Problem:
Circles of radius 3 and 6 are externally tangent to each other and are internally tangent to a circle of radius 9. The circle of radius 9 has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.
Solution:
Let PQ​ be the tangent chord, and let C3​,C6​, and C9​ be the centers of the circles of radii 3,6, and 9, respectively. We see that C6​C9​=3 and C9​C3​=6. Let D3​,D6​, and D9​, respectively, be the feet of the perpendiculars from C3​,C6​, and C9​ to PQ​. Then D3​ and D6​ are points of tangency, and C3​D3​​,C6​D6​​, and C9​D9​​ are parallel. It follows that
C9​D9​=31⋅C3​D3​+2⋅C6​D6​​=5
Now apply the Pythagorean Theorem to right triangle C9​PD9​ to find that
(PQ)2=4(D9​P)2=4[(C9​P)2−(C9​D9​)2]=4(92−52)=224​
The problems on this page are the property of the MAA's American Mathematics Competitions
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