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Given a regular pentagram whose outer vertices lie on a circle of radius 1, a circle interior to and sharing a center with the larger circle will intersect the pentagram in ten places, save for two radii where it is 5, when it intersects the vertices of the inner pentagon and when it inscribes the inner pentagon, and 0 places when it is smaller than the inner pentagon. There are therefore two radii, one in each region of ten intersection points, where the circle is exactly divided into ten equal arcs. What are those two radii?

Beyond knowing that the length of said arc is $\pi r/5$, and that we can find the chord length of an arc of that length to get the radii (using a method that doesn't itself need the radii), I am unsure how to approach the the problem.

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2 Answers 2

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Drop a perpendicular from the centre $O$ to the midpoint $P$ of one pentagram edge, then $OP=\sin\pi/10=\cos2\pi/5$.

At the smaller special radius the aforementioned pentagram edge is a chord subtending $1/10$ of a circle, or $\pi/5$. Let $Q$ be one endpoint of this chord, then $\angle POQ=\pi/10$, $\angle QPO$ is right and the radius $OQ$ is $$\frac{\cos2\pi/5}{\cos\pi/10}=\cot2\pi/5$$ At the larger special radius the chord subtends $3/10$ of a circle or $3\pi/5$ and by a completely analogous calculation the radius is $$\frac{\cos2\pi/5}{\cos3\pi/10}$$

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  • $\begingroup$ Thanks for the help. For those curious, the outer special radius has a numerical value of 0.52573 and the inner radius 0.32492. $\endgroup$ Commented Apr 2 at 7:12
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    $\begingroup$ We can render these as radicals using the standard forms for the cosines. This gives $\sqrt{(5−2\sqrt5)/5}$ for the smaller radius and $\sqrt{(5-\sqrt5)/10}$ for the larger. The ratio of the larger radius to the smaller is $\phi=(1+\sqrt5)/2$. . $\endgroup$ Commented Apr 2 at 22:03
  • $\begingroup$ Note that the result may be simplified to $r=\tan\left(\frac{\pi}{10}\right)$. $\endgroup$ Commented Apr 4 at 18:57
  • $\begingroup$ @JohnWaylandBales Which radius are you saying simplifies to that form, the inner or outer? $\endgroup$ Commented Apr 4 at 21:24
  • $\begingroup$ @AnthonyKhodanian Sorry, I meant the inner, smaller radius. $\cot\left(\frac{2\pi}{5}\right)=\tan\left(\frac{\pi}{10}\right)$ $\endgroup$ Commented Apr 4 at 22:14
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Although the answer which has already been accepted is an excellent answer I thought I would give an alternate answer.

In the highlighted triangles in the diagram we have $|AC|=1, r_1=|AB|=\tan(\pi/10), r_2=|AD|, \angle BAD=\angle ABD=2\pi/5$

Then by Law of Sines,

$$ \frac{r_2}{\sin(2\pi/5)}=\frac{r_1}{\sin(\frac{\pi}{5})}$$

So $$ r_2=\frac{\sin(2\pi/5)}{\sin(\pi/5)}r_1=2\cos\left(\frac{\pi}{5}\right)r_1$$

And

$$ \frac{|AB|}{|AC|} =\frac{r_1}{1}=r_1=\tan\left(\pi/10\right)$$

summarizing, we have

$$ r_1=\tan\left(\frac{\pi}{10}\right)\quad r_2=2\cos\left(\frac{\pi}{5}\right)\tan\left(\frac{\pi}{10}\right) $$

Pentagram diagram

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