[Math] Finding an expression for the diameter of a circle.

Question

The figure shows two circles of radius $1$ that touch at $P$. 1]1 $T$ is a common tangent line; $C_1$ is the circle that touches $C$, $D$, and $T$; $C_2$ is the circle that touches $C$, $D$, and $C_1$; $C_3$ is the circle that touches $C$, $D$, and $C_2$. This procedure can be continued indefinitely and produces an infinite sequence of circles $\{C_n\}$. Find an expression for the diameter of $C_n$.

Answer 1

I'll work with the radii instead of the diameters. Obviously, any diameter is twice the corresponding radius. Now, note from the figure below that the triangle $OAB$ is rectangular at $A$.

Thus, $$ \overline{OA}^{\,2} + \overline{AB}^{\,2} = \overline{OB}^{\,2} $$

So, $$ (\underbrace{1}_{\overline{OA}})^2 + (\underbrace{1 - R_1}_{\overline{AB}})^2 = (\underbrace{1 + R_1}_{\overline{OB}})^2 $$ which gives you

$$ R_1 = \frac{1}{4} $$

Similarly, by making $B$ be the centre of the ever-smaller circles, you should be able to see that $$ (\underbrace{1}_{\overline{OA}})^2 + (\underbrace{1 - 2R_1 - 2R_2 - \cdots - 2R_{n-1} - R_n}_{\overline{AB}})^2 = (\underbrace{1 + R_n}_{\overline{OB}})^2 \qquad(1) $$ for $n > 1$. Now define

$$ S_{n} \equiv \sum_{k\,=\,1}^{n}R_k \qquad (n \ge 1) \qquad(2) $$

so that $(1)$ can be written as $$ 1 + (1 - 2S_{n-1} - R_n)^2 = (1 + R_n)^2 \qquad(3) $$ for $n > 1$. However, $$ (1 - 2S_{n-1} - R_n)^2 = 1 + 4S_{n-1}^2 + R_n^2 - 4S_{n-1} - 2R_n + 4S_{n-1}\,R_n $$ and $(3)$ gives $$ 1 + 1 + 4S_{n-1}^2 + R_n^2 - 4S_{n-1} - 2R_n + 4S_{n-1}\,R_n = 1 + 2R_n + R_n^2 $$ which can be simplified to $$ (1 - 2S_{n-1})^2 = 4(1 - S_{n-1})R_n $$ so

$$ R_n = \frac{1}{4}\frac{(1 - 2S_{n-1})^2}{(1 - S_{n-1})} \qquad (n > 1) \qquad (4) $$

This gives $R_n$ in terms of the radii of only the circles with indices smaller than $n$ so it's very amenable to a recursive calculation. For example, we already have $R_1 = 1/4$. Then $S_1 = 1/4$ and $R_2 = 1/12$. Then $S_2 = 1/3$ and $R_3 = 1/24$. Then $S_3 = 3/8$ and $R_4 = 1/40$, and so on.

Computing a few more values in succession, it's easy to discern the pattern:

$$ S_n = \frac{n}{2(n+1)} \qquad\mbox{and}\qquad R_n = \frac{1}{2n(n+1)} \qquad (n \ge 1) $$

That these satisfy $(2)$ and $(4)$ can be easily proven by induction on $n$.