A thief and a policeman A policeman desperately tries to catch a thief that is $a$ meters away. The thief has the constant velocity $v$, and the policeman has the constant velocity $k\cdot v$, with $k > 1$.
The policeman starts at $(0, 0)$ and the thief at $(0, a)$. The thief never changes its direction (not a very smart thief) and always runs straight to the right. The policeman is a bit smarter and always looks directly at the thief while running. At $(a, a)$, the thief is caught.
For illustration, see this image:

Find $k$.
I found a solution (numerical approximation) but the answer is very unexpected :) I used Excel and the method of small steps. For a very small step and about 
$10^6$ iterations, I got the first six digits of the golden ratio. I want to find a proof but I have no clue at all.
 A: I'll simplify this problem by changing the units and the frame of reference. Let's set the distance unit to the initial distance between the thief and the policeman. (This removes the constant $a$ from consideration.) Let's set the velocity unit to the thief's speed, so now the thief has speed $1$ and the policeman has speed $k$. (This removes the constant $v$ from consideration.) This sets all units for mechanics, including time.
Let's translate and rotate the frame of reference so that the thief begins at the origin $(0,0)$, the policeman starts at position $(1,0)$, and the thief runs up the positive $y$ axis. We use the variables $t$ for time, and $x$ and $y$ for position of the policeman on the Cartesian plane. The initial conditions are then $t=0$, $x=1$, $y=0$. The thief's position at time $t$ is $(0,t)$ (since he has velocity $1$) and the policeman is at $(x,y)$. $x$ and $y$ are functions of $t$, but since $x$ is a strictly decreasing function of $t$ and $y$ is a strictly increasing function of $t$ we can consider any of $t,x,y$ to be a one-to-one function of any of the others.

Our first requirement is that the policeman at point $(x,y)$ is running toward the thief at $(0,t)$. This says of the slope of the tangent line to the policeman's path,
$$\frac{dy}{dx}=\frac{y-t}{x}$$
(Another initial condition coming from that is $\frac{dy}{dx}=0$, which is also obvious from the diagram.) Solving for $t$,
$$t=y-x\frac{dy}{dx}$$
Taking the derivative of that and simplifying,
$$\frac{dt}{dx}=-x\frac{d^2y}{dx^2} \tag{1}$$
Another requirement from the problem is that the policeman's speed is $k$. This gives us
$$\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=k$$
Solving for $\frac{dt}{dx}$ and noting that it must be negative,
$$\frac{dt}{dx}=-\frac 1k\sqrt{1+\left(\frac{dy}{dx}\right)^2}$$
Substituting from equation $(1)$ and simplifying a bit yields
$$kx\frac{d^2y}{dx^2}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}$$
Substituting $u=\frac{dy}{dx}$ and separating the variables gives
$$\frac{du}{\sqrt{1+u^2}}=\frac{dx}{kx}$$
Integrating and noting that $x>0$ gives
$$\sinh^{-1}u=\frac 1k\ln x+C_1$$
Substituting the initial conditions $t=0,\ x=1,\ u=\frac{dy}{dx}=0$ shows us that
$$C_1=0$$
Using that and solving for $u$ gives
$$\begin{align}
u&=\sinh\left(\frac 1k\ln x\right) \\[2ex]
 &=\frac 12\left(e^{(\ln x)/k}-e^{(-\ln x)/k}\right) \\[2ex]
\frac{dy}{dx} &= \frac 12x^{1/k}-\frac 12x^{-1/k}
\end{align}$$
Integrating that,
$$y=\frac{k}{2k+2}x^{(k+1)/k}-\frac{k}{2k-2}x^{(k-1)/k}+C_2$$
Substituting initial conditions $t=0,\ x=1,\ y=0$ tells us that
$$C_2=\frac{k}{k^2-1}$$
So we finally get $y$ as a function of $x$:
$$y=\frac{k}{2k+2}x^{(k+1)/k}-\frac{k}{2k-2}x^{(k-1)/k}+\frac{k}{k^2-1}$$
Our final condition is $x=0$, so substituting that we get our final condition for $y$, namely
$$y_{final}=\frac{k}{k^2-1}$$
In your version of this problem and with my units, $y_{final}=1$, so
$$1=\frac{k}{k^2-1}$$
Solving this for $k$ gives a quadratic equation, whose non-redundant solution is
$$k=\frac{1+\sqrt 5}2$$
I.e. $k$ is the golden ratio.
(I left out some details in this exposition: let me know if you need more detail at any particular point.)
