Pascal's Kissing Circles Triangle

Much like these circles, Pascal apparently came up against Descartes a few times. Happily, they can help each other here; we first need Descartes' circle theorem , which relates the radii of mutually tangent circles, specifically for the case when one of the circles is a straight line. Given two circles of radius $a$ and $b$ , externally tangent to each other and to a straight line, the circle theorem states that the radius $c$ of a third smaller circle tangent to the circles and line is given by $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}+\frac{2}{\sqrt{ab}}$

For example, the " $r$ " in the question can be found by $\frac{1}{r}=\frac{1}{1}+\frac{1}{1}+\frac{2}{\sqrt{1\times 1}}=4$ so that $r=\frac14$ .

Borrowing notation from Pascal's triangle, let $R(n,k)$ denote the $k^\text{th}$ entry of the $n^\text{th}$ row of the triangle (counting both from zero). We then have $R(n,0)=R(n,n)=1\;\;\text{and}\;\;\frac{1}{R(n,k)}=\frac{1}{R(n-1,k-1)}+\frac{1}{R(n-1,k)}+\frac{2}{\sqrt{R(n-1,k-1)R(n-1,k)}}$ for $0<k<n$ .

Computing a few of these, we find the triangle starts as $\begin{matrix} & & & & 1 & & & & \\ & & & 1 & & 1 & & & \\ & & 1 & & \frac{1}{2^2} & & 1 & & \\ & 1 & & \frac{1}{3^2} & & \frac{1}{3^2} & & 1 & \\ 1 & & \frac{1}{4^2} & & \frac{1}{6^2} & & \frac{1}{4^2} & & 1 \end{matrix}$

which Pascal would definitely recognise. This immediately leads to the conjecture that $\frac{1}{R(n,k)}=\binom{n}{k} ^2$

It's easy to see that this works at the ends of the rows, and (as above) for the first few rows. We need to show that $\frac{1}{R(n,k)}=\frac{1}{R(n-1,k-1)}+\frac{1}{R(n-1,k)}+\frac{2}{\sqrt{R(n-1,k-1)R(n-1,k)}}$ still holds.

Substituting in, the right-hand side is $\begin{aligned} \frac{1}{R(n-1,k-1)}+\frac{1}{R(n-1,k)}+\frac{2}{\sqrt{R(n-1,k-1)R(n-1,k)}} &= \binom{n-1}{k-1} ^2+\binom{n-1}{k} ^2+2\binom{n-1}{k-1} \binom{n-1}{k} \\ &=\left(\binom{n-1}{k-1} +\binom{n-1}{k} \right) ^2 \\ &=\binom{n}{k} ^2 \end{aligned}$

which is exactly what we wanted to show.

Finally, to answer the question, $R(10,5)=\frac{1}{\binom{10}{5} ^2}=\frac{1}{252^2}=\frac{1}{63504}$

so the answer is $\boxed{63505}$ .

From this problem , it is learned that the relationship of the radii of three circles each tangents to the other two and to the base line as shown above is given by (see proof below):

$\frac 1{\sqrt{r_3}} = \frac 1{\sqrt{r_1}} + \frac 1{\sqrt{r_2}} \implies \sqrt{r_3} = \frac {\sqrt{r_1r_2}}{\sqrt{r_1}+\sqrt{r_2}}$

Consider the Pascal's triangle of square root of radius $\sqrt{r_{nk}}$ , where $n$ and $k$ are the row and column numbers respectively:

$\begin{array} {rccccccccc} \rm row\ 0: & & & & & 1 & & & \\ \rm row\ 1: & & & & 1 & & 1 & & \\ \rm row\ 2: & & & 1 & & \frac 12 & & 1 & \\ \rm row\ 3: & & 1 & & \frac 13 & & \frac 13 & & 1 \\ \rm row\ 4: & 1 & & \frac 14 & & \frac 16 & & \frac 14 & & 1 \end{array}$

It is obvious from the above that $\sqrt{r_{nk}} = \dfrac 1{\binom nk} \implies r_{nk} = \dfrac 1{\binom nk^2}$ . Therefore $r_{10,5} = \dfrac 1{\binom {10}5^2} = \dfrac 1{252^2} = \dfrac 1{63504}$ $\implies a+b = \boxed{63505}$ .

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Proof: Note that the horizontal distance between the center of the blue circle and the center of the green circle or the length of the full blue line is $\ell_\blue{\rm blue} = \sqrt{(r_1+r_2)^2 - (r_1-r_2)^2} = 2\sqrt{r_1r_2}$ . Similarly, the lengths of the red line and green line are $\ell_\red{\rm red} = 2\sqrt{r_3r_1}$ and $\ell_\green{\rm green} = 2\sqrt{r_2r_3}$ respectively. Now we have:

$\begin{aligned} \ell_\blue{\rm blue} & = \ell_\red{\rm red} + \ell_\green{\rm green} \\ \sqrt{r_1r_2} & = \sqrt{r_3r_1} + \sqrt{r_2r_3} & \small \blue{\text{Divide both sides by }\sqrt{r_1r_2r_3}} \\ \frac 1{\sqrt{r_3}} & = \frac 1{\sqrt{r_2}} + \frac 1{\sqrt{r_1}} \end{aligned}$

Link for the python code

With the relationship between Ford's circles: $1/ \sqrt{r_{middle}} = 1/ \sqrt{r_{left}}+1/ \sqrt{r_{right}}$ . We just evaluate the first rows of the triangle, and then, by hard induction we come to the conclusion that every value in the pascal triangle is replaced by the multiplicative inverse of it´s square.