Limiting distance of a chase of two points? 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?
 A: 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
$$
\dot r=\frac{x\dot x}r+\frac{y\dot y}r=-1-\frac xr=-1-\cos\phi
$$
and
$$
\dot\phi=\frac{\frac{\dot y}x-\frac{\dot xy}{y^2}}{1+\left(\frac yx\right)^2}=\frac{x\dot y-y\dot x}{x^2+y^2}=\frac y{r^2}=\frac{\sin\phi}r\;.
$$
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 $r\to\frac12$.
A: 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.

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 : $x=u+t$
$\frac{dx}{dt}=\frac{du}{dt}+1=\frac{-u}{\sqrt{u^2+y^2}}$
$\frac{du}{dt}=-\frac{u}{\sqrt{u^2+y^2}}-1$
$\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}=\frac{dy}{du}\left(-\frac{u}{\sqrt{u^2+y^2}}-1\right)=\frac{-y}{\sqrt{u^2+y^2}}$
$$y\frac{du}{dy}=\sqrt{u^2+y^2}+u$$
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 $x\to\infty \quad;\quad y\to 0\quad;\quad (t-x)\to\frac12$.
$$|AB[=\sqrt{(x-t)^2+y^2}\to\frac12$$
