There are three spiders and an ant on the edges of a wireframe regular tetrahedron. The spiders all move at the same speed, which is faster than that of the ant. Though the ant can always see the spiders, the ant is invisible. A spider can only detect the ant by being on top of it, in which case the ant is immediately eaten. Show how the spiders can catch the ant.

This is rather similar to an earlier puzzle I posted. The differences are that the cube is now a tetrahedron, the spider's speed increased from $\frac13$(fly's speed) to (fly's speed)$+\varepsilon$, and the fly is now invisible. The first two changes greatly help the spider, but this is still a harder puzzle.

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    $\begingroup$ You should make Spiders and Invisible Ant on a Tesseract and ask if the ant can avoid them :D $\endgroup$ – warspyking Apr 26 '15 at 13:53

Call the vertices $A,B,C,D$. The spiders start themselves off with two spiders at $A$ and one spider at $D$. Then they move around in the following cycle: one spider stays at $A$ and the other two spiders head directly to $B$, then one spider stays at $B$ and the other two spiders head directly to $C$, then one spider stays at $C$ and the other two head directly to $D$, then one stays at $D$ and the other two head directly to $A$, repeat. Eventually they catch the ant: in order to evade the ant must travel across each of the edges $AB, BC, CD, DA$ at least once per cycle (think about what happens if the ant refuses to travel all the way across edge $DA$ for an entire cycle, assuming the ant doesn't start the cycle along edge $DA$).

enter image description here

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  • $\begingroup$ I'm trying to visualize how this guarantees they catch the ant. $\endgroup$ – JLee Apr 25 '15 at 12:43
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    $\begingroup$ This works! Another way to see this: one spider follows the cycle ABCD, one moves back and forth on AC, the other on BD (but the last two move at half speed, ie only move half the time). The ant is forced to also cycle around ABCD, because if he tries to escape the cycle onto the edges AC or BD, he is pushed off by one of the half speed spiders just as the ABCD spider comes by. Eventually, the ABCD spider catches up $\endgroup$ – Mike Earnest Apr 25 '15 at 20:35
  • $\begingroup$ @MikeEarnest For some reason, I understand your explanation, but I can't yet fully get zeb's $\endgroup$ – JLee Apr 25 '15 at 21:56

enter image description here

One spider stands still in A, another spider stands in B. The third spider (let's call him Bob) is the only one moving.
First of all, Bob checks if the ant is between A and B.
If not, he moves to the edge CD (as shown in the picture). Now, the ant has to be on a blue edge, on an orange edge, or on a portion of the green edge, with a probability of $p=\frac1 2$ of being in front or behind Bob.
Now, Bob moves to D, then chooses a random edge to explore between the two blue ones: if the ant is on a blue edge ($p=\frac1 2$), then Bob has a $\frac1 2$ chance of catching it after choosing the path. After all, he got a $p=\frac1 4$ chance of catching the ant starting from the edge CD.
If the ant survived, Bob goes back to CD and does the same thing with orange edges.
If the ant survived again, Bob goes back to CD and repeats with blue edges.
And so on, until he catches the ant (he has a probability of $p=1$ of catching the ant after an infinite time!

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  • $\begingroup$ But during step 6, the ant can sneak from DC into CA. $\endgroup$ – xnor Apr 25 '15 at 9:16
  • $\begingroup$ @xnor I've completely revised the strategy, check this new one! $\endgroup$ – leoll2 Apr 25 '15 at 9:24
  • $\begingroup$ This will not work if the ant can read the spiders' mind. This is not, however, necessarily required in the puzzle. $\endgroup$ – kaine Apr 25 '15 at 12:50
  • $\begingroup$ "the ant can always see the spiders" doesn't mean that it can read spiders' mind. Also, the spider decides randomly its path in the intersection, not before, when it's too late for the ant to switch branch. $\endgroup$ – leoll2 Apr 25 '15 at 12:54

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