mathschallenge.net

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 = (pq)² + (x+y)²
p² + 2pq + q² = p² 2pq + q² + (x+y)²
(x+y)² = 4pq x + y = 2(pq)     (*)

(p+r)² = (pr)² + x²
p² + 2pr + r² = p² 2pr + r² + x²
x² = 4pr 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