# What is the smallest regular $n+1$-gon that contains a regular $n$-gon with unit length sides?

Given a regular $$n$$-gon with sides of unit length, what is the side length of the smallest regular $$n+1$$-gon containing it?

For $$n=3$$ a bit of calculus yields a square of side length $$\cos\frac{\pi}{24}=\frac{1+\sqrt{3}}{2\sqrt{2}}\approx0.965925826289...$$ For $$n=4$$ a bit more calculus and a few more cases to check yield a regular pentagon of side length $$\frac{\sin\tfrac{7\pi}{40}}{\sin\tfrac{3\pi}{10}}=\frac{\sqrt{5-\sqrt{5}+\sqrt{25-10\sqrt{5}}}}{\sqrt{5}}\approx0.936859701734...$$ What is the minimal side length for $$n=5$$? I hope someone, somewhere has already taken the time to publish a list values for small $$n$$. Any reference is welcome, though I'll be satisfied with any effective method to compute the exact values as well.

• The optimal solution polygons do not have the same center it seems. It is not hard to see what is the optimum when they share the center, and that would be an approximation of the optimum; it beats the approximation with inscribed and circumscribed circles. – orangeskid Mar 2 at 0:37

If a regular n-gon is inscribed in a circle of radius $$r$$,, the central angle of a side is $$2\pi/n$$, so the distance of the side from the center $$r_i(n)$$ satisfies $$r_i(n) = r\cos(\pi/n)$$ and the length of the side is $$s_i =2r\sin(\pi/n)$$.

If a regular n-gon is circumscribed around a circle or radius $$r$$, the central angle of a side is $$2\pi/n$$, so the distance of the corner from the center $$s(n)$$ satisfies $$r_c(n) = r/\cos(\pi/n)$$ and the length of the side is $$s_c =2r\tan(\pi/n)$$.

If the sides of a regular n-gon are of unit length, the distance to the vertices $$d(n)$$ satisfies $$\dfrac{1/2}{d(n)} =\sin(\pi/n)$$ so $$d(n) =\dfrac1{2\sin(\pi/n)}$$.

If this is the distance to the side of a regular m-gon (I will set $$m = n+1$$ later), then, if $$s$$ is the side of the m-gon then $$\dfrac{s/2}{d(n)} =\tan(\pi/m)$$ so $$s =2d(n)\tan(\pi/m) =2\dfrac1{2\sin(\pi/n)}\tan(\pi/m) =\dfrac{\tan(\pi/m)}{\sin(\pi/n)}$$.

If $$m = n+1$$, this is $$\dfrac{\tan(\pi/(n+1))}{\sin(\pi/n)}$$.

This, of course, does not take into account the possibility that the outer m-gon might be able to be smaller if it is rotated, so this is an upper bound.

If we use the approximation $$\sin(x) \approx x$$, this is about $$\dfrac{n}{m}$$.

If the side of the m-gon is the same of the side of the n-gon, then $$\tan(\pi/m) =\sin(\pi/n)$$ so $$\dfrac{\pi}{m} =\arctan(\sin(\pi/n))$$ or $$m =\dfrac{\pi}{\arctan(\sin(\pi/n))}$$.