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?
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.