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enter image description here 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.)

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    $\begingroup$ I suspect the close vote came because it probably walks the line of what would be considered a puzzle here (e.g. a mathematical problem with an interesting approach or insight involved in the solution). It could be argued that the solution is a simple application of trigonometry (the method of solving was fairly obvious from first sight, although I did trip up and post the wrong answer initially). But one could argue the 'trick' of erpbtavmvat gur fnzr nccebnpu pna or hfrq sbe rnpu pvepyr gb nyybj bar gb fbyir njnl gur qvfgnapr $q$ zvtug znxr vg fhvgnoyr sbe guvf fvgr. But nice puzzle! $\endgroup$ – Anon Jun 2 at 8:08
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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 $E$.

enter image description here

The following image shows this slightly more clearly:

enter image description here

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.

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    $\begingroup$ This is ............ correct! (IDK why my inbox did not show someone answered.) $\endgroup$ – Culver Kwan Jun 2 at 7:45

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