# What is the smallest number of $45^\circ$–$60^\circ$–$75^\circ$ triangles in non-trivial substitution tiling?

Let base = $$45^\circ$$$$60^\circ$$$$75^\circ$$ triangle.

Over at What is the smallest number of bases that a square can be divided into? it was determined that 23 base were needed to make a $$45^\circ$$$$45^\circ$$$$90^\circ$$ triangle. How about non-trivial dissections of base into similar triangles? Start with base and divide it into smaller copies of base. Ideally the method should be specific to base and wouldn't work with other triangles. Also, at least one of the internal triangles should have no edges parallel to the original triangle.

What are the simplest non-trivial dissections of base into similar triangles?

• For clarification, does "non-trivial" mean all the three triangles at the three corners cannot be parallel triangles to the original one? Or does it mean at least one cannot be parallel? Or at least two? – cr001 Feb 14 at 15:35

## 1 Answer

Hmm... This doesn't exactly fit your criterion in that it's not unique to the $$45^∘–60^∘–75^∘$$ triangle, but the central triangle here has no edges parallel to the edges of the original triangle. The vertices are: $$\begin{array}{ccc} \{0,0\} \\ \{1,0\} \\ \left\{\frac{1}{2} \left(3-\sqrt{3}\right),\frac{1}{2} \left(3-\sqrt{3}\right)\right\} \\ \left\{\frac{1}{22} \left(21-2 \sqrt{3}\right),\frac{1}{22} \left(6+\sqrt{3}\right)\right\} \\ \left\{-\frac{3}{22} \left(-5+\sqrt{3}\right),0\right\} \\ \left\{\frac{1}{22} \left(6+\sqrt{3}\right),\frac{1}{22} \left(6+\sqrt{3}\right)\right\} \\ \left\{\frac{1}{286} \left(237-43 \sqrt{3}\right),\frac{1}{143} \left(21-2 \sqrt{3}\right)\right\} \\ \left\{\frac{1}{286} \left(216-41 \sqrt{3}\right),\frac{1}{22} \left(6+\sqrt{3}\right)\right\} \\ \left\{\frac{1}{286} \left(174-37 \sqrt{3}\right),\frac{1}{286} \left(36+17 \sqrt{3}\right)\right\} \\ \left\{\frac{996-197 \sqrt{3}}{1430},\frac{192+43 \sqrt{3}}{1430}\right\} \\ \left\{\frac{996-197 \sqrt{3}}{1430},\frac{306+73 \sqrt{3}}{1430}\right\} \\ \left\{\frac{1}{130} \left(102-19 \sqrt{3}\right),\frac{318+31 \sqrt{3}}{1430}\right\} \\ \end{array}$$

This is based on the three other four-self-similar-triangle dissections beyond the usual midpoints-of-the-sides one: • I think this answers the question completely (answer is $4$), unless the OP wishes to exclude this and consider it trivial. – YiFan Feb 16 at 1:04
• Yes, looks like that will do it. – Ed Pegg Feb 19 at 18:54