Pythagorean circles

Here’s yet another Catriona Shearer problem:

But let’s start with a related picture, one with only two circles:

Again, the circles have unit radius. What are the dimensions of the rectangle?

The key to figuring this out is that if you have two lines through a point both tangent to a circle, then the line segments from the intersection point to the tangent points are congruent. We know the distance from the top right corner of the rectangle to the top tangent point on the yellow circle is 3. We don’t know the distance from the bottom right corner of the rectangle to the left tangent point on the yellow circle; call it x. Then the length of the diagonal is 3+x. By Pythagoras,

$(3+x)^2 = 4^2+(1+x)^2$

whose solution is x=2. So the height of the rectangle is 3, its width is 4, and its diagonal is 5: the diagonal and two sides make a 3:4:5 triangle, where 3:4:5 is a Pythagorean triple.

You can do the 3-circle problem the same way; here the length of the diagonal is 5+x, the equation is

$(5+x)^2 = 6^2+(1+x)^2$

whose solution is x=3/2. The rectangle then is 2.5 by 6 with diagonal 6.5. Those three numbers are in the proportion 5:12:13, another Pythagorean triple!

Coincidence? No. Of course if x is rational then the lengths of the sides and diagonal are in the proportion of a Pythagorean triple. But in general, with n circles, does x have to be rational? You might expect not. However, the solution for n circles is given by

$(2n-1+x)^2 = (2n)^2+(1+x)^2$

whose solution is

$x = \frac{n}{n-1}$

So the rectangle height:width:diagonal is

$1+\frac{n}{n-1}$ : $2n$ : $2n-1+\frac{n}{n-1}$

or

$\frac{2n-1}{n-1}$ : $\frac{2n(n-1)}{n-1}$ : $\frac{2n(n-1)+1}{n-1}$

which is the integer proportion $2n-1$ : $2n(n-1)$ : $2n(n-1)+1$, scaled by $n-1$. That’s a Pythagorean triple, and in particular it’s a triple in which the hypotenuse is one greater than the longer side. For four circles, for example, the triple is 7:24:25 and the rectangle’s dimensions are 7/3 by 8.