6
$\begingroup$

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.

Sqrt into 45-60-75

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?

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – cr001
    Commented Feb 14, 2019 at 15:35

1 Answer 1

6
$\begingroup$

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.

enter image description here

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:

enter image description here

$\endgroup$
2
  • 1
    $\begingroup$ I think this answers the question completely (answer is $4$), unless the OP wishes to exclude this and consider it trivial. $\endgroup$
    – YiFan Tey
    Commented Feb 16, 2019 at 1:04
  • $\begingroup$ Yes, looks like that will do it. $\endgroup$
    – Ed Pegg
    Commented Feb 19, 2019 at 18:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .