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.

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. You are visitor number 1499 to this page.