# 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 A_{1}A_{2}A_{3}A_{4} with the property that all of its 4 sides, A_{1}A_{2}, A_{2}A_{3}, A_{3}A_{4}, A_{4}A_{1}, are tangent to a given sphere. (The side A_{i}_{j} touches the sphere at point T_{ij}.) Interestingly, the four points of tangency T_{12}, T_{23}, T_{34}, T_{41} 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.

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Copyright © 1996-2017 Alexander Bogomolny

Let there be a 3D quadrilateral A_{1}A_{2}A_{3}A_{4} with the property that all of its 4 sides, A_{1}A_{2}, A_{2}A_{3}, A_{3}A_{4}, A_{4}A_{1}, are tangent to a given sphere. (The side A_{i}_{j} touches the sphere at point T_{ij}.) Interestingly, the four points of tangency T_{12}, T_{23}, T_{34}, T_{41} 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 A_{i} to the sphere be d_{i}, _{i} at vertex A_{i}. Then the center of gravity of any two adjacent vertices, say A_{1} and A_{2}, is the point M with mass _{1} + 1/d_{2}

NA_{1} / d_{1} = NA_{2} / d_{2},

which holds obviously for N = T_{12} making the latter the barycenter of two points A_{1} and A_{2}. Similarly, T_{23}, T_{34}, T_{41} are the centers of gravity of the pairs of the material points placed at the vertices A_{1}, A_{2}, A_{3}, A_{4} that forms the side to which the tangency point belongs. (This, of course, assumes that the masses are 1/d_{1}, 1/d_{2}, 1/d_{3}, 1/d_{4}, respectively.)

The barycenter of the four points can be found by first combining, say, A_{1} and A_{2} and then also A_{3} and A_{4}. This will produce two material points, one at T_{12} with mass _{1} + 1/d_{2}_{34} with mass _{3} + 1/d_{4}._{i} (with masses 1/d_{i}) coincides with the barycenter of the two just constructed material points at T_{12} and T_{34}. In particular, this means that the barycenter of the four points at the vertices of the quadrilateral lies on the segment T_{12}T_{34} joining T_{12} and T_{34}.

But there was no obligation to pair up A_{1} with A_{2} and A_{3} with A_{4}. We would have reached the same point if we started with the barycenters of A_{2} and A_{3} on one hand and A_{4} and A_{1}, on the other. If we did, we would find that the barycenter of the four points lies on the segment T_{23}T_{41}. So one and the same point lies both on T_{12}T_{34} and T_{23}T_{41} 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 T_{12}, T_{23}, T_{34}, T_{41} 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

- P. Winkler,
*Mathematical Puzzles: A Connoisseur's Collection*, A K Peters, 2004, pp. 55-56

### Barycenter and Barycentric Coordinates

- 3D Quadrilateral - a Coffin Problem
- Barycentric Coordinates
- Barycentric Coordinates: a Tool
- Barycentric Coordinates and Geometric Probability
- Ceva's Theorem
- Determinants, Area, and Barycentric Coordinates
- Maxwell Theorem via the Center of Gravity
- Bimedians in a Quadrilateral
- Simultaneous Generalization of the Theorems of Ceva and Menelaus
- Three glasses puzzle
- Van Obel Theorem and Barycentric Coordinates
- 1961 IMO, Problem 4. An exercise in barycentric coordinates
- Centroids in Polygon
- Center of Gravity and Motion of Material Points
- Isotomic Reciprocity
- An Affine Property of Barycenter
- Problem in Direct Similarity
- Circles in Barycentric Coordinates
- Barycenter of Cevian Triangle
- Concurrent Chords in a Circle, Equally Inclined

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Copyright © 1996-2017 Alexander Bogomolny

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