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Consider a regular pentagon of side length $a$. If you form a 5-sided star using the vertices of the pentagon, then you'll get a pentagon inside that star. What is the side length of that pentagon?

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In general, for a n-sided star, what is the side length of the n-sided regular polygon in the star? Take the distance between two adjacent vertices be $a$

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Let $x$ be the length of small n-sided regular polygon in the star & $a$ be the distance between two adjacent vertices, then the angle of spike of regular star polygon is given as $$\alpha=\frac{\pi}{\text{number of vertices (points) in the star}}=\frac{\pi}{n}$$

Now, draw a perpendicular from one vertex of star to the side of the small regular polygon to obtain a right triangle .

Using geometry of right triangle, the length of perpendicular drawn to the side of small regular polygon can be obtained $$=\frac{a}{2}\csc\frac{\pi}{n}-\frac{x}{2}\cot\frac{\pi}{n}$$ Hence, in right triangle, one should have $$\tan\frac{\pi}{2n}=\frac{\frac{x}{2}}{\frac{a}{2}\csc\frac{\pi}{n}-\frac{x}{2}\cot\frac{\pi}{n}}$$ $$x=\frac{a\tan\frac{\pi}{2n}\csc\frac{\pi}{n}}{1+\tan\frac{\pi}{2n}\cot\frac{\pi}{n}}$$ $$x=\frac{a\sin\frac{\pi}{2n}}{\sin\frac{\pi}{n}\cos\frac{\pi}{2n}+\cos\frac{\pi}{n}\sin\frac{\pi}{2n}}$$ $$x=\frac{a\sin\frac{\pi}{2n}}{\sin\left(\frac{\pi}{n}+\frac{\pi}{2n}\right)}$$ $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=\frac{a\sin\frac{\pi}{2n}}{\sin\frac{3\pi}{2n}}}}$$

$$\forall \ \ n=2k+1\ \ (k\in N)$$ Hence for a regular pentagon in the star, setting $n=5$, the side of regular pentagon $$x=\frac{a\sin\frac{\pi}{10}}{\sin\frac{3\pi}{10}}=a\frac{\sin18^\circ}{\cos36^\circ}=a\frac{\frac{\sqrt 5-1}{4}}{\frac{\sqrt 5+1}{4}}=\color{red}{\frac{a}{2}(3-\sqrt 5)}$$

Edited details: If the star regular polygon has $2n$ no. of vertices which is obtained by placing two congruent $n$-sided regular polygons one on the other in symmetrical staggered manner similar to a hexagram then a generalized formula for calculating side $x$ of $2n$-sided regular polygon in the star (having $2n$ no. of vertices & $a$ is the distance between two adjacent vertices) can be derived as follows (see figure below)

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The angle of spike of regular star polygon is given as $$\alpha=\text{interior angle of a n-sided regular polygon}=\frac{(n-2)\pi}{n}$$

Now, draw a perpendicular from one vertex of star to the side of small regular polygon to obtain a right triangle.

Using geometry of right triangle, the length of perpendicular drawn to the side of small regular polygon can be obtained $$=\frac{a}{2}\csc\frac{\pi}{2n}-\frac{x}{2}\cot\frac{\pi}{2n}$$ Hence, in right triangle, one should have $$\tan\frac{(n-2)\pi}{2n}=\frac{\frac{x}{2}}{\frac{a}{2}\csc\frac{\pi}{2n}-\frac{x}{2}\cot\frac{\pi}{2n}}$$ $$\cot\frac{\pi}{n}=\frac{x}{a\csc\frac{\pi}{2n}-x\cot\frac{\pi}{2n}}$$

$$x=\frac{a\csc\frac{\pi}{2n}\cot\frac{\pi}{n}}{1+\cot\frac{\pi}{n}\cot\frac{\pi}{2n}}$$

$$x=\frac{a\cos\frac{\pi}{2n}}{\cos\frac{\pi}{n}\cos\frac{\pi}{2n}+\sin\frac{\pi}{n}\sin\frac{\pi}{2n}}$$ $$x=\frac{a\cos\frac{\pi}{n}}{\cos\left(\frac{\pi}{n}-\frac{\pi}{2n}\right)}$$ $$\bbox[5px, border:2px solid #C0A000]{\color{blue}{x=\frac{a\cos\frac{\pi}{n}}{\cos\frac{\pi}{2n}}}}$$

$$\forall \ \ \ \ n\ge 3\ \ (n\in N)$$ Hence for a regular hexagon in the hexagram (see in the above diagram) , setting $2n=6$ or $n=3$, the side of regular hexagon $$x=\frac{a\cos\frac{\pi}{3}}{\cos\frac{\pi}{6}}=a\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}=\color{red}{\frac{a}{\sqrt3}}$$

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  • $\begingroup$ and when $n$ is even? $\endgroup$
    – B2VSi
    Dec 8 '15 at 13:52
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    $\begingroup$ Good question, for $n$ even integer the star regular polygon doesn't exist i.e. Star polygon with even number of sides or vertices can't be constructed. $\endgroup$ Dec 8 '15 at 13:54
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    $\begingroup$ Nicely done... +1 $\endgroup$
    – Shailesh
    Dec 8 '15 at 14:12
  • $\begingroup$ hmm... I was thinking when $n$ is an integer it could look something like this en.wikipedia.org/wiki/Hexagram $\endgroup$
    – B2VSi
    Dec 8 '15 at 14:22
  • $\begingroup$ @numberphile: Good, the star polygon like hexagram is obtained by placing two congruent regular polygons one on other in symmetric staggered form. But the star polygon obtained this way is different from what you asked in your question. It has odd no. of sides or vertices. Although for even no. of vertices in a star, a similar general formula can be obtained. I can post that formula too if you ask for $\endgroup$ Dec 8 '15 at 14:40

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