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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!

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

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