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I am trying to write a code to draw pentagon star and have stumbled upon a problem. In the attached diagram, what is the ratio of the diameter of outer (red) circle to the inner (blue) circle. The circles are concentric. I've read that it is ~2.61 but would like to develop the relationship myself.

enter image description here

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@Robert Israel : Here is an alternate proof, involving only triangle trigonometry.

With notations displayed on the figure below, we have to find the ratio $\dfrac{OA}{OB}.$

  • by reasoning in right triangle $OCA \ $: $ \ \widehat{OAB}=90°-72°=18°.$

  • by reasoning in right triangle $OCB \ $: $ \ \widehat{OBC}=90°-72°/2=54°$, thus $\widehat{ABO}=180°-54°=126°.$

Thus, sine law in triangle $OAB$ gives:

$$\tag{1}\dfrac{OA}{OB}=\dfrac{\sin \widehat{ABO}}{\sin \widehat{OAB}}=\dfrac{\sin 126°}{\sin 18°}=\dfrac{\cos 36°}{\cos 72°}$$ (exact value); thus $\dfrac{OA}{OB}\approx 2.618.$

Knowing that $\cos 36°=\dfrac{1+\sqrt{5}}{4}$, it is not difficult to conclude that the exact value can also be written$1+\Phi$ (where $\Phi:=\dfrac{1+\sqrt{5}}{2}$ is the golden ratio). enter image description here

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  • $\begingroup$ Knowing cos 36 and sin 18 seems esoteric, but you can calculate, say, sin 18 by using the identity cos 3x = sin 2x. cos 3x = 4cos^3 x - 3 cos x can be derived from seeing cos 3x + cos x = cos (2x+x) + cos (2x-x). Or you can use geometry to derive things: mathworld.wolfram.com/GoldenTriangle.html has more information. $\endgroup$
    – aschultz
    Oct 16, 2018 at 19:42
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Hint: everything can be done with analytic geometry.
If the outer radius is $R$, then in a cartesian coordinate system the vertices on the outer circle are at $[0,R], \ldots$. Find the equations of the lines, see where they intersect...

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