Consider the four triangles with vertices at centre of sphere; contact of sphere with facet plane; meet of facet plane with vertical line. These are similar, so congruent. Now divide each by a horizontal perpendicular from contact point to vertical line. The corresponding smaller triangles are congruent. Therefore their vertical sides are equal, so their horizontal edges lie in the same horizontal plane, and the contact points are coplanar. An alternative algebraic proof computes the radius, which turns out to be infinite --- would that be more convincing? On reflection, it might be easier to describe in plaintext! WFL On 4/7/16, Andy Latto <andy.latto@pobox.com> wrote:
On Wed, Apr 6, 2016 at 10:21 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
By symmetry, the centre of your sphere lies on the "bi-altitude" line joining the mid-points of near and far tetrahedron edges: choose this line as our vertical. All four planes lie at the same angle to the it, therefore their points of tangency to the sphere all lie on the same horizontal plane.
I don't see why this follows. If you translated one of the planes "upwards", it would stil remain at the same angle to the vertical line, but the point of tangency would change.
Or to put it another way, two of the planes intersect our vertical line at one, lower, point, and two intersect it at a different, higher, point. So there is no symmetry that says that the points of tangency should have the same z-coordinate.
Or to put it a third way, the same proof shows there is no sphere inside the tetrahedron tangent to all four planes, and the fallacy is in the assertion that all four points of tangency lie in the same horizontal plane.
I think your claim is true, but your proof seems faulty, and I'm having trouble patching it up.
Andy
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