Actually, we

*could*do this question using Descartes' circle theorem, but that's probably a bit overkill. Instead, let's tackle it with basic primary school geometry, and a bit of

*induction*and

*telescoping*! Consider the following figure:

*Problem*: order the radii of all the little yellow circles up from the bottom of the center as a sequence \(\{r_n\}\) and show that \begin{align}r_n=\frac{1}{2n(n+1)}.\end{align} When doing proofs for general results like this, it can often really really help to work out a few baby cases. So, let's start by showing that \(r_1=\frac{1}{4}\).

As you can see from the diagram below, we've got a (vertical) trapezium with two right angles opposite each other and with the parallel sides of lengths \(1\) and \(r_1\). Thus, by chopping off a rectangle at the bottom of this trapezium, we're left with a right-angled triangle with sides of lengths \(1-r_1,1\) and \(1+r_1\).

And so, Pythagoras' theorem tells us that \begin{align} (1-r_1)^2+1^2&=(1+r_1)^2 \\ \Rightarrow 1&=4r_1.\end{align} Oookay, that was fair straight forward. Let's try this again for \(r_2\):

So, we see that by repeat this trapezium trick to calculate \(r_2\) we get: \begin{align} (1-2r_1-r_2)^2+1^2&=(1+r_2)^2\\ \Rightarrow 1&=(1+r_2)^2-(1-2r_1-r_2)^2\\ &=4(r_1+r_2)(1-r_1).\end{align} Substituting in \(r_1=\frac{1}{4}\), we show that \(r_2=\frac{1}{12}\). Now, if we keep going with this trapezium trick (*), we see that to compute \(r_{n+1}\), we get: \begin{align} (1-2(r_1+\ldots+r_n)-r_{n+1})^2+1^2&=(1+r_{n+1})^2\\ \Rightarrow 1&=(1+r_{n+1})^2-(1-2(r_1+\ldots+r_n)-r_{n+1})^2\\ &=4(r_1+\ldots+r_{n+1})(1-r_1-\ldots-r_{n}).\end{align} Thus, \begin{align} r_{n+1}=\frac{1}{4-4(r_1+\ldots+r_n)}-(r_1+\ldots+r_n).\end{align} So, we've now got this recursive expression for \(r_{n+1}\) and recursion should really make us think:

*induction*.

We're going to use a slightly stronger form (**) of mathematical induction than what you're used to seeing. Don't worry - the logic is the same. =) Okay, here goes...

Let's start off with a proposition: let the proposition \(p(n)\) be the statement that \(r_n=\frac{1}{2n(n+1)}\), and we're going to prove these statements for every \(n\in\mathbb{N}\).

Base case: \(n=1\)... well, we've actually already proven this above. We showed that \(r_1=\frac{1}{4}=\frac{1}{2\times1\times 2}\).

(Strong) induction step: now, we're going to do something a little different to showing that "\(p(k)\Rightarrow p(k+1)\)". We're actually going to show that \begin{align}p(1),p(2),\ldots, p(k)\text{ all being true }\Rightarrow p(k+1)\text{ is true.}\end{align} Take a moment to think about that logic, and I hope that you see that it's the same type of logic as you usually use in induction, except that it

*feels*stronger because you're assuming more things for the induction step.

Okay....

*breathe out*.

So, let's assume that \(p(1),\ldots,p(k)\) are all true. Now, we showed before that \begin{align} r_{k+1}=\frac{1}{4-4(r_1+\ldots+r_k)}-(r_1+\ldots+r_k).\end{align} And now, invoking all of our assumptions, we can try to work out what \(r_1+\ldots+r_k\) is equal to, and hence what \(r_{k+1}\) is equal to: \begin{align} r_1+\ldots+r_k&=\frac{1}{2}\left(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\ldots+\frac{1}{k\times (k+1)}\right)\\ &=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{k}+\frac{1}{k}-\frac{1}{k+1}\right)\\ &=\frac{k}{2(k+1)}.\end{align} W000t! Telescoping! And substituting this in, we get that: \begin{align}r_{k+1}&=\frac{1}{4-\frac{2k}{k+1}}-\frac{k}{2(k+1)}\\ &=\frac{k+1}{2(k+2)}-\frac{k}{2(k+1)}\\&=\frac{(k+1)^2-k(k+2)}{2(k+1)(k+2)}=\frac{1}{2(k+1)(k+2)},\end{align} which is precisely what we want. I.E.: the proposition \(p(k+1)\) is true!

Therefore, by the principle of (strong) mathematical induction, the propositions \(p(n)\) are true for all \(n\in\mathbb{N}\).

=)

*: Bah, I didn't feel like drawing the diagram.

**: strictly speaking, it's not stronger - it's the same strength! It's just that it

*seems*stronger.

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