3D Quadrilateral - a Coffin Problem

Unlike a triangle, a quadrilateral is not a rigid shape. There are unequal quadrilaterals with the same side lengths: holding one of the sides as the base, the three other sides, while joined, can be shifted left or right. In 3D - a three-dimensional space - there appears an additional degree of freedom to modify a quadrilateral. In 3D, a quadrilateral can be folded (partially or completely) along a diagonal so that in 3D a quadrilateral is not necessary flat, i.e., a 3D quadrilateral need not have coplanar vertices. (Points are coplanar if they lie in the same plane.)

So let there be a 3D quadrilateral A₁A₂A₃A₄ with the property that all of its 4 sides, A₁A₂, A₂A₃, A₃A₄, A₄A₁, are tangent to a given sphere. (The side A_ij touches the sphere at point T_ij.) Interestingly, the four points of tangency T₁₂, T₂₃, T₃₄, T₄₁ are necessarily coplanar.

Remark

This problem is one from a collection of "coffin problems" gathered by Tanya Khovanova. Years ago, the high school graduates who sought admission to the Mathematics Department of Moscow State University had to pass 4 entrance exams of which the first two were in mathematics - written and oral. While the written examination was one and the same for all candidates and was administered to all of them on the same date and at the same time, the oral exam was a one-on-one affair between a student and an examiner who had the sole responsibility and the freedom of giving the student problems of his or her liking. Many an examiner used the opportunity to sieve out unwanted (mostly Jewish) candidates by offering them simply sounding problems with very uncommon or non-trivial solutions. The problem entered the folklore as the "coffin problems" - groby in Russian. The above is one such coffin.

Solution

Copyright © 1996-2008 Alexander Bogomolny

The simplest approach to solving this problem is by using the notion of barycenter - the center of gravity or the center of mass - of the system of four material points located at the vertices of the given quadrilateral. The idea is to employ one of the basic properties of the center of gravity, as has been done in proving Ceva's theorem and elsewhere.

which holds obviously for N = T₁₂ making the latter the barycenter of two points A₁ and A₂. Similarly, T₂₃, T₃₄, T₄₁ are the centers of gravity of the pairs of the material points placed at the vertices A₁, A₂, A₃, A₄ that forms the side to which the tangency point belongs. (This, of course, assumes that the masses are 1/d₁, 1/d₂, 1/d₃, 1/d₄, respectively.)

The barycenter of the four points can be found by first combining, say, A₁ and A₂ and then also A₃ and A₄. This will produce two material points, one at T₁₂ with mass 1/d₁ + 1/d₂ and the other at T₃₄ with mass 1/d₃ + 1/d₄. We may conclude that the barycenter of four points A_i (with masses 1/d_i) coincides with the barycenter of the two just constructed material points at T₁₂ and T₃₄. In particular, this means that the barycenter of the four points at the vertices of the quadrilateral lies on the segment T₁₂T₃₄ joining T₁₂ and T₃₄.

But there was no obligation to pair up A₁ with A₂ and A₃ with A₄. We would have reached the same point if we started with the barycenters of A₂ and A₃ on one hand and A₄ and A₁, on the other. If we did, we would find that the barycenter of the four points lies on the segment T₂₃T₄₁. So one and the same point lies both on T₁₂T₃₄ and T₂₃T₄₁ implying that the two segments intersect. But any two straight lines (or straight line segments) define a plane which includes the two lines. And this is all that is needed to prove that the four points of tangency T₁₂, T₂₃, T₃₄, T₄₁ are coplanar.

Barycenter and Barycentric Coordinates

1. 3D Quadrilateral - a Coffin Problem
2. Barycentric Coordinates
3. Barycentric Coordinates: a Tool
4. Barycentric Coordinates and Geometric Probability
   • Stick Broken Into Three Pieces (Trilinear Coordinates)
   • Stick Broken Into Three Pieces (Cartesian Coordinates)
5. Ceva's Theorem
6. Determinants, Area, and Barycentric Coordinates
7. Maxwell Theorem via the Center of Gravity
8. Medians in a Quadrilateral
9. Three glasses puzzle
10. Van Obel Theorem and Barycentric Coordinates

Copyright © 1996-2008 Alexander Bogomolny