XXXIX OM - III - Problem 6

Calculate the maximum volume of a tetrahedron contained within a hemisphere of radius 1.
Let us denote the given hemisphere by $H$, the whole sphere by $K$, and the circle forming the flat part of the boundary of $H$ by $E$. Let $O$ be any tetrahedron contained in $H$. Its interior is the intersection of four open half-spaces determined by the planes of the faces. The center of the sphere $K$ does not belong to the interior of the tetrahedron; hence it does not belong to at least one of these half-spaces. Therefore, there is a face $S$ of the tetrahedron $Q$ such that the opposite vertex of the tetrahedron and the center of the sphere $K$ lie on opposite sides of the plane of this face (or, in the limiting case, the center of the sphere lies on this plane).

Thus, $Q$ is contained in the smaller of the two parts (or in one of the two equal parts) into which the plane of the face $S$ divides the sphere $K$. Therefore, the height of the tetrahedron $Q$ dropped onto the face $S$ has a length not greater than $1$, so the volume of $Q$ does not exceed one third of the area of this face. The area of the face $S$, being a triangle, is not greater than the area of an equilateral triangle inscribed in the circle which is the section of the sphere $K$ by the plane of this face (see Note); and even more so, it is not greater than the area of an equilateral triangle inscribed in the great circle $E$. The latter area equals $3\sqrt{3}/4$. Therefore, the volume of $Q$ is less than or equal to $\sqrt{3}/4$.

In all the estimates mentioned, we can achieve equality by taking $O$ to be a tetrahedron whose base is an equilateral triangle inscribed in $E$, and the fourth vertex is placed at the farthest point from $E$ on the hemisphere $H$ (at the "north pole").

The number $\sqrt{3}/4$ is the sought maximum.

Note: For completeness, we will provide an elementary proof of the fact that among all triangles inscribed in a fixed circle of radius $r$, the one with the largest area is the equilateral triangle.

Let $ABC$ be any such triangle. By placing point $C$ at the midpoint of the larger of the two arcs into which points $A$ and $B$ divide the circle circumscribed around triangle $ABC$, we obtain an isosceles triangle $ABC$, acute or right, with an area not less than that of triangle $ABC$. Therefore, in searching for a triangle with the maximum area (inscribed in a given circle), we can limit our attention to the class of isosceles triangles without obtuse angles (inscribed in this circle).

If now $ABC$ is a triangle from this class ($|AC| = |BC|$), then denoting by $x$ the distance from side $AB$ to the center of the circle, we get the relationships: $|AB| = 2\sqrt{r^2 - x^2}$, $h_C = r + x$ (the height dropped from vertex $C$), and thus

We have used here the inequality between the geometric mean and the arithmetic mean of four numbers: $a_1 = a_2 = a_3 = r + x$, $a_4 = 3r - 3x$. This inequality becomes an equality if and only if these numbers are equal, i.e., for $x = r/2$. It remains to note that the last equality means that triangle $ABC$ is equilateral.