# Locus of Points at Constant Distance from the Edges of a Trihedron

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Choose a constant and a trihedron with apex and edges , , . How do you find the locus of points such that the sum of the distances of to the faces of the trihedron equals ?

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First, construct the points , , , on the edges , , such that the distances of these points to the opposite faces are .

By construction, each point on the triangle , , has this property, namely, with on the triangle, the distances of this point to the faces of the trihedron are , , .

Let , , be the areas of the corresponding faces. Triple the volume of the tetrahedron . The volume is now , so all the are equal; let their common value be . But the volume is also . So .

Let , , be the points symmetric to , , with respect to the point , forming a second trihedron. Now each point lies inside one of the trihedral angles determined by the six edges of the trihedrons, so the general solution is the convex hull of the six points , , , , , , resembling a triangular antiprism.

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Contributed by: Izidor Hafner (April 2017)
Open content licensed under CC BY-NC-SA

## Details

This problem was posed in [1, p. 220]. For the original solution, see [1, p. 225].

Reference

[1] V. V. Prasolov and I. F. Sharygin, Problems in Stereometry (in Russian), Moscow: Nauka, 1989.

## Permanent Citation

Izidor Hafner

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