## Tetrahedrons and spheresGiven an arbitrary tetrahedron, is there a sphere tangent to each of its edges? (See fig.1)
By a tetrahedron we obviously mean an arbitrary pyramid based on a triangle. It is clear that the answer need not always be positive for an arbitrary tetrahedron. But then - for which? How can this be verified in a simple way? There are many simple ways, so let's take a closer look at three of them. For instance, for an arbitrary tetrahedron the following conditions are equivalent: - There is a sphere tangent to each of its edges.
- The circumferences inscribed into its faces are pairwise tangent (see fig.2).
- There are four pairwise tangent spheres with centres in the vertices of the tetrahedron.
- For each pair of skew edges of the tetrahedron the sum of their lengths is the same, i.e. AB+CD = AC+BD = AD+BC (fig.3).
The above criteria seem to be quite convenient and useful, but they have to be proved first, of course. All the implications "downwards", i.e. (1) => (2) => (3) => (4) are rather obvious. It thus remains to prove the missing one: (4) => (1). The proof is based on the following simple fact: if a circumference is inscribed into an angle, then the segments PA and PB are congruent (fig.4). Consequently, if a circumference is inscribed into a triangle ABC, we get three pairs of congruent segments (fig.5).
Observe that the inverse is also true; if on the three sides of a triangle ABC we have points P, Q, R and the points divide the sides of the triangle into pairwise congruent segments (fig.6), then these points are tangential for the circumference inscribed into ABC (fig.5). Indeed, if, for instance, the point R were slightly moved towards A, then Q must move similarly towards A, since the segments AR and AQ are supposed to be congruent. But then the point P would have to be moved - for the same reason - towards B on one hand and towards C on the other, which is clearly impossible.
Now, take an arbitrary triangle ABC and let P,Q and R be points on its sides tangential for the inscribed circumference. Take moreover a triangle ABD such that AD+BC = AC+BD. On the sides AD and BD lay off segments shown on fig.7. Then DQ''=DP'', which implies that the circumference inscribed into ABD is tangent to its sides at the points R, Q' and P'. Hence we infer that AB is a straight line perpendicular to the straight line FG (fig.8). This property will be preserved even if we turn the triangle ABD around the straight line AB and the triangles ABC and ABD land up in different planes. But then the straight lines perpendicular to these two planes and passing through the points F and G, respectively, will intersect (fig.9).
And that's about all, since this implies that if we have a tetrahedron satisfying condition (4), then the straight lines perpendicular to its faces and passing through the centres of circumferences inscribed into these faces will intersect at one common point. And this point, being equidistant from all the edges of the tetrahedron, will be the centre of the required sphere. And now a problem - which at first glance may seem unrelated to the facts we have just discussed: Given a hyperbola with foci F and G, prove that the centres of circumferences inscribed into triangles FGA, where A is any point on one of the branches of the hyperbola, lie on a straight line tangent to this branch at its vertex P (fig.10). |