Can You Solve This Astronomical Enigma?

Riddler Express

Two cicada broods with periods $A$ and $B$ years have just emerged together. They interfere whenever one brood emerges exactly one year after the other (a multiple of $A$ and a multiple of $B$ differing by $1$). With $A$ and $B$ coprime and both at most $20$, what is the longest a pair can go before its first interference?

Solution

Interference first happens in the earliest year $\max(aA, bB)$ for which a multiple of $A$ and a multiple of $B$ are one year apart, $|aA - bB| = 1$. Because $A$ and $B$ are coprime, such multiples always exist, and we want the pair that pushes this first collision as late as possible. Scanning all coprime pairs up to $20$, the winner is $A = 17$, $B = 19$: $$17 \times 9 = 153, \qquad 19 \times 8 = 152, \qquad |153 - 152| = 1,$$ so the first interference is in year $$\boxed{153.}$$ The two largest coprime periods available are also nearly equal, which forces their multiples to take the longest to drift to within a single year of each other.

The computation

For each coprime pair, find the earliest year where a multiple of $A$ sits one off a multiple of $B$, and take the largest such year.

from math import gcd
def first_interference(a, b):
    return min(max(x, y) for x in range(0, a*b + 1, a)
               for y in range(0, a*b + 1, b) if abs(x - y) == 1)
pairs = [(a, b) for a in range(1, 20) for b in range(a, 21) if gcd(a, b) == 1]
print(max((first_interference(a, b), a, b) for a, b in pairs))  # (153, 17, 19)

Riddler Classic

Three planets orbit a star in circles: planet $A$ at radius $3$ with period $3$ years, $B$ at radius $4$ with period $4$, $C$ at radius $5$ with period $5$. They are collinear right now. When are the three planets next collinear (the star need not be on the line)?

Solution

Planet $i$ sits at $\big(r_i\cos\omega_i t,\ r_i\sin\omega_i t\big)$ with $\omega_i = 2\pi/T_i$. Three points are collinear exactly when the signed area of their triangle vanishes, that is when $$\begin{vmatrix} r_1\cos\omega_1 t & r_1\sin\omega_1 t & 1\\ r_2\cos\omega_2 t & r_2\sin\omega_2 t & 1\\ r_3\cos\omega_3 t & r_3\sin\omega_3 t & 1 \end{vmatrix} = 0.$$ With $(r,\omega) = (3, \tfrac{2\pi}{3}), (4, \tfrac{\pi}{2}), (5, \tfrac{2\pi}{5})$ this expands to $$15\sin\frac{4\pi t}{15} - 12\sin\frac{\pi t}{6} - 20\sin\frac{\pi t}{10} = 0.$$ At $t = 0$ all three sit on the same axis, so that is one solution; the next positive root is $$t \approx \boxed{7.77 \text{ years.}}$$

The computation

Track the three planets’ actual positions and scan for the first time after $t = 0$ when the signed area of their triangle changes sign.

import numpy as np
rw = [(3, 2*np.pi/3), (4, np.pi/2), (5, 2*np.pi/5)]
def area(t):
    (x1, y1), (x2, y2), (x3, y3) = [(r*np.cos(w*t), r*np.sin(w*t)) for r, w in rw]
    return (x2 - x1)*(y3 - y1) - (x3 - x1)*(y2 - y1)
ts = np.linspace(1e-5, 9, 4_000_000)
v = area(ts)
print(ts[:-1][np.diff(np.sign(v)) != 0][0])    # ~7.767