tag:blogger.com,1999:blog-6772272495978689480.post8311554725907238256..comments2018-04-10T08:31:52.417-04:00Comments on MAA Books: Beautiful MathematicsMathematical Association of Americahttp://www.blogger.com/profile/10559021045290192742noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6772272495978689480.post-63754732039517714622011-12-14T12:54:30.775-05:002011-12-14T12:54:30.775-05:00Cute. I make it $2(1+\sqrt 6)$. Consider the tet...Cute. I make it $2(1+\sqrt 6)$. Consider the tetrahedron whose vertices are the centres of the spheres. It has edge length 2. By a standard formula, it's inradius is therefore $1/\sqrt 6$. The base of the large tetrahedron is distance 1 further from the centroid of everything than the base of the small one, and so the large one is larger by a factor $1+1/\sqrt 6\over 1/\sqrt 6 = 1+\sqrt 6$.TheOtherJimSimonshttps://www.blogger.com/profile/06660176721991238873noreply@blogger.com