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Solution to Problem 258
Correct solutions came from Sunil Sangwal; Philippe Fondanaiche, France; Dan Dima, Romania; Al Zimmermann, USA; Bill Webb, USA; Cee Franklin, USA; David Stigant, USA; Farid Lian, Colombia; Juan Carlos Marivela, Spain; Ron Welch, USA; John Snyder, USA; Jerome Cherry, Canada; Lou Cairoli, USA.
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The centers of the spheres sit at the
vertices of a tetrahedron, as shown on the left; the center of the central
sphere is at X, lying above E on the face ABC of the
tetrahedron. As each sphere has radius one, each tetrahedral vertex
is two units from each of the others; Pythagoras applied to triangle AFB
gives AF = √3. The area of triangle ABC is therefore (BC)(AF)/2
= √3, which is three times the area of triangle BEC;
therefore √3 = (EF)(BC)/2 and EF =
1/√3. From triangle DEF we get DF2 = ED2
+ EF2, 22 = ED2 +
(1/√3)2, and ED = 2√6/3. Finally, DX
= DE - XE = √(BF2
+ EF2 + EX2), which, after
substituting and simplifying, yields EX = 1/√6.
Therefore DX = BX = AX = CX =
3/√6. The radius of the small sphere in the middle is
3/√6 - 1, or approximately .2247.
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