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Two days ago I recalled a problem I was given a long time ago. The problem is:

Four ants are placed on the vertices of a square with side 1. The ants start moving, each directed towards its left neighbour. What is length of each ant's path up to their meeting.

So obviously they will meet in the center of the square and each ant will have travelled the same path. I assumed that on each step each ant travels $1/n$ (in straight direction) of the distance to its neighbour. That way the path is $\frac{1}{n} + \frac{\sqrt{n^2-2n+2}}{n^2} + \frac{\sqrt{n^2-2n+2}^2}{n^3} + \dots = \sum_{i=1}^{\infty} {\frac{\sqrt{n^2-2n+2}^i}{n^i}=1}$ So I fanally each ant's path is of length equal 1.

That's ok but I do remember that when I was told the problem for first time somebody told me a quite logic solution not involving any convergence. Can anybody of you come up with that beatiful solution?

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Previously: Four turtles/bugs puzzle –  Rahul Mar 20 at 22:37
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1 Answer 1

The tangent vector of each ant at every time forms an angle of $3/4\pi=135°$ with the radius (can you see why? Because they are facing their neighbors!).

(In particular, the ants walk exactly along logarithmic spirals, directed towards the center. But we don't need to know this to perform the calculation. )

This means that $v_r$, the component of the velocity directed towards the center (the one that interests us) is $-|v|/\sqrt{2}$. For simplicity we take $|v|=1$, so $v_r=-1/\sqrt{2}$. In particular, it is constant.

So supposing it starts at $r_0=1/\sqrt{2}$, the half-diagonal of the unit square: $$ r(t) = r_0 + v_rt = 1/\sqrt{2} - 1/\sqrt{2}\,t = 1/\sqrt{2}\,(1-t). $$

So we get to the origin ($r=0$) at $t=1$.

Now $|v|$ is constant, so the length of the path simply $L=|v|t$. Since the ants run for a time $t=1$, we get $L=1\cdot 1=1$.

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