## Three Circles

#### Problem

Three touching circles have a common tangent.

If the radii of the circles in decreasing order of size are `p`, `q`, and `r`, prove that the following relationship holds:

1 p | + | 1 q | = | 1 r |

#### Solution

Consider the diagram.

Using the Pythagorean Theorem we get:

(`p`+`q`)^{2} = (`p``q`)^{2} + (`x`+`y`)^{2}`p`^{2} + 2`pq` + `q`^{2} = `p`^{2} 2`pq` + `q`^{2} + (`x`+`y`)^{2}

(`x`+`y`)^{2} = 4`pq` `x` + `y` = 2(`pq`) (*)

(`p`+`r`)^{2} = (`p``r`)^{2} + `x`^{2}`p`^{2} + 2`pr` + `r`^{2} = `p`^{2} 2`pr` + `r`^{2} + `x`^{2}`x`^{2} = 4`pr` `x` = 2(`pr`)

Similarly, `y` = 2(`qr`).

From (*), 2(`pr`) + 2(`qr`) = 2(`pq`).

Dividing by 2(pqr) gives the desired result, | 1 p | + | 1 q | = | 1 r |

Problem ID: 175 (May 2004) Difficulty: 3 Star