# 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.)

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 Aij touches the sphere at point Tij.) 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 problems entered the folklore as the "coffin problems" - groby in Russian. The above is one such coffin.

Winkler tells the story and and outlines the solution in Khovanova's writeup, with a mention that the problem was offered at the 1975 entrance exams. Honsberger (1978) expands on the solution and mentions that the problem has been given at an elementary contest in Romania.

Solution 1

Solution 2 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 Aij touches the sphere at point Tij.) Interestingly, the four points of tangency T12, T23, T34, T41 are necessarily coplanar.

### Solution

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.

The barycenter of a system of points does not change if any number of points is replaced with their barycenter.

Now, the tangents from a point to a sphere are all the same length. Let the length of the tangents from Ai to the sphere be di, i = 1, 2, 3, 4. Place mass 1/di at vertex Ai. Then the center of gravity of any two adjacent vertices, say A1 and A2, is the point M with mass 1/d1 + 1/d2 that satisfies

NA1 / d1 = NA2 / d2,

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 Ai (with masses 1/di) 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.

Observe that the theory of material points allows negative masses. So the proof works even in case when some of the side lines of the quadrilateral touch the sphere at the side extensions.

### References

1. R. Honsberger, Mathematical Morsels, MAA, 1978, pp. 199-201
2. P. Winkler, Mathematical Puzzles: A Connoisseur's Collection, A K Peters, 2004, pp. 55-56 ### Barycenter and Barycentric Coordinates
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