The bigger circle has a radius of $1$, and it is tangent to the two straight lines that forms an angle of $120$ degrees. The smaller circle is tangent to the two straight lines and the big circle. What is the radius of the small circle?
(To the person who casted close vote, geometric puzzles are on topic.)
The answer is:
$$\frac{\frac{2}{\sqrt{3}}-1}{1+\frac{2}{\sqrt{3}}}\approx0.0717$$
The solution may be found by:
Letting the centre of the large circle be $B$ and the centre of the small circle be $A$. Let the radius of the small circle be $a$. Let the $120$ degree angle be the point $C$. Let the distance from the perimeter of the small circle to $C$ be $d$. Let the point where the right tangent meets the large circle be $D$ and the point where the tangent meets the small circle be $E$.
The following image shows this slightly more clearly:
Then we:
Form the right-angled triangle $CBD$. (We know it is right-angled because the tangent to a circle is always perpendicular to the radius at that point - this is also easy to see intuitively). By the fact that $\sin{60}=\frac{\sqrt{3}}{2}$ we have:
$$\frac{\sqrt{3}}{2}=\frac{1}{1+2a+d}$$
Then form the right-angled triangle $CAE$; we similarly obtain:
$$\frac{\sqrt{3}}{2}=\frac{a}{a+d}$$
We then have two simultaneous equations to solve.
Specifically, we have:
$$1+2a+d=\frac{2}{\sqrt{3}}$$ $$a+d=\frac{2a}{\sqrt{3}}$$ $$\therefore~~~~d=a\left(\frac{2}{\sqrt{3}}-1\right)$$ $$\therefore~~~~1+a+\frac{2a}{\sqrt{3}}=\frac{2}{\sqrt{3}}$$ $$\therefore~~~~a=\frac{\frac{2}{\sqrt{3}}-1}{1+\frac{2}{\sqrt{3}}}$$
Which is what we wanted to show.
