The tetrahedron SABC has the following property: there exist five spheres, each tangent to the edges SA,SB,SC,BC,CA,AB, or to their extensions. a) Prove that the tetrahedron SABC is regular. b) Prove conversely that for every regular tetrahedron five such spheres exist.
Problem
Source: IMO 1962, Day 2, Problem 7
Tags: geometry, 3D geometry, tetrahedron, sphere, incenter, IMO, IMO 1962
23.04.2014 20:33
a) Let S be the sphere tangent to the edges from the inside, and let SX be the sphere tangent to the edges of the face opposite to vertex X, and the extensions of the edges from X. Consider say S and SS. If we cut these spheres by plane ABC, in both cases we get a circle on plane ABC tangent to the segments AB,BC,CA - this is the incircle of △ABC. Hence S and SS touches the edges AB,BC,CA in the same points. Now cut the two spheres by plane SAB. We get the incircle and S-excircle of △SAB. We have proven that S and SS touch AB in the same point, so the incircle and S-excircle of △SAB touch AB at the same point. This implies that SA=SB. Similarly, we get that any two edges are the same length, therefore the tetrahedron is indeed regular. b) Look at the regular tetrahedron SABC. By symmetry, the center O of the regular tetrahedron SABC is of equal distance d from all its edges, so taking S(O,d) works. Let the incenter of say △SAB be I, the S-excenter of △SAB be J. Consider the homothety with center S that maps I to J. This homothety maps the incircle ω of △SAB to the S-excircle of △SAB; since ω⊂S, we have ω′⊂S′ and therefore S′ also touches edge AB. By rotational symmetry, S′ also touches edges BC and CA. Moreover, because S touched edges SA,SB,SC, S′ will touch the extensions of SA,SB,SC. This proves that SS=S′ is up for the job. By symmetry, so do SA,SB,SC exist.