# Maximizing the size of n similar equilateral triangles from a rectangle

I have 3 rectangles of greenhouse sheeting material. They are each 12 feet long and 4 feet high.

I want to use this material to clad a 3/4 icosahedron dome structure. What that means is that I will need to create 15 equilateral triangles out of the sheets (A full icosahedron has 20 facets). I want the triangles to be as large as possible, and thus the length of the 2x4 struts making up the sides of the triangles to be as long as possible.

So my task is to find the maximum strut length.

For simplifying my problem I wanted to consider the 3 sheets as a unitary sheet of 144 square feet (12 * 4 * 3), even though that might not work in the final analysis.

My reasoning process is as follows, but my answer doesn't work. What am I doing wrong?

1) Divide 144 square feet by 15 = 9.6 square feet per triangle.

2) Plug in 9.6 to into the formula for finding the area of an equilateral triangle, and solve for x:

$$9.6 = \frac{x^2\sqrt{3} }{4}$$

Solving, I get a max strut of 4.7 feet. Since 0.7 * 12 = 8.4, the max strut length would be 56.4 inches.

3) Intuitively it seem to me that every rectangle of the material that has a width of 112.8 inches of material has room for two triangles within it. Since the total length in inches is 432 inches, that means that I can fit 3.8 rectangles, or roughly speaking, 7.6 triangles, quite a bit from the 15 facets I need.

Where did I go wrong? From my algebra classes long ago, it seems like setting up a system of equations might be a better approach?

Edit: To clarify, I think that complementary right triangles can be "stitched" into being considered a single facet of the icosehedron.

• Does each triangle have to be cut as one piece from one of the rectangular sheets, or can you stitch together pieces to form a triangle? May 4, 2019 at 1:07
• Your area calculation ignores the requirement of getting whole triangles out of the square. You will have a number of pieces left over. If you are willing to put small pieces together you can attain the $56.4$ inches. May 4, 2019 at 1:16

Packing problems are generally hard. What you have to do is choose a general configuration, then use geometry to compute the largest triangle that you can get. In this case there is an obvious configuration that respects your three $$4 \times 12$$ sheets.
This gets you five triangles out of each rectangle. The side is $$4$$ feet, limited by the three sides along the bottom edge of the rectangle. You might do a bit better by bending the outer two triangles down, allowing you to lengthen the side. You can do that until point H goes out the top of the rectangle. I think that gains a trivial amount.