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I don't know if I have my phrasing correct here. I am basically trying to get the radius around the parameter of a regular polygon. From here on I will just use a pentagon for a concrete example.

This problem is most easily seen in polar coordinates but does not need to be as such.

So we have a pentagon inscribed in a circle with radius R = 1. Therefore from the origin, O to any of the corners R = 1. Then the distance from the origin to a midpoint (the distance from O to A) is simply cos(36). Now I just need to know all the other points between the two.

These basics and my question are more easily explained in this diagram: enter image description here

Now I want to rotate all around the pentagon in polar space sweeping theta. What is the distance, P from the origin to the perimeter?

My approach was to look at that important segment from O to A because this forma a nice right angle with the outside edge. Then after finding φ (which is my biggest concern) I can get what that distance is because I made a right triangle with segment OA segment P and the perimeter segment. I have the length of OA, |OA| = cos(36) and I have the angle of the corner at the origin. This is 36-φ. Then with a little trig voila $P = \frac{\cos(36)}{\cos(36-φ)}$.

Something I haven't covered yet is how to get φ. φ is the angle distance from the closest corner where phi should never get over 36 in this example. Right now I am getting phi by,

$$φ = |(\frac{\theta}{72}) - 36|$$ Wolfram Equation

$$P = \frac{\cos(36)}{\cos(36 - φ)}$$

I feel like this is a bit of overkill to try to solve this problem. Is there a better way to go about this problem? I can also list the n-sided version for these equations

$$a = \frac{360}{n}$$

$$b = \frac{360}{2n}$$

$$φ = |(\frac{\theta}{a}) - b|$$

$$P = \frac{\cos(b)}{\cos(b - φ)}$$

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  • $\begingroup$ Welcome to MSE: Please use MathJax. $\endgroup$ – José Carlos Santos Aug 9 '17 at 20:26
  • $\begingroup$ You can take $0^\circ \le \phi < 72^\circ$ , and reverse its direction to start from min. $\theta$. So you can put $\phi = \theta mod 72^\circ$. Then $\cos(\phi-36^\circ)= \cos(36^\circ-\phi)$ and what you did is correct. $\endgroup$ – G Cab Aug 9 '17 at 21:12
  • $\begingroup$ Thank you. This approach is a little bit simpler. This addresses my concern with simplifying $\phi$. I was having trouble with that negative sign that seemed to complicate things. $\endgroup$ – user2716722 Aug 10 '17 at 12:09
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In your image suppose other vertices for the triangle on the right side are B, C because you have already drawn O vertex there. Now some mathematical calculations.

$AB=\sin(36)$
$OB=r$
$PA=P\cos(36-\phi)$
$OA=\cos(36)$

Now apply: $a^2+b^2=c^2$

$OA^2+PA^2=PO^2$
$P^2\cos^2(36-\phi)+\cos^2(36)=P^2$

Solving you get:

$P=\sqrt{\frac{\cos^2(36)}{\sin^2(36-\phi)}}$

Is this the correct solution? Please let me know.

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    $\begingroup$ This approach is correct and enlightening approach to the problem. It however does not simplify the equation for P or address the acquisition of $\phi$. $\endgroup$ – user2716722 Aug 10 '17 at 12:16

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