A number of geometry puzzles play upon a surprising fact about spheres that have had holes bored through them. Imagine you have a bead that is one inch (about 2.5 centimeters) high and that you drill a hole exactly through the middle of this so that the remaining part of the sphere is only half an inch high. Now imagine that an enormously large drill has been used to bore a hole though the Earth so large that the part of the Earth that is left behind is only half an inch high. Amazingly, the residual volumes of these two holey spheres, the drilled bead and the drilled Earth, are exactly the same! It just happens that even though the Earth is vastly larger than the bead, the drill has to take out proportionately more in order to make the height of the hole the same, so that the volume left doesn't depend separately on the initial size of the sphere or of the hole, but only on their relation, which is forced by requiring the hole to be half an inch long. This fact enables the following poem-problem to have a solution even though its seems as if not enough information has been provided:
Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six inches long.
Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough
Having already learned the secret that the volume that remains of a drilled sphere doesn't depend on the initial size of the sphere, we can cheat and give a kind of meta-argument that is much shorter than the geometric proof. The volume left behind of any sphere with a six-inch-long hole through it must be the same as the volume left behind of a six-inch-diameter sphere with a hole of zero diameter drilled through it! This is equal to 4/3(π 33), or approximately 113 cubic inches.