Can You Roll The Perfect Bowl?

The Riddler for January 31, 2020, inspired by a viral bowls shot. The Express finds the shallowest angle at which a rolling bowl can thread between two others; the Classic maximises the volume of a pyramid built from identical thin triangular tiles.

Riddler Express

Each bowl is a sphere of radius $1$. Your opponent’s two red bowls are separated by a gap of $3$, so their centres are $5$ apart. Your green bowl travels in a straight line; let $\theta$ be the angle between that line and the line joining the two red centres. What is the minimum $\theta$ that lets your bowl pass between the reds without touching either?

The Riddler, FiveThirtyEight, January 31, 2020(original post)

Solution

Follow the centre of the green bowl. To clear a red bowl without touching it, the green centre must stay at least $1 + 1 = 2$ from that red centre. So the green centre traces a straight line that must keep distance at least $2$ from each of the two red centres $A$ and $B$, which are $5$ apart.

The shallowest such line just grazes both reds, sitting at distance exactly $2$ from $A$ and from $B$. By symmetry it passes through the midpoint $M$ of $AB$, with $AM = 2.5$. Drop the perpendicular from $A$ to the line: it has length $2$ (the clearance) and meets the line at a right angle, since a sphere’s radius is perpendicular to a tangent at the point of contact. In the right triangle with hypotenuse $AM = 2.5$ and the side of length $2$ opposite the angle the path makes with $AB$, $$\sin\theta = \frac{2}{2.5} = 0.8, \qquad \boxed{\theta = \arcsin 0.8 \approx 53.13^\circ}.$$ Any larger angle threads the gap with room to spare; any smaller angle drives the green centre within $2$ of a red centre, a collision.

The computation

Re-encode the geometry rather than the formula: send the green centre along a line through the midpoint at angle $\theta$, measure the smallest distance from that line to each red centre, and find the least $\theta$ for which both distances reach the clearance $2$.

import numpy as np
A, B = np.array([-2.5, 0.0]), np.array([2.5, 0.0])   # red centres, 5 apart
def clears(theta_deg):
    t = np.radians(theta_deg)
    d = np.array([np.cos(t), np.sin(t)])             # path direction at M=origin
    dist = lambda P: abs(P[0]*d[1] - P[1]*d[0])      # distance from P to the line
    return dist(A) >= 2 and dist(B) >= 2
thetas = np.linspace(0, 90, 900001)
print(min(th for th in thetas if clears(th)))        # ~53.13

The smallest passing angle is about $53.13^\circ$, matching $\arcsin 0.8$.

Riddler Classic

The Magna-Tiles are identical isosceles triangles with one $30^\circ$ angle and two $75^\circ$ angles. Twelve of them, with their $30^\circ$ corners at the centre, lie flat as a regular dodecagon. Using fewer (between three and eleven) of them in the same way builds a pyramid on a regular-polygon base. To maximise the volume enclosed by the pyramid, how many tiles should you use?

The Riddler, FiveThirtyEight, January 31, 2020(original post)

Solution

Put $n$ tiles together with their $30^\circ$ corners meeting at the apex. Their $30^\circ$ angles total $30n$ degrees; at $n=12$ this fills $360^\circ$ and the figure lies flat, and for $n<12$ it folds up into a pyramid whose base is a regular $n$-gon.

Fix the tile. Let its short side (opposite the $30^\circ$ apex) be $s$ and its two equal legs be $L$. By the law of sines, $s/\sin 30^\circ = L/\sin 75^\circ$, so $L = s\,\sin 75^\circ/\sin 30^\circ$, a constant of the tile. The $n$ short sides form the base, a regular $n$-gon of side $s$ and circumradius $$R = \frac{s}{2\sin(\pi/n)}.$$ The legs are the slant edges from the apex to the base vertices, all of length $L$, so the pyramid’s height is $h = \sqrt{L^2 - R^2}$ and the volume is $$V(n) = \frac13 \,\underbrace{\tfrac12\, n R^2 \sin\!\tfrac{2\pi}{n}}_{\text{base area}} \,\sqrt{L^2 - R^2}.$$ As $n$ grows the base widens ($R$ up) but the pyramid flattens ($h$ down to $0$ at $n=12$, where $R=L$), so the volume peaks at an interior $n$. Evaluating $V(n)$ for $n = 3, \ldots, 11$, the maximum is at $$\boxed{n = 10}$$ (about $73$ cubic inches for a $3$-inch short side). For very large pancakes of $N$ tiles the best fraction approaches $\sqrt{2/3} \approx 81.6\%$ of them, close to $10/12$.

The computation

Build each pyramid from the tile geometry and read off the maximiser: compute $R$, the height $\sqrt{L^2-R^2}$ and the volume for every $n$ from $3$ to $11$.

import math
s   = 3.0                                            # short side (inches)
leg = s * math.sin(math.radians(75)) / math.sin(math.radians(30))
def volume(n):
    R = s / (2 * math.sin(math.pi / n))
    h = math.sqrt(leg**2 - R**2)                     # apex height
    area = 0.5 * n * R*R * math.sin(2 * math.pi / n)
    return area * h / 3
best = max(range(3, 12), key=volume)
print(best, round(volume(best), 1))                  # 10 73.1

The volume is largest at $n = 10$ tiles (about $73$ cubic inches), as boxed.