# Special triangles in convex polygons

Given identical 30-60-90 triangles, what is the convex polygon with the highest number of sides that I can build from them?

This seems a very easy task by first look, but I’m totally stuck right now. The only thing I’m pretty sure of is that the polygon can at most have 12 sides, because a 150 degrees angle is the maximum convex angle I can find combining triangles. Obviously this does not guarantee that a dodecagon can be filled with 30-60-90 triangles.

Any suggestion? I was tempted to post it in Math, so if you think it is not pertinent I will shut the question down.

• Did you mean "regular" polygon? It would be trivial to assemble an irregular polygon – Hugh Oct 23 '18 at 22:14
• The regular polygon with the highest number of sides that I can build is an hexagon indeed. I mean a general convex (I specified it in the title but didn’t add it in the text, just edited) polygon. – Francesco Arnaudo Oct 23 '18 at 22:17
• If say six — I can't think of anything better. – Hugh Oct 23 '18 at 23:07
• "equal-sided 30-60-90 triangles" is an impossibility unless you allow non-zero curvature. If you mean "identical 30-60-90 triangles" you should say that. – Nij Oct 24 '18 at 5:56
• At the end of second paragraph, is "decagon" a typo for "dodegacon" ? – Evargalo Oct 24 '18 at 7:02

Here is a convex dodecagon made of $$50$$ of those triangles.

Can it be done with fewer?

• I didn’t even thought that it could be done with such a small number of triangles! Great work! – Francesco Arnaudo Oct 24 '18 at 6:31

It is possible to do better than a hexagon, if an irregular polygon is acceptable.

One can construct an irregular dodecagon with $$180$$ identical 30-60-90 triangles as shown below. The dodecagon would have sides of alternating lengths, with ratio $$4:\sqrt{12}$$; as a specific size for the 30-60-90 triangle is not given.

In the diagram below, the larger white triangles are not unit 30-60-90 triangles — they are actually made up of four, smaller 30-60-90 triangles, which is demonstrated in the top-left (shaded in pink). The image also includes two regular hexagons (of different sizes), a bunch of equilateral triangles, and some rectangles (all constructed from 30-60-90 triangles). It is not possible to get more than 12 sides because the minimum external angle that can be created from 30-60-90 triangles is 210 degrees.

It is also possible to construct an equilateral triangle or a hexagon. On reflection (and thanks to @Hugh's comment) you probably can't make a square — but can get relatively close to a square. By taking $$\sqrt{3}$$ by $$1$$ squares, each made from two triangles, you could stack them $$n$$ by $$\frac{n}{\sqrt{3}}$$ which, for sufficiently large $$n$$, would approach $$1:1$$.

• Well done, @Penguino, I didn't think of making a dodecagon that big! You say that it's possible to make a square, but I don't think that's a perfect square...? Please correct me if I'm wrong — regardless of you using two different sizes of triangle, all four sides contain a small "long leg" and a small "short leg". Other than that, the bottom and top edges contain: 2 big "hypotenusen" and 1 big "long leg". The left and right sides contain: 2 big "long legs" and 1 big "hypotenuse". Which, doesn't add up: $2n + 2n + n\sqrt{3} \neq 2n + n\sqrt{3} + n\sqrt{3}$. – Hugh Oct 24 '18 at 2:51
• Yep - you are right - I sketched to soon :) – Penguino Oct 24 '18 at 2:55
• I've just suggested an edit, it corrects some grammar and adds math formatting. :) – Hugh Oct 24 '18 at 3:21
• Feel free to edit Hugh. – Penguino Oct 24 '18 at 3:26
• Great answer indeed! Did you find a ‘filling algorithm’ while building the figures, or just tried casual combinations? Btw, regarding the square, there are two triangles in the bottom left of the picture that are hypotenuse/long cathetus-sided, which is not allowed :) – Francesco Arnaudo Oct 24 '18 at 6:29

• That is not quite regular. If the hypotenuse of the triangle has length $1$, some sides have of the dodecahedron have length $7$, others have length $4\sqrt{3}\approx6.928$. – Jaap Scherphuis Oct 24 '18 at 3:04