Sum of the Squares of the Sides of a Projected Regular Tetrahedron

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Let be a regular tetrahedron with side . Project orthogonally to a plane and let , , …, be the lengths of the projections of the six sides of in . This Demonstration shows that .

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Proof:

Consider the same problem for a cube with side ; let , , …, be the lengths of the projections of the 12 sides of to .

If is a unit vector in a 3D coordinate system, let its angles to the axes be , , . Then . In our case, is perpendicular to and the axes are one corner of the cube.

The lengths of the projections of the edges are , , , each appearing four times in . Now, , so .

Each pair of opposite faces of the cube is projected as a pair of congruent parallelograms.

Suppose is inscribed in and that .

The edges of coincide with the six face diagonals of . A face of is projected as a parallelogram. In a parallelogram, the sum of the squares of the diagonals equals the sum of the squares of the sides. Therefore, the sum of the squares of the projections of two opposite edges of equals the sum of the squares of the projections of two pairs of opposite edges of . Finally then, .

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


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Details

This problem was posed in [1, pp. 20, 28].

Reference

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



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