Gil Kalai's blog post https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ poses a riddle about a chase of two points.

Point A chases point B at unit velocity. Point B heads right at unit velocity. At $t = 0$, $A(0) = (0,1)$ and $B(0) = (0,0)$.

Question: what is the limiting distance $||B(t) - A(t)||$ as $t$ goes to infinity?

  • $\begingroup$ My gut tells me that it's 1... $\endgroup$
    – Sambo
    Jul 15, 2018 at 3:43
  • $\begingroup$ I tried writing out the differential equations, which failed because they are not linear and I'm not very good at that. Per your guess, it's definitely lower than 1 (since B is running away at a sub-optimal direction and the distance starts at 1). It will also not reach 0. $\endgroup$
    – mich
    Jul 15, 2018 at 3:44
  • $\begingroup$ The distance between the points is not 1, but $\sqrt2$. I would say that the distance is $\sqrt2-1$. $\endgroup$
    – Moti
    Jul 15, 2018 at 5:59
  • $\begingroup$ @Moti: The distance is $1$. $\endgroup$
    – joriki
    Jul 15, 2018 at 6:07
  • $\begingroup$ You are right... but I still think that the distance is as stated. $\endgroup$
    – Moti
    Jul 15, 2018 at 6:23

2 Answers 2


The problem becomes more managable if you transform to a coordinate system where $\mathbf B$ is at rest. Then $\mathbf A$ has velocity

$$ \pmatrix{\dot x\\\dot y}=-\pmatrix{\frac xr+1\\\frac yr}\;. $$

Now transform to polar coordinates $r=\sqrt{x^2+y^2}$ and $\phi=\arctan\left(\frac yx\right)$, with

$$ \dot r=\frac{x\dot x}r+\frac{y\dot y}r=-1-\frac xr=-1-\cos\phi $$


$$ \dot\phi=\frac{\frac{\dot y}x-\frac{\dot xy}{y^2}}{1+\left(\frac yx\right)^2}=\frac{x\dot y-y\dot x}{x^2+y^2}=\frac y{r^2}=\frac{\sin\phi}r\;. $$

So we have

$$ \frac{\mathrm dr}{\mathrm d\phi}=\frac{\dot r}{\dot\phi}=-r\cdot\frac{1+\cos\phi}{\sin\phi}=-r\frac{\cos\frac\phi2}{\sin\frac\phi2}\;. $$

Dividing by $r$ and integrating both sides yields

$$ \log r=C-2\log\sin\frac\phi2\;. $$

The initial value is $r\left(\frac\pi2\right)=1$, which yields $C=-\log 2$, so

$$ r=\frac1{2\sin^2\frac\phi2}\;. $$

In the limit $t\to\infty$ we have $\phi\to\pi$, and thus $r\to\frac12$.

  • $\begingroup$ You mixed up $\phi$ with $x$ towards the end $\endgroup$
    – Dylan
    Jul 15, 2018 at 8:25
  • $\begingroup$ @Dylan: Indeed I did, thanks -- fixed. $\endgroup$
    – joriki
    Jul 15, 2018 at 8:37

There is a mismatch of notations between your wording and the wording in https://gilkalai.wordpress.com/2018/06/29/test-your-intuition-35-what-is-the-limiting-distance/ . I will refer to the notations from the link instead of the notations in the above question because the wording in the link is more clear.

The figure below shows the result of numerical simulation.

enter image description here

The result isn't $1$ but $\frac12$. This is proved thanks to analytical solving.

The coordinates of the moving points are B$(x,y)$ and A$(t,0)$.

Since the absolute velocity of B is $=1$ in the direction of $\overrightarrow{BA}$ : $$\begin{cases} \frac{dx}{dt}=\frac{t-x}{\sqrt{(t-x)^2+y^2}} \\ \frac{dy}{dt}=\frac{-y}{\sqrt{(t-x)^2+y^2}} \end{cases}$$ $\frac{dy}{dx}=\frac{-y}{t-x}$

Change of variable : $x=u+t$




$$y\frac{du}{dy}=\sqrt{u^2+y^2}+u$$ This is a fist order homogeneous ODE easy to solve. The general solution is : $$u=y\sinh\left(c+\ln(y)\right)$$ The initial conditions $t=0$ , $x=0$ , $y=1$ , $u=x-t=0$ which implies $c=0$ thus : $$u=y\sinh\left(\ln(y)\right)=y\frac{y-\frac{1}{y}}{2}=\frac{y^2-1}{2}$$ $$t-x=-u=\frac{1-y^2}{2}$$ When $x\to\infty \quad;\quad y\to 0\quad;\quad (t-x)\to\frac12$. $$|AB[=\sqrt{(x-t)^2+y^2}\to\frac12$$

  • $\begingroup$ I think in the displayed equation right before "This is a separable ODE" you're missing a factor $\sqrt{u^2+y^2}$? The one in the denominator of the right-hand side of the previous equation. (By the way, my comment "The distance is $1$" under the question referred to Moti's comment that the (original) distance between the points is $\sqrt2$ and not $1$; I wasn't talking about the limit distance there.) $\endgroup$
    – joriki
    Jul 15, 2018 at 11:04
  • $\begingroup$ This was fixed meanwhile. Referring of the wording of the problem from the link, the initial distance is 1. I don't consider the wording of the mich's question. $\endgroup$
    – JJacquelin
    Jul 15, 2018 at 11:20

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