TITLE: A convex polyhedron floats in the sea. Can $90\%$ of its volume be below water level and more than half of its surface be above water level?
QUESTION [8 upvotes]: An object in the shape of a convex polyhedron floats in the sea. Could it happen that $90\%$ of its volume is below the water level and more than half of its surface is above the water level?

As I was told, this is a PURE mathematical problem.
There are shapes with infinite area and finite volume, but they are not convex. Roughly speaking, if you have a face, then the area of the other faces is always greater than the area of this one. The idea behind the idea is this: project all the other faces onto this plane. The projection will cover our face, since the polyhedron is convex. And the projected area is not more than the area of the projected faces.
However, I can't prove it mathematically, I thought to consider a pyramid or a cone, for example: consider a regular pyramid floating downward with its top. Let the area of its lateral surface be $S$. Then the area of the base is equal to $S\cos α$, where $α$ is the angle of inclination of the lateral faces to the base.

REPLY [4 votes]: Take a right square pyramid with side $24$ and height $5$.
It is easy to see the distance between apex and any side of its base square equals to $\sqrt{(24/2)^2+5^2} = 13$. From this, one can deduce the lateral surface area is $\frac{13}{13 + (24/2)} = 52\%$ of total surface area of the pyramid.
Flip the pyramid upside down and push the apex $5 \times (0.9)^{\frac13}$ unit under water. $90\%$ of the volume will be below water while the surface area below water is $52\% \times (0.9)^{\frac23} \sim 48.47\% < 50\%$.