Consider a cube of side $2$ feet containing 65 flies. Show that at a given point in time, there are at least $9$ flies such that the distance between any $2$ of them is no larger than $1.74$ feet.
So, I know that for $9$ of the flies to line up across the longest line of the cube would be it's diagonal $=\sqrt{2^2+2^2}$ or $2.828...$. But I'm not sure if this is something I'll need to determine the answer $1.74$ feet. Would someone be able to give me some pointers as to what to do to do this proof?
Tile the cube into $2\times 2\times 2$ smaller cubes of side length $1$'. Since $8\cdot 8<65$, one of the cubes must contain at least $9$ flies. The diagonal of such a cube is $\sqrt 3$'.