### Instruction: Polyhedron $P$ is inscribed in a sphere of radius $36$ (meaning that all vertices of $P$ lie on the sphere surface). What is the least upper bound on the ratio $$\frac{\text{volume of }P}{\text{surface area of }P}~?$$In other words, what is the smallest real number $t$ such that $$\frac{\text{volume of }P}{\text{surface area of }P} \le t$$must be true for all polyhedra $P$ that can be inscribed in a sphere of radius $36$?### Response: I need to relate the volume and surface area of an arbitrary polyhedron inscribed in a sphere.
One thing I know is that the volume of any polyhedron is one-third times the product of its surface area and its inradius, which is the radius of the largest sphere that can fit inside the polyhedron and touch all its faces.
So I can write $$\frac{\text{volume of }P}{\text{surface area of }P} = \frac{1}{3} \times \text{inradius of }P.$$
Now I want to maximize this ratio, which means I want to maximize the inradius of $P$.
The inradius of $P$ is at most the radius of the sphere that circumscribes $P$, which is given as $36$.
So the inradius of $P$ can be anything from $0$ to $36$, and the ratio can be anything from $0$ to $12$.
The least upper bound of this ratio is then $12$, since any larger value would not be an upper bound for all possible $P$.
Therefore, the smallest $t$ that satisfies the inequality is $t = 12$.