The trivial but wrong way is to go for 1 unit in some direction, then trace the circumference of a unit-radius circle. The road will surely be found this way, but the path length is \(1+2\pi\), which can be improved upon.

Every possible position for the road corresponds to a tangent line to the unit circle centered at the starting location. Therefore, this is a problem about finding a shortest path that touches all tangent lines to a unit circle. To make this more concrete, let us transform the space into polar form as below.

We see that in polar form, the unit circle becomes the line \(r=1\) and each tangent line touching the circle at angle \(\theta_0\) becomes the graph \(r=\sec(\theta-\theta_0)\). The trivial solution corresponds to a path that traces \(r=0\) to \(r=1\) at \(\theta=0\), followed by \(\theta=0\) to \(\theta=2\pi\) at \(r=1\). This is not efficient.

In fact, any continuous path starting at \(r=0\) “reaches the road” (i.e. touches every tangent line) as long as it touches \(r=\sec(\theta)\) somewhere on \(\theta\in [0,\pi/2)\) and somewhere on \(\theta\in (3\pi/2,2\pi]\), and it stays above \(r=1\) between those two points. And up to shifts in \(\theta\), these are the only possible solutions.* Note that the path doesn’t need to be closed, like the trivial path was. So we are left to find the shortest such path (e.g. the one in green above).

Whereas in rectangular form, the shortest path between two points is a straight line, in polar form, the shortest path between two points is along a secant graph. It is a fairly easy matter to show that the shortest path must take the following form:

1. Start from \((r,\theta)=(0,\phi)\) for some \(\phi\in [0,\pi/2)\);
2. Drop onto the graph \(r=\sec(\theta-2\phi)\), between the points \((r,\theta)=(\sec(\phi),\phi)\) and \((r,\theta)=(1,2\phi)\);
3. Take the path \(r=1\) from \(\theta=2\phi\) to some \(\theta=2\pi-2\psi\);
4. Drop onto the graph \(r=\sec(\theta-(2\pi-2\psi))\), between the points \((r,\theta)=(1,2\pi-2\psi)\) and \((r,\theta)=(\sec(2\pi-\psi),2\pi-\psi)\).

Finally, we need to solve two minimizations, which will find the best \(\phi\) and \(\psi\). \(\psi\) can actually be solved by inspection, but formally:

\(\min_{\phi\in[0,\pi/2)} \sec(\phi) + \tan(\phi) + \pi - 2\phi\)
solution at \(\frac{\sin(\phi)}{\cos^2(\phi)} + \sec^2(\phi) = 2 \Rightarrow \sin(\phi)=1/2 \Rightarrow \phi=\pi/6\).

The solved path has total length \(\tan(\pi/4) + \pi – \pi/2 + \sec(\pi/6) + \tan(\pi/6) + \pi – \pi/3 = 1+\sqrt{3}+7\pi/6\), and is plotted below. This is about 12% shorter than the trivial path.

Consider an equilateral triangle ABC with edge length 1. At each vertex is an object that is capable of movement at exactly speed 1. Beginning at time 0, each of the three objects moves toward its initial adjacent neighbor object, as in a game of pursuit. Of course, by symmetry, the objects will meet at the incenter of ABC. The question: how far will they have traveled?

The paths followed by the objects at first seem non-trivial, but in fact turn out to be nice. We will come back to this, but for the problem, the exact paths are irrelevant. We just need to find the amount of time it takes the objects to meet. The key part of the solution is to realize that, due to symmetry, the three objects always define an equilateral triangle at any time. Furthermore, their direction of motion is always along the edges of such a triangle, at speed v=1.

What we have is then a shrinking and rotating equilateral triangle that eventually decays to a point. Let us characterize this process.

At time \(t\), let \(s(t)\) be the length traveled by one of the objects. Let \(l(t)\) be the edge length of the triangle. Let \(A(t)\) be the area of the triangle. Then we have several relationships:

• \(\frac{ds}{dt}=1\)
• \(A(t)=\frac{\sqrt{3}}{4} l(t)^2\), and \(\frac{dA}{dt} = \frac{\sqrt{3}}{2} l(t) \frac{dl}{dt}\)
• By geometry, \(A(t+\Delta t) – A(t) = -3 (l(t) – \frac{ds}{dt} \Delta t) \frac{\sqrt{3}}{4} \frac{ds}{dt} \Delta t + o({\Delta t}^2)\), thus \(\frac{dA}{dt} = -3 \frac{\sqrt{3}}{4} l(t)\)

Combining: \(-\frac{3}{4}\sqrt{3} l(t) = \frac{\sqrt{3}}{2} l(t) \frac{dl}{dt}\)
\(\frac{dl}{dt} = -\frac{3}{2}\). Integrating with boundary conditions, we get \(-\frac{3}{2}t_f = -1\), \(t_f = \frac{2}{3}\), and since speed is 1, \(s(t_f) = t_f = \frac{2}{3}\). This is how far the object travels.

Finally, since the process is self-similar at every step along the path, the path of the object must be a logarithmic spiral. Furthermore, since the linear speed on the path is constant, the angular speed must be exponentially increasing (in angle) toward the center of the spiral.

To solve for this spiral explicitly, we have \(\frac{dr}{dt}=\frac{dr}{d\theta} \frac{d\theta}{dt} = \frac{\sqrt{3}}{3} \frac{dl}{dt} = -\frac{\sqrt{3}}{2}\).
\(\frac{dr}{d\theta} = -k \exp(-k \theta)\), so \(\frac{d\theta}{dt} = \frac{\sqrt{3}}{2k} \exp (k\theta)\), and \(\theta(t) = -\frac{1}{k} \ln(-\frac{\sqrt{3}}{2}t + C)\). Plugging into \(r(\theta) = \exp(-k\theta)\) and applying the boundary condition \(r(\theta_0) = \frac{\sqrt{3}}{3}\), we get \(r(t) = -\frac{\sqrt{3}}{2}t + \frac{\sqrt{3}}{3}\).
Next, \(\int_{\theta_0}^\infty r(\theta) d\theta = \frac{\sqrt{3}}{3k} = \frac{2}{3}\), so \(k=\frac{\sqrt{3}}{2}\) and \(\theta(t) = -\frac{2\sqrt{3}}{3} \ln(-\frac{\sqrt{3}}{2}t + \frac{\sqrt{3}}{3})\).