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.

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/a^{2} + 1/(1-a)^{2} + 1/r^{2} + 1 =
0.5 * (1/a + 1/(1-a) + 1/r - 1)^{2}.

Setting

z = a(1-a),

we get

1/a^{2} + 1/(1-a)^{2} = (1-2z)/z^{2},
and 1/a + 1/(1-a) = 1/z,

so

2(1 + 1/r^{2} + (1-2z)/z^{2}) = (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

A^{2} = p(p-1)(p-(a+r))(p-(1-a+r)) = (1+r)r(1-a)a = (1+r)r * z =
r^{2}.

Here we used z=r/(1+r). Thus, A^{2} = x^{2}/4 =
r^{2}, so x=2r.

- Back to Tanya Khovanova's welcome page
- Back to Math "coffins" welcome page
- Back to The main problem list
- Problems submitted by others

Last revised August 2003