All edges of a triangular pyramid are equal to $a$. Find the maximum area that the orthogonal projection of this pyramid onto a plane can have.

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Answer: $a^{2} / 2$. The projection of the tetrahedron can be a triangle or a quadrilateral. In the first case, it is the projection of one of the faces, so its area does not exceed $\sqrt{3} a^{2} / 4$. In the second case, the diagonals of the quadrilateral are projections of the edges of the tetrahedron, so the area of the orthogonal projection, equal to half the product of the lengths of the diagonals by the sine of the angle between them, does not exceed $a^{2} / 2$; equality is achieved when a pair of opposite edges of the tetrahedron are parallel to the given plane. It remains to note that $\sqrt{3} a^{2} / 4<$

$a^{2} / 2$.