Fun Q6: Side length of the pentagon in a five sided star? 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?

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 $a$
 A: 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) 


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 
$$x=\frac{a\cos\frac{\pi}{3}}{\cos\frac{\pi}{6}}=a\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}=\color{red}{\frac{a}{\sqrt3}}$$
