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# Not sure where to start on this one

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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 $$36$$?

Jun 14, 2021

So for this question, formulas for volume and surface areas of spheres are essential.  The volume of P is defined as $$\frac{4}{3}πr^3 = \frac{4}{3}π36^3$$, and Surface area of P is $$4πr^2$$, which in this case is equivalent to $$4π36^2$$.  Once we put this into the inequality (volume of P)/(surface area of P) ≤ t, we have, after simplification 12≤t.  This, the smallest real number t should be 12.