# Is it possible to partition a square into finitely many triangles such that no two triangles share a side?

Here is a problem from Miklos Schweitzer competition from 2017.

Is it possible to partition a square into finitely many triangles such that no two triangles share a side?

I believe it is impossible. I tried proving that the polygon formed by triangles (without sharing any side) must have a concave angle ($>\angle180^{\circ}$), but... or maybe just making concave polygons with the triangles...

Help guys!

• It suffices to show that you can't bisect a square into two noncontiguous triangles. – Allawonder Jan 8 '18 at 20:53
• See KöMaL. I'm not sure this is close enough to warrant locking it until the contest closes? Better safe than... – Jyrki Lahtonen Jan 8 '18 at 21:25