Can You Tell When The Snow Started?

The Riddler for April 3, 2020. The Express is a social-distancing geometry problem: two walkers swap places on a wide path while staying six feet apart. The Classic is the famous snowplow problem, where the plow’s speed falls as the snow deepens.

Riddler Express

You and someone walking toward you are each in the middle of a wide sidewalk, $12$ feet apart. You pass while always staying at least $6$ feet apart, and you each end up back in the middle, at the other’s starting spot. The other person mirrors your path. What is the shortest total distance you could walk? Extra credit: what if the other person refuses to leave the centre line, so all the dodging is up to you?

The Riddler, FiveThirtyEight, April 3, 2020(original post)

Solution

Because the other person mirrors you, their position is always the point-reflection of yours through the midpoint $M$ of the path. So the distance between you two is exactly twice your distance from $M$, and staying $6$ feet apart means staying at least $3$ feet from $M$ at every moment. You start $6$ feet from $M$ and must reach the other’s start, the antipodal point also $6$ feet from $M$.

The problem reduces to the shortest path between two points $6$ feet from a centre, on opposite sides, that avoids the forbidden disk of radius $3$ around $M$. That path leaves your start along a tangent to the disk, hugs the circle for a short arc, then leaves along a tangent to the destination. Each tangent has length $\sqrt{6^2 - 3^2} = 3\sqrt3$ (the leg of a right triangle with hypotenuse $6$ and one side $3$), and the radius to each tangent point makes angle $\arccos(3/6) = 60^\circ$ with the line to $M$. The two endpoints are $180^\circ$ apart, so the arc spans $180^\circ - 60^\circ - 60^\circ = 60^\circ = \tfrac{\pi}{3}$, of length $3\cdot\tfrac{\pi}{3} = \pi$. Adding the two tangents and the arc, $$2\cdot 3\sqrt3 + \pi = \boxed{6\sqrt3 + \pi \approx 13.53}\ \text{feet}.$$

For the extra credit the other person walks the straight centre line while you do all the dodging, so the symmetry is gone: you must skirt a $6$-foot exclusion disk that moves along the path as they advance. The shortest route is again two tangents and a connecting curve, but the curve now wraps a moving circle, and finding the tangent points reduces to a transcendental equation best solved numerically. The shortest distance you must walk is about $17.45$ feet, roughly four feet more than when your counterpart shares the dodging.

The computation

Encode the geometry of the main problem from the distance and radius alone: the shortest path is two tangents of length $\sqrt{6^2-3^2}$ plus the arc of the radius-$3$ circle between the tangent points.

import numpy as np
D, r = 6.0, 3.0                          # endpoint distance, exclusion radius
tangent = np.sqrt(D**2 - r**2)           # straight run out to each tangent point
half = np.degrees(np.arccos(r / D))      # 60 deg: radius-angle at each tangent point
arc = np.radians(180 - 2*half) * r       # endpoints are 180 deg apart
print(2*tangent + arc, 6*np.sqrt(3) + np.pi)   # 13.5339  13.5339

The tangent-arc-tangent length computed from the geometry is $6\sqrt3 + \pi \approx 13.53$ feet, as boxed.

Riddler Classic

One morning it starts snowing at a constant rate. At noon a snowplow begins to clear the road, and its speed is inversely proportional to the depth of the snow: twice as much snow means half the speed. In its first hour the plow travels 2 miles; in its second hour, only 1 mile. When did it start snowing?

The Riddler, FiveThirtyEight, April 3, 2020(original post)

Solution

Measure time $t$ from the moment the snow began. Snow falls at a constant rate, so its depth is proportional to $t$, and the plow’s speed (inversely proportional to depth) is $\frac{dx}{dt} = \frac{c}{t}$ for some constant $c$. Let $t_s$ be how long the snow had already fallen when the plow set out at noon. The distances in the two hours are $$\int_{t_s}^{t_s+1}\frac{c}{t}\,dt = c\ln\frac{t_s+1}{t_s} = 2, \qquad \int_{t_s+1}^{t_s+2}\frac{c}{t}\,dt = c\ln\frac{t_s+2}{t_s+1} = 1.$$ Dividing the first by the second removes $c$ and gives $\ln\frac{t_s+1}{t_s} = 2\ln\frac{t_s+2}{t_s+1}$, that is, $$\frac{t_s+1}{t_s} = \left(\frac{t_s+2}{t_s+1}\right)^2 \;\Longrightarrow\; t_s^2 + t_s - 1 = 0.$$ The positive root is $$t_s = \frac{\sqrt5 - 1}{2} = \frac{1}{\varphi} \approx 0.618 \text{ hours} \approx 37 \text{ minutes}.$$ So the snow began about $37$ minutes before noon, at roughly $$\boxed{11{:}23 \text{ a.m.}}$$ the golden ratio making a quiet appearance in the melting snow.

The computation

Solve the cubic from the ratio of the two integrals and turn the result into a clock time.

import sympy as sp
ts = sp.symbols('ts', positive=True)
root = sp.solve((ts + 1)**3 - ts*(ts + 2)**2, ts)[0]
print(root, float(root))                       # (sqrt(5)-1)/2 ~ 0.618 hours
minutes = float(root) * 60
print(f"start at 12:00 minus {minutes:.1f} min")  # ~37.1 min before noon