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 $a$
Best Answer
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 $$x=\frac{a\sin\frac{\pi}{10}}{\sin\frac{3\pi}{10}}=a\frac{\sin18^\circ}{\cos36^\circ}=a\frac{\frac{\sqrt 5-1}{4}}{\frac{\sqrt 5+1}{4}}=\color{red}{\frac{a}{2}(3-\sqrt 5)}$$
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 $$x=\frac{a\cos\frac{\pi}{3}}{\cos\frac{\pi}{6}}=a\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}=\color{red}{\frac{a}{\sqrt3}}$$