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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
radicalp
 + 
1
radicalq
 = 
1
radicalr

Solution

Consider the diagram.


Using the Pythagorean Theorem we get:

(p+q)2 = (pminusq)2 + (x+y)2
p2 + 2pq + q2 = p2 minus 2pq + q2 + (x+y)2
(x+y)2 = 4pq implies x + y = 2radical(pq)     (*)

(p+r)2 = (pminusr)2 + x2
p2 + 2pr + r2 = p2 minus 2pr + r2 + x2
x2 = 4pr implies x = 2radical(pr)

Similarly, y = 2radical(qr).

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

Dividing by 2radical(pqr) gives the desired result,  
1
radicalp
 + 
1
radicalq
 = 
1
radicalr
Problem ID: 175 (May 2004)     Difficulty: 3 Star

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