Problem: Circles with centers A,B and C each have radius r, where 1<r<2. The distance between each pair of centers is 2 . If B′ is the point of intersection of circle A and circle C which is outside circle B, and if C′ is the point of intersection of circle A and circle B which is outside circle C, then length B′C′ equals
Answer Choices:
A. 3r−2
B. r2
C. r+3(r−1)
D. 1+3(r2−1)
E. none of these
Solution:
Let D and E be the points of intersection of B′C′ with AB and AC, respectively, and let B′F be the perpendicular drawn from B′ to AC. Then ED∣∣CB by symmetry, which implies ∠AED, and hence ∠B′EF. is 60∘.
Applying the Pythagorcan theorem to △B′FC yields the equality B′F=r2−1. Now B′C′=B′E+ED+DC′=2(B′E)+ED=2(B′E)+EA.
Therefore.
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