# Proving an intuitive property for families of non-overlapping convex polygons on the plane

Let $$\mathscr{F}$$ be a finite family of non-overlapping convex polygons on the plane. I'd like to prove that

$$\exists F \in \mathscr{F}\exists F'\subset \mathbb{R}^2 \mathscr{F}\cup \{F'\}$$ is a family of non-overlapping convex polygons, $$F'$$ is congruent to $$F$$ and shares one side with it.

For now, I'm trying to prove the result for triangles, but with no success. This is quite an intuitive proposition, especially for triangles. Indeed one expects to always find a triangle in the "external region" with enough space around to construct a congruent triangle on one of its sides.

Observation If $$P\in\mathscr{F}, |P\cap ConvHull(\mathscr{F})|\geq 2$$ then $$\mathscr{F}$$ respects the thesis.

After Hagen von Eitzen answered I went back to thinking about my original problem, i.e. what happens if the polygons are all congruent triangles. I was not able to adapt his answer, which I very much liked, to this special case.

• So what you need is that for some $F$, two of its vertices are part of the boundary of the covex full of $\cup \mathcal{F}$. Is that right? Aug 26, 2019 at 18:27
• Should $F\cup\{F'\}$ be $\mathscr{F}\cup\{F'\}$? Aug 26, 2019 at 18:28
• nevermind, my statement above is false (found an easy counterexample) Aug 26, 2019 at 18:30
• I doubt the added observation(or rather don't see it easily) because two vertices on the boundary of the convex hull does not mean that the edge connecting them is an edge of the convex hull ... Aug 26, 2019 at 18:36
• I'm pretty sure the original statement is false. Imagine a sequence of isoceles triangles, each with the unique side facing down and the 'point' facing up, alternating between wide short 'planks' and tall thin 'spikes'. Offset the planks upwards a little bit from the baseline to pack them closer to the spikes. Then clearly none of the triangles in this arrangement can have a congruent triangle attached. Now, bend this into a 'circle'... (in fact, a triangle is probably good enough.) Aug 26, 2019 at 18:47

Complete this image to a circle.

The big triangles cannot be used, except perhaps with their base edge. The smaller triangles filling the gaps can obviously not be used. The out-pointing small triangles cannot be used either because we can obviously not use their inner edge, and by choosing the height smaller than the height of the gaps below them, their other two edges cannot be used either.

So what about the base edges of the big triangles? All we need to ensure is that the height of the triangles exceeds the diameter of the $$n$$-gon formed by theses edges (so make them significantly taller than in the sketch) and we are done!

After fine-tuning the details, the following seems to be a minimal working example reachable with this method (21 triangles):

The blue triangles are wlog. equilateral. Hence we need $$\gamma<60^\circ$$, so $$\beta>60^\circ$$ and $$\alpha<60^\circ$$. Moreover, $$n\cdot(\alpha-\epsilon)=360^\circ$$. Thus we certainly need $$n>6$$. With $$n=7$$ as in the image, we can pick $$\alpha$$ anywhere strictly between $$51\frac37^\circ$$ and $$60^\circ$$ (which make $$\epsilon<8\frac 57^\circ$$).

• you can also just fill in the central disk of the circle with one large polygon, so that's not a difficult consideration either Aug 26, 2019 at 18:58
• Thank you, this answers it Aug 26, 2019 at 19:05
• @DanRust I wanted to use only triangles. Then again, I could just split your inner polygon into triangles ... Aug 26, 2019 at 19:21
• I think that forming a 7gon (so using 21 triangles) should be the smallest configuration achievable with my method (because it needs the "gap" angles $<60^\circ$) Aug 26, 2019 at 19:35
• I'm very happy you added the details. Would you have any idea on how to tackle the problem where the polygons considered are cll congruent triangles? I have been quite unsuccessful in adapting your answer to this case. Aug 28, 2019 at 20:46