ 1 \pm i\sqrt{4\lambda(2-\lambda) - 1}\right\}$ From this we deduce that $u \; = \; a \hspace{2cm} v \; = \; a\sqrt{\tfrac{\lambda}{2-\lambda}} \hspace{2cm} w \; = \; \tfrac{a}{\sqrt{2-\lambda}}$ From this we deduce that $\lambda = \tfrac{v^2}{w^2}$ and hence that $u^2 + v^2 \; = \; 2w^2$ Since $u^2 + v^2$ is even, $u$ and $v$ must have the same parity. If we define nonnegative integers $p,q$ by setting $u+v = 2p$ , $u-v=2q$ , we deduce that $p^2 + q^2 = w^2$ . We therefore need to consider Pythagorean triples.

• The first case is $p=1$ , $q=0$ , $w=1$ , which leads to $u=v=w=1$ . This triangle is equilateral, and so should be ignored.
• The second case is $p=4$ , $q=3$ , $w=5$ , which leads to $u=7$ , $v=1$ . A triangle with sides $7,1,5$ does not exist, so we ignore this case as well.
• The third case is $p=12$ , $q=5$ , $w=13$ , which leads to $u=17$ , $v=7$ . A triangle with sides $17,7,13$ does exist, so it must be the desired triangle with shortest perimeter. This makes the answer $17 + 7 + 13 = \boxed{37}$ .