When Will You Cross Your Path?

Anita the ant walks $1$ inch in a straight line, rotates counterclockwise by an angle $\varphi$, walks $2$ inches, rotates by $\varphi$ again, walks $3$ inches, and so on. At some point her path crosses back over her very first $1$-inch segment. Given that $\varphi$ is the smallest angle for which this ever happens, how long was the segment along which she first crossed the initial segment?

The Fiddler, Zach Wissner-Gross, October 3, 2025(original post)

Solution

Place the start at the origin with the first segment along the positive $x$-axis, so segment $1$ is $\{(x,0):0\le x\le1\}$ and the vertices are $$P_0=(0,0),\quad P_k=P_{k-1}+k\bigl(\cos(k-1)\varphi,\ \sin(k-1)\varphi\bigr).$$ For small $\varphi$ the spiral opens outward and never returns. As $\varphi$ grows, the first segment able to swing back onto the initial one is the fourth. At the smallest such $\varphi$ the crossing is marginal: segment $4$ meets the initial segment exactly at its far endpoint $(1,0)=P_1$. The line of segment $4$ passes through $P_1$ when $\bigl(P_3-P_1\bigr)\times(\cos3\varphi,\sin3\varphi)=0$, which telescopes to $$\sum_{j=2}^{3} j\,\sin\bigl((4-j)\varphi\bigr)=2\sin2\varphi+3\sin\varphi =\sin\varphi\,(4\cos\varphi+3)=0,$$ so $\cos\varphi=-\tfrac34$. At that angle the crossing happens along the fourth segment, of length $\boxed{4\text{ inches}}$.

The computation

Walk the actual spiral: build the vertices, and for a sweep of angles find the first segment that crosses the interior of the initial unit segment. The smallest crossing angle is $\arccos(-\tfrac34)$, and the crossing segment is the fourth.

import numpy as np
from math import cos, sin, acos, pi, degrees
def verts(phi, K):
    P = [(0., 0.), (1., 0.)]
    for k in range(2, K + 1):
        h = (k - 1)*phi; P.append((P[-1][0] + k*cos(h), P[-1][1] + k*sin(h)))
    return P
def crosses(A, B, eps=1e-9):             # interior crossing of {y=0, 0<x<1}?
    (x1, y1), (x2, y2) = A, B
    if y1 == y2: return False
    t = -y1 / (y2 - y1)
    return eps < t < 1 - eps and eps < x1 + t*(x2 - x1) < 1 - eps
for phi in np.linspace(.001, pi - .001, 60000):    # smallest crossing angle
    P = verts(phi, 16)
    hit = next((k for k in range(2, 16) if crosses(P[k-1], P[k])), None)
    if hit: print(round(degrees(phi), 3), "deg, seg", hit); break   # 138.593, 4

Extra Credit

What was the measure of the angle $\varphi$?

Solution

From the marginal condition $\cos\varphi=-\tfrac34$, $$\varphi=\arccos\!\Bigl(-\tfrac34\Bigr)\approx 2.4189\text{ rad}\approx\boxed{138.59^\circ}.$$ (The exact angle is paywalled at the source; this closed form is my own.)

The computation

Pin the angle by bisection: shrink the bracket to the smallest $\varphi$ at which the fourth segment just crosses the initial one. It converges to $\arccos(-\tfrac34)$.

def seg4_crosses(phi):
    P = verts(phi, 4); return crosses(P[3], P[4])
lo, hi = 2.0, 3.0
for _ in range(80):
    mid = (lo + hi) / 2
    lo, hi = (lo, mid) if seg4_crosses(mid) else (mid, hi)
print(round(degrees(hi), 3), round(degrees(acos(-0.75)), 3))   # 138.59  138.59