# Thread: Help, Constructing a perfect Hexagram?

1. Hello. I'm trying to construct an object that contains a perfect hexagram (perfect as in all extruding points are equidistant from center), like the Star of David. I need to be mathematically accurate with my dimensions, and I need help.

If inscribed in a rectangle, what would be the generalized ratio of the rectangle's length to its width. For example, an equilateral triangle's excribed rectangle has a ratio of : So what would it be for a hexagram?

2.

3. Start with a regular hexagon, and then connect vertices to form a hexagram.

4. Originally Posted by mathman
Start with a regular hexagon, and then connect vertices to form a hexagram.
Ok. So maybe it would be easier to ask the dimensions of a hexagon's exscribed quadrilateral.

5. The hexagram will consist of 12 equilateral triangles, which includes the six that form the hexagon in the middle and another 6 that make up the points of the star. The rectangle that encloses it will have one side which is equal to 3 of the triangle bases end to end, and the other side will be 4 triangles high. So if the base of each of the 12 triangles is x, the rectangle is 3x by 2x*(square root of 3).

6. Originally Posted by Harold14370
The hexagram will consist of 12 equilateral triangles, which includes the six that form the hexagon in the middle and another 6 that make up the points of the star. The rectangle that encloses it will have one side which is equal to 3 of the triangle bases end to end, and the other side will be 4 triangles high. So if the base of each of the 12 triangles is x, the rectangle is 3x by 2x*(square root of 3).
Great thinking! Thanks!

It turns out I miscalculated the generalized ratio. If an equilateral triangle has sides of a length of , then by the Pythagorean theorem the height is .

So a hexagram is 4 triangle heights by 3 triangle widths. So if a triangle width is 1. Then the hexagram is high, and 3 wide.

So generally, the ratio width:height is (???)

7. Originally Posted by brody
It turns out I miscalculated the generalized ratio. If an equilateral triangle has sides of a length of , then by the Pythagorean theorem the height is .
The equilateral triangle has 3 equal angles of 60 degrees. When you bisect it, you have a 30-60 right triangle, and the ratio of sides is 1, 2, and . So I think you were right in the first place.

8. Originally Posted by Harold14370
Originally Posted by brody
It turns out I miscalculated the generalized ratio. If an equilateral triangle has sides of a length of , then by the Pythagorean theorem the height is .
The equilateral triangle has 3 equal angles of 60 degrees. When you bisect it, you have a 30-60 right triangle, and the ratio of sides is 1, 2, and . So I think you were right in the first place.
Hmm... I'm not sure. Those might be the general ratios, but in terms of the side length , the altitude .

Tri1.jpg

By the Pythagorean theorem, becomes .

So the corresponding ratios are...

...which should be correct. So I was right in both cases. I just showed the second in terms of the triangle's side length.

9. Originally Posted by brody
[
Hmm... I'm not sure. Those might be the general ratios, but in terms of the side length , the altitude .

By the Pythagorean theorem, becomes .
Show your work. is not equal to

10. Originally Posted by Harold14370
Originally Posted by brody
[
Hmm... I'm not sure. Those might be the general ratios, but in terms of the side length , the altitude .

By the Pythagorean theorem, becomes .
Show your work. is not equal to
Oops. You seem to be correct. Such silly mistakes in such simple math.

I accidentally tried simplifying by square root which is obviously not true.

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