Problem:
Circles C1β and C2β each have radius 1 , and the distance between their centers is 21β. Circle C3β is the largest circle internally tangent to both C1β and C2β. Circle C4β is internally tangent to both C1β and C2β and externally tangent to C3β. What is the radius of C4β?
Answer Choices:
A. 141β
B. 121β
C. 101β
D. 283β
E. 91β
Solution:
Let r be the radius of C4β. Let the point Pkβ be the center of Ckβ for 1β€kβ€4. Let A be the point of intersection of circles C1β and C4β and let β be the tangent line to P1β at A. By symmetry, P3β is the midpoint of P1βP2ββ, and P1βP3β=41β. Also by symmetry, P4β lies on the perpendicular bisector of P1βP2ββ. The radius P1βAβ is perpendicular to β. Likewise the radius P4βAβ is perpendicular to β, so P1β,P4β, and A all lie on the segment P1βAβ with P4β between P1β and A. Then P1βP4β=1βr.

The radius of C3β is 43β. Then P3βP4β=43β+r, and β³P1βP3βP4β is a right triangle with P1βP3β=41β, P3βP4β=43β+r, and P1βP4β=1βr. By the Pythagorean Theorem,
(41β)2+(43β+r)2=(1βr)2
Expanding gives
161β+169β+23rβ+r2=1β2r+r2
which gives r=(D)283ββ.
The problems on this page are the property of the MAA's American Mathematics Competitions