I heard this cute math puzzle recently (thanks Piotr :)

Three ants start at the vertices of a unit equilateral triangle. Ant 1 always faces ant 2, 2 always faces 3, and 3 always faces 1. They all start walking at the same time towards their target neighbor, at one unit per second. What happens?

Now I'll wait while you figure that one out.

In the meantime, here's another nice symmetry-related thing that Joanna showed me: The surface area on a sphere in between any two parallel slices through the sphere is only dependent on the vertical distance between the slices. So if you slice off the top mile of the Earth, and take out a one mile slice at the equator, those sections have the same surface area.

That's a little surprising, until you realize that since sphere are perfectly symmetric (they have constant curvature in any direction), the circumference of the slice you're taking is shrinking as you move away from the equator at exactly the same rate that it's flattening out, so to speak. That is, three inches "above" (3 inches along the axis of rotation) the equator translates to three inches on the ground, but three inches "below" the north pole translates to a looong way on the ground. And this exactly offsets the fact that the Earth is much bigger around at the equator than the pole.

...(ant spoiler alert)...

Ok back to the ants. By symmetry, they will collide at the center of the triangle. The question is, how long does it take for them to get there? They're spiraling inwards as they turn to keep facing each other, so they don't take a straight route there. But, by symmetry, at every point in time they're still in an equilateral triangle formation, essentially starting from scratch on a slightly smaller and rotated triangle! That means that if you figure out what's going on at the first instant, the same thing applies at every later instant.

From there it's easy. The inward distance from each vertex to the center of the triangle is √3/3. And the component of each ant's velocity that is pointing inward is cos(30)=√3/2. So they will collide in the center in 2/3 of a second.

Update: arithmetic corrected... again.

Three ants start at the vertices of a unit equilateral triangle. Ant 1 always faces ant 2, 2 always faces 3, and 3 always faces 1. They all start walking at the same time towards their target neighbor, at one unit per second. What happens?

Now I'll wait while you figure that one out.

In the meantime, here's another nice symmetry-related thing that Joanna showed me: The surface area on a sphere in between any two parallel slices through the sphere is only dependent on the vertical distance between the slices. So if you slice off the top mile of the Earth, and take out a one mile slice at the equator, those sections have the same surface area.

That's a little surprising, until you realize that since sphere are perfectly symmetric (they have constant curvature in any direction), the circumference of the slice you're taking is shrinking as you move away from the equator at exactly the same rate that it's flattening out, so to speak. That is, three inches "above" (3 inches along the axis of rotation) the equator translates to three inches on the ground, but three inches "below" the north pole translates to a looong way on the ground. And this exactly offsets the fact that the Earth is much bigger around at the equator than the pole.

...(ant spoiler alert)...

Ok back to the ants. By symmetry, they will collide at the center of the triangle. The question is, how long does it take for them to get there? They're spiraling inwards as they turn to keep facing each other, so they don't take a straight route there. But, by symmetry, at every point in time they're still in an equilateral triangle formation, essentially starting from scratch on a slightly smaller and rotated triangle! That means that if you figure out what's going on at the first instant, the same thing applies at every later instant.

From there it's easy. The inward distance from each vertex to the center of the triangle is √3/3. And the component of each ant's velocity that is pointing inward is cos(30)=√3/2. So they will collide in the center in 2/3 of a second.

Update: arithmetic corrected... again.

## 6 comments:

The puzzle is reminiscent of a Niles Pierce ACM 95a riddle and 95b problem set, except there it was a square with four ladybugs.

ah, yeah it generalizes nicely to arbitrary bugs on regular n-gons :)

"By symmetry, they will collide at the center of the triangle."

... if they collide. I mean, you show later how long it takes for them to collide, so obviously they do, but one could certainly imagine symmetric solutions that don't involve collision (either in a finite amount of time, or even in the limit).

ah, yes, good point. it's a simple enough situation that i glossed over that much :)

Isn't the distance from the vertex to the center Sqrt(3)/3, not Sqrt(3)/4?

damnit, you're right. that's like the fourth time i've failed at the arithmetic of this problem :)

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