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The post consists of initial ideas on top and the proof at the bottom.

I think the key idea is to perform the process opposite to what you describe about subdividing into triangles. Indeed, our problem can be thought of a plane graph where all ants are moving in the counterclockwise direction except one on the outer circle which goes clockwise.

Now notice that it seems you can actually combine several areas into one, it should look the same or worse on the outside, and you get at the end two ants on the same circle moving in the opposite directions.

"The same or worse" part is conjectural, but the intuition is that assuming you have just two circles with common edge, "the best" you can do is to let one go through the edge then immediately after that let the next one proceed. If you think about how that affects the outside world, it's very similar to having a single ant going around the whole stuff.

The proof. Let's introduce some notation. First, I'll be considering ants on the plane which together form something with an outer circle. Now let's assume the ants have decided upon the schedule and they started their driving. The outer ant is drunk driver that goes in a clockwise direction.

We will prove that the crash will necessarily occur before the drunk driver makes a full cycle.

Crucial lemma. Suppose I have two areas A and B combined by an edge and a schedule of driving that includes drunk driver making a full cycle. Then it's possible to choose one of the areas, A or B, and the modification of route of drunk driver that makes a full cycle around the chosen area.

The proof requires some pictures, I'll try lemma is established by drawing a picture (update: an answer that refers to explain in words if there a paper with the solution was posted. the answer is interestthe same so you can read the corresponding description there). After you have the lemma, you're done, because you see that there must be a simple face (minimal area) with a sober and drunk ants which must crash.
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Update: I think I proved it, see

The post consists of initial ideas on top and the proof at the bottom.

I think the key idea is to perform the process opposite to what you describe about subdividing into triangles. Indeed, our problem can be thought of a plane graph where all ants are moving in the counterclockwise direction except one on the outer circle which goes clockwise.

Now notice that it seems you can actually combine several areas into one, it should look the same or worse on the outside, and you get at the end two ants on the same circle moving in the opposite directions.

"The same or worse" part is conjectural, but the intuition is that assuming you have just two circles with common edge, "the best" you can do is to let one go through the edge then immediately after that let the next one proceed. If you think about how that affects the outside world, it's very similar to having a single ant going around the whole stuff.

Let's introduce some notation. First, as above, I'll be considering ants on the plane which together form something with an outer circle. Now let's forget about assume the outer ant for a while. The remaining ants have decided upon the schedule and they keep guardstarted their driving.

Let's say the question is to find an The outer ant who makes is drunk driver that goes in a plot, by going counterclockwiseclockwise direction.His goal,

We will prove that the plot, is to make at least one crash will necessarily occur before the drunk driver makes a full loopcycle. Here's a crucial

Crucial lemma:

Lemma. Suppose I have two areas A and B combined by an edge and a successful plot is madeschedule of driving that includes drunk driver making a full cycle. Then a successful plot can be made either for area it's possible to choose one of the areas, A or for area Bthat consists of a half of original plot , and crossing the edge at modification of route of drunk driver that makes a full cycle around the right timechosen area.

The proof requires some pictures, I'll try to explain in words laterif there is interest. After you have the lemma, you're done, because you see that there must be a simple face (minimal area) with two a sober and drunk ants going in different directionswhich must crash.
show/hide this revision's text 4 finished writing the proof modulo picture

Update: I think I proved it, see at the bottom.

I think the key idea is to perform the process opposite to what you describe about subdividing into triangles. Indeed, our problem can be thought of a plane graph where all ants are moving in the counterclockwise direction except one on the outer circle which goes clockwise.

Now notice that it seems you can actually combine several areas into one, it should look the same or worse on the outside, and you get at the end two ants on the same circle moving in the opposite directions.

"The same or worse" part is conjectural, but the intuition is that assuming you have just two circles with common edge, "the best" you can do is to let one go through the edge then immediately after that let the next one proceed. If you think about how that affects the outside world, it's very similar to having a single ant going around the whole stuff.

Let's introduce some notation. First, as above, I'll be considering ants on the plane which together form something with an outer circle. Now let's forget about the outer ant for a while. The remaining ants have decided upon the schedule and they keep guard.

Let's say the question is to find an ant who makes a plot, by going counterclockwise. His goal, the plot, is to make at least one full loop. Here's a crucial lemma:

Lemma. Suppose I have two areas A and B combined by an edge and a successful plot is made. Then a successful plot can be made either for area A or for area B that consists of a half of original plot and crossing the edge at the right time.

The proof requires some pictures, I'll try to explain in words later. After you have the lemma, you're done, because you see that there must be a simple face (minimal area) with two ants going in different directions.
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