Consider the $13$ triangles formed by three consecutive marked points. We want to maximize the area of these triangles. First, it can be shown that when we fix a segment with endpoints on the circle, and add a third point on the circle to form a triangle, the maximum height of such a triangle, and thus the maximum area of the triangle, must be when the third point is equidistant from the endpoints on the segment and on the circle. Now we take a regular $13$-gon formed by the $13$ marked points, and inscribed in the given circle. Since each of the vertices of the regular $13$-gon is equidistant from those on each side of it, the triangles formed by consecutive marked points have the maximum area, given a fixed base length (which is the distance between vertices two apart from each other). But now we take any other configuration of $13$ marked points. This configuration must have a distance between vertices two apart from each other that is less than the distance formed by a regular $13$-gon, and thus must have a triangle with a lesser area than those formed by the regular $13$-gon. Thus the greatest possible area of the smallest triangle is the area of the triangle formed by three consecutive points on a $13$-gon. It remains to prove the area of the triangle formed by three consecutive points on a regular $13$-gon inscribed in a circle of radius $13$ is less than $13$. The base of this triangle is less than $13$, since $2 \cdot \frac{360}{13} ^ {\circ} < 60 ^{\circ}$. The height of this triangle is less than $2$, since if we draw a triangle with the same base and third point at the center of the triangle, the two heights of that triangle add to $13$. The height of the new triangle must be greater than $11$, because otherwise the base would be greater than or equal to $2\sqrt{48}$ which is absurd, so the height of the triangle formed by the three consecutive points is less than $2$. Since the base and height are both smaller than that of a triangle of area $13$, this triangle has a smaller area than $13$, Q.E.D.