15. Each edge of a triangular pyramid (tetrahedron) is equal to $a$. Find the maximum area that the orthogonal projection of this pyramid onto a horizontal plane can have.
15. The orthogonal projection of a tetrahedron can be either a triangle or a quadrilateral. In the first case, a regular tetrahedron is projected into a triangle that coincides with the projection of one of its faces. The largest projection in this case will be when the plane of the face is parallel to the projection plane. The projection will be equal to $\frac{a^{2} \sqrt{3}}{4}$, i.e., the area of a regular triangle. In the second case, a regular tetrahedron is projected into a quadrilateral. To obtain the projection with the largest area, the tetrahedron should be positioned so that its vertices coincide with four vertices of a cube, and its edges coincide with the diagonals of the cube's faces, i.e., the regular tetrahedron should be inscribed in a cube with a side length of $\frac{a \sqrt{2}}{2}$, assuming the edge of the tetrahedron is $a$. Then the projection of the tetrahedron will be a square, the area of which is $\frac{a^{2}}{2}$. Clearly, $\frac{a^{2}}{2} > \frac{a^{2} \sqrt{3}}{4}$, i.e., the orthogonal projection of a regular tetrahedron, when it is projected into a square, will be the largest.