One solution for this problem.

I like this problem the best in 7th IGO. See the picture for an easy generalization with cut and jigsaw proof.

buratinogigle wrote: I like this problem the best in 7th IGO. See the picture for an easy generalization with cut and jigsaw proof. Dear my teacher, I absolutely agree with you.

dungnguyentien wrote: One solution for this problem. It would be $x+y+z=y+(z+x)>y+t>2a$ instead of $2c$ right?

According to the figure, three equilateral triangles with side lengths $a,b,c$ have one common vertex and do not have any other common point. The lengths $x, y$, and $z$ are defined as in the figure. Prove that $3(x+y+z)>2(a+b+c)$. Proposed by Mahdi Etesamifard