# Solution to Problem 21 submitted by Amit Sahai and Yevgeniy Dodys

## Problem 21

Four circles on a plane are such that each one is tangent to the three others. The centers of three of them lie on a line. The distance from the center of the fourth one to this line is x. Find x, if the radius of the fourth circle is r.

## Solution

Assume the radius of the large circle is 1, and those of the two smaller circles are a and (1-a), while the fourth circle has radius r. By Soddy's formula (which, incidentally, has a very amusing history),

1/a2 + 1/(1-a)2 + 1/r2 + 1 = 0.5 * (1/a + 1/(1-a) + 1/r - 1)2.

Setting

z = a(1-a),

we get

1/a2 + 1/(1-a)2 = (1-2z)/z2, and 1/a + 1/(1-a) = 1/z,

so

2(1 + 1/r2 + (1-2z)/z2) = (1/r + 1/z -1)2.

Solving, we get z = r/(1+r). Finally, look at the triangle whose vertices are the 3 centers of the circles (excluding the outer circle of radius 1). Its 3 sides are 1, a+r, (1-a+r), while the height towards the side which is 1 is exactly the needed x. Now, on the one hand, the area of this triangle is A=x*1/2 = x/2 (height times base over 2). On the other hand, from Heron's formula, the semiperimeter p=1+r, so

A2 = p(p-1)(p-(a+r))(p-(1-a+r)) = (1+r)r(1-a)a = (1+r)r * z = r2.

Here we used z=r/(1+r). Thus, A2 = x2/4 = r2, so x=2r.

Last revised August 2003