6
$\begingroup$

It is know that a square can be dissected into other square such that no two of the squares have the same size. This is the simplest dissection of that kind:

enter image description here

Is it also possible to dissect an equilateral triangle into equilateral triangles such that no two of them have the same size?

$\endgroup$
3
$\begingroup$

I think the following argument holds:

Suppose that $ABC$ is the great equilateral triangle with $A$ pointing upwards and that there exists a finite partition of $ABC$ into equliateral triangles such that every two triangles have different size lengths.

First note that in the vertex $A$ the angle must be filled with a small equilateral triangle $AXY$. There are a number of equilateral triangles $T_1,..,T_k$ in the partition with one side contained in $XY$ and the vertices pointing downwards. Among these there is one with the smallest side length: $D_1E_1F_1$ (with $D_1,E_1$ on $XY$).

The angles formed by the triangles $(T_j)$ in $D_1,E_1$ must be filled near their vertices with two equilateral triangles, one of which has smaller side length than $D_1E_1$. Denote with $A_1$ this vertex and $A_1B_1C_1$ this smaller triangle.

I think that now we can continue recursively to find always a smaller triangle:

  • the triangle $A_1B_1C_1$ behaves just like the initial triangle $AXY$ and we will find one smallest triangle pointing downwards with side contained in $B_1C_1$

  • this triangle has two neighbor triangles and one of them is smaller, which we denote by $A_2B_2C_2$, and so on.

We can continue this procedure indefinitely because all the triangle sides are different.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ opinions? is this right? $\endgroup$ – Beni Bogosel Mar 10 '13 at 20:55
  • 1
    $\begingroup$ To me the proof seems perfectly valid. And I think its a lot more elegant than the other one, too. $\endgroup$ – Dominik Mar 11 '13 at 15:26
  • $\begingroup$ @Dominik: The article proves a more general thing, I guess, that's why it is more complicated. $\endgroup$ – Beni Bogosel Mar 11 '13 at 19:15
3
$\begingroup$

This paper claims it is impossible.

$\endgroup$
  • $\begingroup$ Is my proof valid? $\endgroup$ – Beni Bogosel Mar 11 '13 at 6:32
2
$\begingroup$

Although as Ross Millikan points out, such a dissection is impossible, it is of interest to note that, provided we count upwardly oriented triangles as different from downwardly oriented, there is a dissection into 15 triangles. That is, none of the upwardly oriented triangles are same sized, nor any of the downwardly oriented.

http://arxiv.org/abs/0910.5199

[Note that site allows a free PDF download.]

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.