Shortest Distance Between Pair of Beetles on a Cube

Question

This is a problem from the AMT (an Australian paper) from 2011:

The first thing I thought was when the first beetle was directly over the line FH or the second beetle directly over AC, the distance would be at a minimum. I could create a function (somehow utilizing the different rates and actually finding a way to get that distance as a function) and find that minimum, but this paper didn't require any complicated formulas or extra "stuff", so Pythagoras was probably the only formula you needed to know. I was a bit stuck as to how I should continue, trying out just simple cases and seeing where that would take me, however, that didn't work. Does anyone have any insights/methods to do this problem?

Answer 1

Refer to the figure:

$\hspace{4cm}$

Let $HJ=s, AK=2s$. Then the distance $KJ$ is: $$d(s)=\sqrt{LO^2+JO^2+KL^2}=\sqrt{\left(\frac{80\sqrt{55}}2-2s\right)^2+\left(\frac{80\sqrt{55}}2-s\right)^2+(40\sqrt{110})^2}.\\ (d^2(s))'=-4\left(40\sqrt{55}-2s\right)-2\left(40\sqrt{55}-s\right)\Rightarrow 10s=240\sqrt{55} \Rightarrow s=24\sqrt{55}\\ (d^2(s))''=10>0 \ \Rightarrow \text{min}$$ Hence: $$d(24\sqrt{55})=440.$$

Answer 2

This is a hint to make the straightforward attack a bit less work.

Note that in the Eucliden distance formula for the distance between the two beetles, the difference in their $z$ coordinates (taking the $z$-axis parallel to $BF$) is constant ($z_1 - z_2 = 40 \sqrt{110}$ always). The distance is minimized when the square of the distance is minimized. The $z$ contribution to the squared distance ($(z_1 - z_2)^2 = 1600 \cdot 110$) is constant and a constant offset does not alter the location of a minimum. Consequently, you can pretend the two beetles are in the plane (both having $z = 0$) and find the shortest separation distance there, then displace them in the $z$ direction to find out their shortest separation distance on their original paths.