What is the maximum value that the area of the projection of a regular tetrahedron with an edge of 1 can take?

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If the projection is a quadrilateral, then the length of each of its diagonals is no more than 1.

## Solution

Consider two possible cases. 1. The projection of the tetrahedron is a triangle. Then the length of each of its sides is no more than 1 (the length of a segment can only decrease when projected). Let two of the sides of the triangle be a and b, and let C be the angle between these sides. From the formula for the area of a triangle, it follows that the area of the projection is $\mathrm{ab} * \sin (\mathrm{C}) / 2$, which does not exceed $1 / 2$. 2. The projection of the tetrahedron is a quadrilateral. The diagonals of this quadrilateral are projections of some edges of the tetrahedron, so the length of each diagonal is no more than 1. Let the lengths of the diagonals be $\mathrm{d}_{1}, \mathrm{~d}_{2}$, and let $\mathrm{U}$ be the angle between these diagonals. From the formula for the area of a quadrilateral, it follows that the area of the projection is $\mathrm{d}_{1} \mathrm{~d}_{2} * \sin (\mathrm{U}) / 2$, which does not exceed $1 / 2$. It remains to verify that there is a plane whose projection has an area exactly $1 / 2$. It is not difficult to check that a plane parallel to a pair of skew edges of the tetrahedron has the required property: the projection of the tetrahedron onto this plane is a square with a diagonal of 1.

## Answer

0.50

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