Place a circle of the largest possible radius inside a cube.

#
Let $a$ be the length of the edge of a cube. A section of the cube by a plane passing through its center and orthogonal to one of the diagonals is a regular hexagon. The radius of the inscribed circle of this hexagon is $\frac{a \sqrt{6}}{4}$, so a circle of radius $\frac{a \sqrt{6}}{4}$ can fit inside the cube. We will show that a circle of a larger radius cannot fit inside the cube. First, note that it is sufficient to consider circles centered at the center of the cube. Indeed, if a circle of radius $R$ is contained within the cube, then the circle symmetric to it with respect to the center of the cube is also contained within the cube. From the convexity of the cube, it follows that a circle of radius $R$, centered at the center of the cube and lying in a plane parallel to the plane of the original circle, is also contained within the cube. Consider a circle of radius $R$ centered at the center of the cube and a sphere of the same radius and the same center. We are only interested in the case where $R > a / 2$ and the considered circle lies inside the cube. In this case, six spherical segments lie outside the cube. The radii of the circles lying at their bases are $r = \sqrt{R^2 - (a / 2)^2}$, so $r$ increases as $R$ increases. Consider the cones whose vertices are at the center of the cube and whose bases are the circles at the bases of the spherical segments. If the plane $\Pi$, containing the considered circle, intersects one of these cones, then part of the circle passes through the spherical segment and therefore partially lies outside the cube. Thus, we need to prove that if $R > \frac{a \sqrt{6}}{4}$, then the plane $\Pi$ intersects one of the cones. The plane $\Pi$ divides the rays emanating from the center of the cube and directed to the centers of the faces into two triplets (each triplet lies on one side of the plane $\Pi$). Consider the plane $\Pi'$, which passes through the center of the cube and is perpendicular to one of the diagonals and divides these rays into the same two triplets. In the plane $\Pi'$, there is a circle of radius $\frac{a \sqrt{6}}{4}$ that lies entirely inside the cube. It is easy to verify that the plane $\Pi'$

is tangent to three cones (corresponding to the triplet of rays that are the axes of these cones) along three rays $OX, OY, OZ$. The rays $OX, OY, OZ$ lie strictly inside the cones corresponding to the circle of radius $R > \frac{a \sqrt{6}}{4}$.

Thus, these rays lie on one side of the plane $\Pi$, since the axes of the corresponding cones lie on one side of this plane. The planes $\Pi$ and $\Pi'$ have a common point (the center of the cube), so they intersect along some line. The rays $OX, OY$, and $OZ$ form angles of $120^{\circ}$ with each other, so no line can divide the plane $\Pi'$ in such a way that these rays lie in the same half-plane. Therefore, the plane $\Pi'$ intersects one of the cones if $R > \frac{a \sqrt{6}}{4}$.