# Stereographic Projection of a Coffin Problem

Hubert Shutrick came up with an elegant solution to one of the *coffin problems*:

So 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.

One solution based on the idea of center of gravity (barycenter) appeared elsewhere.

Hubert Shutrick's solution draws on the conformality (angle preservation) propety of the stereographic projection.

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Copyright © 1996-2018 Alexander Bogomolny

So 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

Stereographic projection converts the coffin problem into a problem in Euclidean geometry that is interesting in its own right. Since each vertex of the skew quadrilateral along with two adjacent tangency points defines a circle on the sphere by the plane of the sides that meet in the vertex; these four circles A, B, C, D can be projected down onto the plane and A is tangent to B is tangent to C is tangent to D is tangent to A (because, any successive two share a point but lie in different planes). Show that the points where they touch lie on a circle. It is not difficult to prove by just chasing angles but, of course, follows directly from the elegant solution of the original problem.

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Copyright © 1996-2018 Alexander Bogomolny