# Spider and fly on a cube

A spider and a fly play a game with a cube of side length $$s=1$$ and with a positive real number $$d$$.

• First, the spider picks its starting point $$S$$ somewhere on the surface of the cube.

• Then the fly picks its point $$F$$ somewhere on the surface of the cube.

• Then the spider crawls a distance of $$d$$ from point $$S$$ towards the fly in point $$F$$. The fly does not move.

The spider wins the game, if it manages to reach the point $$F$$.
Otherwise the fly wins the game.

Question: For which values of $$d$$ does the spider have a winning strategy?

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This puzzle is (vaguely) related to: $$~$$ A crawling spider and a cautious fly

• My spider (and fly) sense is picking up some ambiguity in this use of the word "towards". Usually it means "along the shortest connecting line", but that's probably not how it's used here? – Bass Jan 11 at 11:57
• @Bass: No, no, no. The spider can move in any way it likes, and in particular along the shortest connecting line. – Gamow Jan 11 at 12:07
• Why can't the spider always use d=10, or any other larger number? I believe you meant to have some restriction to minimize d. – JPhi1618 Jan 11 at 20:06

I think Oray got the right answer. Here are some drawings to illustrate the solution.

This image shows fold-out nets of the cube with various points in the middle. The first has the centre of a face in the middle, the second the midpoint of an edge, and the last a cube corner.

The fold outs are such that when you draw a straight line from the marked point in the middle to any other point on the cube, the whole line will lie within the figure, and shows you the shortest path to that point.

The first figure has a radius of $$2$$, which is the best.
The second figure has a radius of $$\sqrt{145}/6 \approx 2.0069$$.
The last has a radius of $$\sqrt(5) \approx 2.236$$

It is fairly obvious that the diametrically opposite points are at least a distance of $$2$$ apart, but the only way I can think to prove it is to split it into various cases depending on how many faces the path crosses. In each case, regardless of how you unfold the cube, diametrically opposite points will be either horizontally or vertically a distance of $$2$$ apart on the unfolded cube, making the path between them at least $$2$$.

This is not a proof but basic intuition;

The value of $$d$$ should be (considering they are wise and choose not random points but the best possible point where they have the advantage)

$$2s=2$$

where it makes

our spider wins the game always.

with the strategy of

that the spider chooses the center point of any surface on the cube and that will make our fly will choose the longest distance from there which will be located in the center of the surface on the other side of the cube. If spider chooses other points on the cube except the centers of the surfaces of the cube, the longest distance will be more than $$2$$.

Lastly,

the longest distance will be only possible if spider chooses any corner on the cube, where our fly will choose opposite corner with the shortest distance between fly and spider as $$\sqrt{5}s$$.

Actually, with some logic, and an intelligent spider/fly we can further reduce the value for d from 2s down to 1.6s(rounded up).

First, our theoretical cube would likely be sitting flat on another surface. this allows us to remove one side of the cube as a viable point for either S or F. Second the intelligent spider would position itself on the middle of the top surface, This makes the furthest viable point be one of the 4 bottom corners.

So the fly wanting to be as far as possible from the spider would land there. Now can use Pythagorean Theorem to determine the Distance from any bottom corner to the center of the top. The result would be s²+(3s)²=2d².

Since we were given s=1 When we solve for d that becomes √(1²+(3)²)/2=d That resolves to 1.58113=d

IF however the spider were not intelligent and simply picked the worst possible place, and the spider picked the best possible place, then that returns to spider on corner, fly on opposite corner. As covered above that's s²+(2s)²=d² Which ultimate resolves to d=2.236 (rounded down).

• "First, our theoretical cube would likely be sitting flat on another surface." Maybe it's likely, but it could be hanging by a string. – LarsH Jan 12 at 21:18