# 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?

• Pretty sure it is true, but do i have to build an example? – the4seasons Sep 24 '15 at 12:57
• 3-dimensional or n-dimensional? – Ben Frankel Sep 24 '15 at 13:10
• So, without the flavor, this problem is: Define a convex polyhedron such that a plane can bisect it in such a way that >90% of the volume is on one side of the plane and >50% of the surface area is on the other. Right? I'm fairly sure any regular polyhedron would not qualify so we must find an irregular shape. – Engineer Toast Sep 24 '15 at 13:51
• @Engineer Toast: Yes, yes, and yes. – Gamow Sep 24 '15 at 14:48
• Is the lake made of water? And does Archimedes principle apply? – ghosts_in_the_code Sep 25 '15 at 5:58

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

• Great answer. I wonder if it floats stably, or would flip up? – Dr Xorile Sep 24 '15 at 17:19
• great answer, I also tried to use a pyramid but only tried one with a bigger height and that one didn't work so I give up haha. The math checks out but I would suggest expanding the answer a bit explaining how you calculated those values. – Ivo Beckers Sep 24 '15 at 20:39
• A flat enough pyramid behaves basically like two squares, the top of which is out and the bottom of which is submerged except for a lip around the edge. The volume submerged is $(h'/h)^3$, where $h'$ is the height that's submerged and $h$ is the height of the pyramid. – Dr Xorile Sep 24 '15 at 21:37
• The proportion of the slanted parts of the area that is submerged is $(h'/h)^2$, which will always be between $0.93=0.9^(2/3)$ and 1 given the volume constraint. The area of the bottom (slanted bit) is $\sqrt{h^2+(w)^2}/w$ times the top area (where $w$ is half the width). This can be made arbitrarily close to 1 by making $w$ bigger, so simply choosing $w$ big enough will get you there. – Dr Xorile Sep 24 '15 at 21:46

Lateral thinking solution:

Professor Erasmus lives on a planet that is made entirely of water except for the small floating patch of earth that supports his house. The Professor-Erasmus-polytope is about as large as the planet, so that most of its surfaces skim along the top of the water, while most of its volume remains below the surface.

Artist's conception:

• Possibly the best answer I've seen in a while :D – dmg Sep 24 '15 at 13:13
• I think this would have to be pretty well balanced for the displacement of water. Either way, I think the poor prof's house is doomed to flood :D – Set Big O Sep 24 '15 at 13:15
• Ooh, displacement is a problem... Maybe the polytope is made of a permeable material, like sponge. But sponge isn't homogenous if you count the little holes... – Kevin Sep 24 '15 at 13:16
• @Kevin: Displacement is in this case indeed a problem, but a sponge is not a convex polyhedron. A sponge would also have most of its surface under water. – jarnbjo Sep 24 '15 at 13:24
• The answer sort of still works if the polytope is an impermeable solid. But the polytope would be much larger than Erasmus' planet before displacement. Effectively all the water "beads" on a single surface as it is attracted towards the polytope's center of gravity. But this only works if you define the "water line" as "the surface of the imaginary sphere that Erasmus' squashed planet is a spherical segment of" – Kevin Sep 24 '15 at 13:35