AoPS - Problem

Source: IMO LongList 1967, Poland 2

Tags: geometry, 3D geometry, sphere, tetrahedron, circumcircle, IMO Shortlist, IMO Longlist

orl

16.12.2004 20:57

Prove this proposition: Center the sphere circumscribed around a tetrahedron which coincides with the center of a sphere inscribed in that tetrahedron if and only if the skew edges of the tetrahedron are equal.

Let $ABCD$ be the tetrahedron. $I$, its incenter, is projected on the four planes of the faces in the circumcenters of the faces. Let $O_A,O_B,O_C,O_D$ be the circumcenters of $BCD$ and so on. We clearly have $d(O_D,BC)=d(O_A,BC)\Rightarrow \angle BAC=\angle BDC$. From this equality and the like we find that the sum of the palne angles around each vertex is $\pi$ (call this assertion $(*)$). Now take the faces $BCD,CAD,ABD$ and rotate them around $BC,CA,AB$ until they are on the same plane as $ABC$. It is now easy to see from $(*)$ that $ABC,CBD,ADB,DAC$ are congruent, and the conclusion follows.