# Maximum number of circles tangent to two concentric circles

In a recent math contest, the following question arose:

Two concentric circles of radii 1 and 9 make a ring. In the interior of this ring $n$ circles are drawn without overlapping, each being tangent to both of the circles of the ring. What is the largest possible value for $n$?

I solved it like this:

The radius of each of the small circles must be $(9-1)/2=4$. I connected the centres of two of the small circles together, and also to the centre of the large circle. Let the central angle be $\theta$. The triangle formed has side lengths $4+1=5$, $4+1=5$, and $4+4=8$. This can be split in two to get two 3-4-5 triangles. Since these triangles are right, we can solve for $\theta$:

\begin{align} \sin\frac{\theta}{2}&=\frac{4}{5}\\ \frac{\theta}{2}&=\arcsin{\frac{4}{5}}\\ \theta&=2\arcsin\frac{4}{5}. \end{align}

Now the answer to the question is simply $\left\lfloor\frac{2\pi}{\theta}\right\rfloor=3$.

However, this contest was a no calculator contest, and thus I was not able to compute $\arcsin(4/5)$ for the answer. How do you solve this question without a calculator?

## 3 Answers

You don't need to compute $\arcsin \frac45$ exactly, because you're going to be taking the floor anyway.

You just need to know that the angle opposite the side of length $4$ in a $(3,4,5)$ right triangle is somewhere between $45^\circ$ (the threshold between $3$ and $4$ circles) and $60^\circ$ (the threshold between $2$ and $3$ circles).

You can figure this out by comparing the sides $(3,4,5)$ to the sides $(s,s,\sqrt2 s)$ and $(s, \frac{\sqrt3}{2}s, 2s)$ that we'd see in triangles with a $45^\circ$ and $60^\circ$ degree angle, respectively.

• Wow that was fast! It's noticing these things that makes me bad at geometry. – ericw31415 Apr 24 '18 at 2:08

Depending on what is meant by non-overlaping circles, there is another possible way to get an answer of $3$. • This would be a particularly evil question if the radii $1$ and $9$ were chosen so that only $2$ circles could be drawn in the "usual" way. (I think $1$ and $15$ would do the trick.) – Misha Lavrov Apr 24 '18 at 2:16
• @MishaLavrov Yes, $1$ and $15$ would do it. – John Wayland Bales Apr 24 '18 at 2:29
• I'm not sure how you could argue that the two circles on top aren't overlapping. The circles themselves have non-empty intersection, as do their interiors. – Carmeister Apr 24 '18 at 7:13
• @Carmeister The two circles on top are tangent. One could interpret "overlapping" as meaning "having two points in common". – Servaes Apr 24 '18 at 8:45

Note that $\displaystyle \frac{\sqrt{2}}{2}<\frac{4}{5}<\frac{\sqrt{3}}{2}$. We have $\displaystyle \sin45^\circ<\sin\frac{\theta}{2}<\sin60^\circ$.

$90^\circ<\theta<120^\circ$.

Hence $\displaystyle 4>\frac{360^\circ}{\theta}>3$.