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Are right angled triangles the only triangles that can be divided into 2,4,8,16- - (2^n)smaller triangles, each being the same shape as the original? In other words is it the only one where infinite descent occurs?

Terence Coates
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    Every triangle $T$ can be partitioned into 4 smaller triangles whose vertices are made up by midpoints and vertices of $T$. – Crostul Feb 08 '21 at 21:23
  • @Crostul Each can be partitioned into $4$, yes. But haw about partitioning into $2$...? – CiaPan Feb 08 '21 at 21:34
  • Maybe I'm wrong, but I think right-angled angled triangles can be divided into $2,3,4,5,...$ similar triangles. In fact, no matter how you draw a new line inside that joins a vertex to a side of any of the existing right-angles triangles (without passing through a side), two new right-angles are created and they are both necessarily similar to all the others. – Adam Rubinson Feb 08 '21 at 21:34

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Suppose we have a triangle $T$ specified by a triple of angles $(\alpha, \beta, \gamma)$. Now, clearly the only way to divide $T$ into two triangles is by cutting through one of the vertices of $T$ associated with one of the three angles. So consider cutting $\alpha$ into 2 smaller angles $\alpha_1$ and $\alpha_2$ such that $\alpha_1 + \alpha_2 = \alpha$, yielding the triangles specified by the angle triples $(\alpha_1, \beta, \pi - \beta - \alpha_1)$ and $(\alpha_2, \gamma, \pi - \gamma - \alpha_2)$. If those two triangles are to be similar to the original one, we must have $\alpha_1 = \gamma$ and $\alpha_2 = \beta$ and thus $\alpha = \beta + \gamma$ which yields $\alpha = \pi/2$. This implies that $T$ is right-angled.

A note on dividing into 4 triangles: This can be done with any triangle as pointed out by Crostul's comment. Indeed by subdividing the triangles gotten by this process we can divide any triangle into $4^n$ subtriangles similar to the original one (for any $n \in \mathbb N$).

Watercrystal
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