# Find the distance between the center of an unit circle and an internally tangent circle, which's tangents meet at the outer's center in a 72° angle.

There is a circle c, that has a center O, and a circle d that's internally tangent to it. If there are two tangents of d that meet at O, making a $72°$ angle, what's the shortest distance between the d and O?

I used GeoGebra to make this picture, and I can find an approximate answer (it's approximately $0.26$), but some quick tests with bigger numbers revealed more decimals. The orange line segment's length is the shortest distance I want to find.

• This constructible operation is a near duplication of the cube. $2^{1/3}-1=0.25992 ...$. – Oscar Lanzi Aug 24 '16 at 10:27

Let $m:=\tan(36^\circ)=\sqrt{5-2\sqrt 5}$.

We may suppose that we have $$x^2+y^2=1\tag1$$ $$mx-y=0\tag2$$ $$mx+y=0\tag3$$

We want to find a circle $(x-a)^2+y^2=r^2$ where $0\lt a\lt 1, 0\lt r\lt 1$ which touches each of $(1),(2),(3)$.

First, $r=1-a$.

Second, since the distance between $(a,0)$ and $(2)$ is $r$, we have $$r=\frac{|am-0|}{\sqrt{m^2+1^2}}$$

From these, we get

$$a=\frac{4}{4+\sqrt{10-2\sqrt 5}},\qquad r=1-\frac{4}{4+\sqrt{10-2\sqrt 5}}$$

Therefore, the distance we want is $$a-r=\color{red}{\frac{8}{4+\sqrt{10-2\sqrt 5}}-1\approx 0.25961618368249972459552471788}$$

Continue the line that goes through the two centers. The radius of the small circle is $r$. We have $d+r+r=1$ and $\sin(36^\circ)=r/(d+r)$. When you solve this system, the solution is $$d=\frac{1}{1+2\frac{\sin(36^\circ)}{1-\sin(36^\circ)}}\approx0.2596$$