mathematics - The Erasmus polyhedron

Professor Erasmus has constructed a special convex polyhedron from perfectly homogeneous material, which he modestly calls the "Professor-Erasmus-polyhedron". The professor claims that he can put the polyhedron into the lake next to his house, so that the polyhedron floats

• with 90% of its volume below the water level, and


• more than 50% of its surface area above the water level.

Has the professor once again made a mathematical blunder, or does such a polyhedron indeed exist?

Answer

Ok, i'll take a stab at it.

Take a cone of radius 100 and height 10. place it point first into the water, so that .34 units stick out of the water. At that point, 90% of the volume (94397.4 of 104720) is underwater, and less than half of the surface area (29462 of 62988.5) is under water.

Oh fine, polyhedron, i'll change it to a

pyramid of base 100x100 height 10. Point first, if .34 units stick out of the water. Volume (30048 of 33333), surface (9516 of 20198)

I came to this idea from the idea that a really flat pyramid would have the surface area of the base be close to 1/2 of the total surface area. Then I just had to work out where 90% of the volume would be, and check if that area is under 50%.

Let's take height as a ratio of width (xw). So, we know that the height underwater is .966 of the total height, which means the width underwater is also .966 of the original width.
So, the surface area of the part underwater is $2 * .966w * \sqrt{\frac{(.966w)^2}{4} + (.966xw)^2}$
And of course the full surface area is $w^2 + 2 * w * \sqrt{\frac{w^2}{4} + (xw)^2}$

Plugging that all in gives an X < .2883