Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Recently some old notes of mine have gotten me to thinking about the problem of subdividing a triangle into $N$ smaller triangles, all congruent to one another. A little thought shows the following are possible values of $N$:

$\bullet$ If $N$ is a perfect square then any triangle can be subdivided into $N$ such smaller triangles

$\bullet$ If $N$ is a sum of two squares, say $N = a^2+b^2$ then a right triangle with side lengths in the ratio $a:b:\sqrt{a^2+b^2}$ can be subdivided into $N$ smaller triangles

$\bullet$ If $N=3$ then a $30-60-90$ triangle admits a decomposition into 3 congruent triangles

$\bullet$ Furthermore, by iterating the subdivisions indicated above, one can also obtain such subdivisions for any $N$ of the form $3^k\cdot m^2$

Question 1: Are the values of $N$ listed above the only ones possible if one requires the subtriangles be similar to the original triangle?

Question 2: In each of the examples above, the subtriangles formed are always similar to the original triangle. Does the answer to Question 1 change if one no longer requires that subtriangles be similar to the original triangle. (For example, if one does not require that the subtriangles are congruent to the original triangle, then an equilateral triangle may be subdivided into 3 congruent copies of a 30-30-120 triangle)

share|improve this question
    
For question 2, you could take two copies of a right triangle divided into $N$ congruent right triangles and combine them to form an isosceles triangle divided into $2N$ congruent right triangles. –  Robert Israel Jun 8 '11 at 2:42
add comment

2 Answers

up vote 8 down vote accepted

See http://www.michaelbeeson.com/research/papers/SevenTriangles.pdf and http://www.michaelbeeson.com/research/papers/TriangleTiling.pdf

Beeson conjectures that the cases you list, together with $2n^2$ (a special case of $m^2+n^2$ and $6n^2$ (barycentrically subdivide an equilateral triangle and then split each of the resulting 30-60-90 right triangles into $n^2$ similar triangles) are the only ones possible, without the constraint that the original triangle and the smaller ones be similar. He proves that no subdivision is possible for $n=7$, $11$, $19$, and $23$.

share|improve this answer
add comment

Regarding the first question, yes, those are the only options. Proof can be found in:

S. L. Snover, C. Waiveris, and J. K. Williams. Rep-tiling for triangles. Discrete Math., 91(2):193–200, 1991.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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