30 August 2021 · Problems Maths

Radiodrome prodrome (27/08/21 Riddler Classic)

Original puzzle

If the field length is twice the goal line length, the chaser must run $\dfrac{1+\sqrt{17}}{4}\approx 1.28078$ times as fast as the target.

Fig 1: The Chase!

The blue curve is the result of a chaser travelling at constant speed and always directly towards the instantaneous position of the target. The name for such a curve is the radiodrome or pursuit curve.

For simplicity, we select units such that the length of the field is $1$ and the length is $2$. Actually, let's keep it general; suppose the interception happens at some y-coordinate $L$.

Let the chaser's position be $(x(t),y(t))$ with $(x(0),y(0))=(0,0)$. The target's position is $(1,t)$. This means the target moves at speed $1$ while the chaser moves at $v$.

Here are two methods to solve for $v$ such that $(x(L),y(L))=(1,L)$, indicating the interception occurs at $t=L$.

Method #1: Calculus-free approach

We first consider the chaser's vertical motion. We know the chaser, travelling vertically at speed $\dot{y}(t)$, must travel a total of $L$ in the vertical direction within the time period $t\in[0,L]$. This fact is a surprise tool that will help us later.

Now, consider $g(t)$ which is the length of the green line. Notice that $g(0)=1$ but $g(L)=0$, meaning that over $t\in[0,L]$, the line has on net contracted by $1$. This contraction is the result of the chaser's and target's movements:

Fig 2: The red triangle is similar to the blue triangle, scaled down by $v$.

And so, $g(t)$'s net contraction of $1$ must be the chaser's contraction $\color{blue}vL$ minus the target's elongation $\color{red}\dfrac{L}{v}$:

$$\begin{align*}\therefore 1 &= vL - \frac{L}{v}\\\therefore 0&=v^2L-v-L\\\therefore v &= \color{red}{\boxed{\frac{1+\sqrt{1+4L^2}}{2L}}}\end{align*}$$

noting that we ignore the other root as that will result in a negative $v$.

For a sanity check, we can express this equation also as $$v=\frac{1}{2}\Big(\frac{1}{L}+\sqrt{\frac{1}{L^2}+1}\Big)$$which clearly monotonically increases as $L$ decreases. This confirms our intuition that the faster the chaser runs, the sooner they'll intercept the target.

With our actual problem where $L=2$, we have the aforementioned solution that $$\color{red}{\boxed{v=\frac{1+\sqrt{17}}{4}\approx 1.28078}}$$

Method #2: Solving for the radiodrome exactly

We can also arrive at our answer by first finding the expression $y(x)$ that describes the radiodrome curve.

As mentioned before, the slope of the green line, which is also the tangent of the radiodrome (blue line), is $$\frac{dy}{dx}=y'=\frac{t-y}{1-x}$$

Multiplying both sides by $(1-x)$ and taking the derivative wrt. $x$:

$$\begin{align*}y'&=\frac{t-y}{1-x}\\y'(1-x)&=t-y\\\therefore y''(1-x)-y'&=\frac{dt}{dx}-y'\\\therefore y''\cdot\dot{x}(t)&=\frac{1}{1-x}\end{align*}$$

We can deduce $\dot{x}(t)$ given that the tangent line has slope $\dfrac{t-y}{1-x}$:


Substituting this into the above equation, we obtain a differential equation that is purely geometric (ie. time-independent): $$\frac{y''}{\sqrt{1+y'^2}}=\frac{1}{v(1-x)}$$

The LHS is simply $\frac{d}{dx}\sinh^{-1}(y')$. Hence we integrate both sides wrt. $x$: $$\sinh^{-1}(y')=-\frac{1}{v}\ln(1-x)+C_1$$

To determine the constant $C_1$, we use the initial condition $y'=0$ when $x=0$, indicating that the chaser's initial movement is purely horizontal. Substituting $y'=0, x=0$ into the above yields $C_1=0$. Manipulating the above further:

$$\begin{align*}\sinh^{-1}(y')&=-\frac{1}{v}\ln(1-x)\\\therefore y'&=\sinh\Big(-\frac{1}{v}\ln(1-x)\Big)\\&=\frac{1}{2}\Big(\exp\big(-\frac{1}{v}\ln(1-x)\big)-\exp\big(\frac{1}{v}\ln(1-x)\big)\Big)\\&=\frac{1}{2}\Big((1-x)^{-1/v}-(1-x)^{1/v}\Big)\\\therefore y&=\frac{1}{2}\int \Big((1-x)^{-1/v}-(1-x)^{1/v}\Big)dx\\&=-\frac{1}{2}\Big(\frac{(1-x)^{1-1/v}}{1-1/v}-\frac{(1-x)^{1+1/v}}{1+1/v}\Big)+C_2\end{align*}$$

To determine the constant $C_2$, we use the other boundary condition, which is that the interception occurs at $(1,L)$. Substituting $x=1, y=L$ into the above yields $C_2=L$. Now, to determine the value of $v$, we can use our initial condition of $(0,0)$ again by substituting $x=0,y=0$ into the above equation:

$$\begin{align*}0 &= L - \frac{1}{2}\Big(\frac{1}{1-1/v}-\frac{1}{1+1/v}\Big)\\&=L-\frac{1}{2}\Big(\frac{v}{v-1}-\frac{v}{v+1}\Big)\\&=L-\frac{v}{v^2-1}\\\therefore 0 &= v^2L-v-L\\\therefore v&=\frac{1+\sqrt{1+4L^2}}{2L}\end{align*}$$

which is the same expression for $v$ as in Method #1.

Having kept the solution general means we have some fun numbers showing up. For instance, if the field is square and $L=1$, we have $v=\frac{1+\sqrt{5}}{2}$, highlighting yet again how the golden ratio appears at the most surprising times.

Fig 3: A golden chase