37th Iranian Mathematical Olympiad

Assume that the circle mentioned in the problem is the unit circle on the complex plane. Then we can say every vertex $A_i$ is equivalent to a complex number $z_i$ such that $|z_i|=1$. On the other hand we have $d(A_i,A_j)=|z_i-z_j|$. So we should prove that $$\sum _{1\le i<j\le k}|z_i-z_j{|}^2\le k^2.$$ We know that $$|z_i-z_j{|}^2=(z_i-z_j)({\bar{z}}_i-{\bar{z}}_j)=|z_i{|}^2+|z_j{|}^2-({\bar{z}}_i{z}_j+{\bar{z}}_j{z}_i)=2-{\bar{z}}_i{z}_j-{\bar{z}}_j{z}_i.$$ So we have $$\sum _{1\le i<j\le k}|z_i-z_j{|}^2=2(\genfrac{}{}{0ex}{}{k}{2})-\sum _{1\le i<j\le k}({\bar{z}}_i{z}_j+{\bar{z}}_j{z}_i).$$ and $$\begin{array}{rl}\sum _{1\le i<j\le k}({\bar{z}}_i{z}_j+{\bar{z}}_j{z}_i)& =\sum _{1\le i\ne j\le k}{\bar{z}}_i{z}_j=\sum _{1\le i\le k}{\bar{z}}_i(z_1+\cdots +z_k-z_i)\\ & =\left(\sum _{1\le i\le k}{\bar{z}}_i\right)\left(\sum _{1\le i\le k}{z}_i\right)-k={\left|\sum _{1\le i\le k}{z}_i\right|}^2-k.\end{array}$$ Therefore, $$\sum _{1\le i<j\le k}|z_i-z_j{|}^2=k^2-k-\left({\left|\sum _{1\le i\le k}{z}_i\right|}^2-k\right)=k^2-{\left|\sum _{1\le i\le k}{z}_i\right|}^2.$$ Since ${\left|{\sum }_{1\le i\le k}{z}_i\right|}^2\ge 0$, we have $$\sum _{1\le i<j\le k}|z_i-z_j{|}^2\le k^2.$$ Equality holds whenever ${\sum }_{1\le i\le k}{z}_i=0$, in other words, when the circumcenter and the centroid of $A_1{A}_2\dots A_k$ coincide.