MHB What is the Shortest Path for a Spider to Capture a Fly on a Cube?

[Ackbach]

Here is this week's POTW:

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A spider at corner $S$ of a cube of side length $1$ inch wishes to capture a fly at the opposite corner $F$. The spider, who must walk on the surface of the solid cube, wishes to find the shortest path.

• Find a shortest path with the aid of calculus.
• Find a shortest path without calculus.

Choose the spider's starting point as the origin, and let the fly be at the corner $(1,1,1)$ inches.

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[Ackbach]

No one answered this week's POTW. What follows is my solution.

Spoiler

First, we use calculus. There are six faces to the cube: $x=0, x=1, y=0, y=1, z=0,$ and $z=1$. I will use each of these equations to represent the face lying in the plane described by that equation. Construct the straight line from the origin to $(1,1,1)$. By symmetry, we may rotate the cube about this line $2\pi/3$ radians, and obtain the same situation as before. We may also reflect the cube about the plane $y=x$, and obtain the same situation as before. This means we may restrict our attention to two faces - arbitrarily I choose the $y=0$ and $z=1$ faces, and only paths on those faces. It seems reasonable to assume that the shortest path desired would be a straight segment from the origin to some point on the edge in common between these two planes, and then another straight segment from that point to $(1,1,1)$. To find the desired point $(x,0,1)$, we represent the distance from the origin to that point as $d_1=\sqrt{x^2+1}$, and then the distance from that point to $(1,1,1)$ would be $d_2=\sqrt{(1-x)^2+1}=\sqrt{x^2-2x+2}$. The total distance, therefore, is given by
$$d=\sqrt{x^2+1}+\sqrt{x^2-2x+2}.$$
Differentiating and setting the result equal to zero results in
$$\frac{x}{\sqrt{x^2+1}}+\frac{x-1}{\sqrt{x^2-2x+2}}=0 \quad \implies \quad
\frac{x^2}{x^2+1}=\frac{x^2-2x+1}{x^2-2x+2}.$$
Here we must be careful - we may have introduced extraneous solutions by squaring. Cross-multiplying yields
\begin{align*}
x^4-2x^3+2x^2&=x^4+x^2-2x^3-2x+x^2+1 \\
0&=-2x+1 \\
x&=\frac12.
\end{align*}
So, then, the solution is to reach the midpoint of an adjoining cube edge, and then to go directly to the opposite corner.

Without calculus, we could simply argue by symmetry that this must be the solution. There is a symmetry from the first half of the journey to the second half (you could think of path reversal), so that they should be symmetric. This leads to the same answer.