What is Anita's 𝝋?

It’s time for you to check Anita’s work. What was the measure of angle $\varphi$?

Ok, so as we saw in the prior Classic Fiddler problem, Anita will be crossing the $1$-inch segment while she is on her $4$-inch line segment, that is when she is traveling from $$X_3 = (1+2\cos \varphi+3\cos 2\varphi, 2\sin \varphi + 3\sin 2\varphi)$$ to $$X_4 = (1 + 2 \cos \varphi + 3 \cos 2\varphi + 4\cos 3\varphi, 2 \sin \varphi + 3 \sin 2\varphi + 4 \sin 3 \varphi).$$ We have already somewhat loosely located $\varphi \in ( \frac{3\pi}{4}, \frac{4\pi}{5} ).$

We see that the line from $X_3$ to $X_4$ is given by $$y - 2 \sin \varphi - 3 \sin 2\varphi = \tan 3\varphi \left( x - 1 - 2 \cos \varphi - 3 \cos 2\varphi \right).$$ Additionally we note that since for large angles (e.g., $\phi = \pi$) the crossing point is near $(0,0)$ and since the crossing point for the smaller $\phi = \frac{4\pi}{5}$ being closer towards the $(1,0)$ endpoint of the initial segment, it stands to reason that for the optimal angle $\varphi,$ that the crossing point would be precisely at the endpoint $(1,0).$ In this cas we would have $$ - 2 \sin \varphi - 3 \sin 2 \varphi = \tan 3\varphi \left( -2 \cos \varphi - 3 \cos 2\varphi \right).$$ Since $$\tan 3 \varphi = \frac{ 3 \tan \varphi - \tan^3 \varphi }{ 1 - 3 \tan^2 \varphi } = \tan \varphi \frac{ 3 - \tan^2 \varphi }{ 1 - 3 \tan^2 \varphi } = \tan \varphi \frac{ 4 \cos^2 \varphi - 1}{ 4 \tan^2 \varphi - 3},$$ $$-2 \cos \varphi - 3 \cos 2\varphi = 3 - 2 \cos \varphi - 6 \cos^2 \varphi,$$ and $$2 \sin\varphi + 3 \sin 2\varphi = 2 \tan \varphi \cos \varphi ( 1 + 3 \cos \varphi ),$$ the above equality is equivalent (which much algebraic expansion and reduction) to \begin{align*} 0 &= \tan \varphi \frac{4 \cos^2 \varphi - 1}{4 \cos^2 \varphi - 3} ( 3 - 2\cos \varphi - 6 \cos^2 \varphi) + 2 \tan \varphi \cos \varphi (1 + 3 \cos \varphi) \\ &= \frac{\tan \varphi}{4 \cos^3 \varphi - 3} \left( (4 \cos^2 \varphi - 1) ( 3 - 2 \cos\varphi - 6 \cos^2 \varphi ) + 2 \cos \varphi ( 4 \cos^2 \varphi - 3 ) (1 + 3 \cos \varphi) \right) \\ &= - \frac{\tan \varphi (4 \cos \varphi + 3)}{4 \cos^2 \varphi - 3},\end{align*} so we have that $\varphi = \cos^{-1} \left(-\frac{3}{4}\right) \approx 2.41885840578\dots.$

Anita's Walk

Anita the ant is going for a walk in the sand, leaving a trail as she goes. First, she walks $1$ inch in a straight line. Then she rotates counterclockwise by an angle $\varphi$, after which she walks another $2$ inches. She rotates counterclockwise an angle $\varphi$ again, after which she walks $3$ inches. She keeps doing this over and over again, rotating counterclockwise an angle $\varphi$ and then walking $1$ inch farther than she did in the previous segment.

At some point during her journey, she crosses over her initial $1$-inch segment. By “cross over,” I am including the two end points of that first segment. Anita realizes that $\varphi$ was the smallest possible angle such that she crossed over her $1$-inch segment. (Among the ants, she’s known for her mathematical prowess.) How long was the segment along which she first crossed over the $1$-inch segment? Your answer should be a whole number of inches.

Let's let Anita's position on the $xy$-plane with origin centered at her trail's beginning after completing her line segment of length $n$, be $X_n = (x_n, y_n)$. In particular we then see that $$x_n = \sum_{k=1}^n k \cos ((k-1) \varphi) \,\,\text{and}\,\, y_n = \sum_{k=1}^n k \sin ((k-1) \varphi).$$ One thing we notice immediately is that as $n \to \infty,$ $X_n$ does not converge and in fact $\|X_n\| \to \infty.$ Therefore, if there is crossing over the initial segment it would have to happen in the first handful of steps, rather than after some extremely large cycle.

We see that if $\phi=\pi,$ then obviously Anita would de facto double back into her first line segment in her $n=2$ step, since she would start retracing her steps immediately. Let's try to see whether or not there can be some $\phi \lt \pi$ such that Anita crosses the first line segment in her $n=3$ step. In this case we see that $X_2 = (1 + 2\cos \phi, 2\sin \phi)$ and $X_3 = (1 + 2 \cos \phi + 3 \cos 2\phi, 2\sin \phi + 3\sin 2\phi),$ so the line connecting $X_2$ and $X_3$ is given by $$y - 2\sin \phi = \tan 2\phi \left( x - 1 - 2 \cos \phi \right).$$ In particular, if we assume that Anita crossed at some point $(h, 0)$ on the initial line segment, that is with $0 \leq h \leq 1,$ then we would get the equality \begin{align*}0 &= 2 \sin \phi \left( 1 - \frac{ \cos \phi \left( 2 \cos \phi + 1 - h \right) }{ \cos 2\phi } \right)\\ &= \frac{2 \sin \phi}{\cos 2\phi} \left( \cos 2\phi - \cos \phi \left( 2\cos \phi + 1 - h \right) \right) \\ &= \frac{2 \sin \phi}{\cos 2\phi} \left( 2 \cos^2 \phi - 1 - \left(2 \cos^2 \phi + (1-h) \cos \phi\right) \right) \\ &= -\frac{2\sin \phi}{\cos 2\phi} \left( 1 + (1-h) \cos \phi \right)\end{align*} For $h \in (0,1],$ we have $-\frac{1}{1-h} \lt -1,$ so there are no solutions of $\phi \in \mathbb{R}$ with $1 + (1-h) \cos \phi = 0.$ Therefore, besides the case of $\phi = \pi$ as we saw before, there are no solutions where Anita crosses the $1$-inch segment on her $n=3$ step.

Let's look at the case of say $\phi = \frac{4\pi}{5}.$ In this case, we see that $X_3 = (0.309017\dots, -1.677599\dots)$ and $X_4=(1.545085\dots, 2.126627\dots),$ so the $n=4$ line segment crosses the $1$-inch line segment at $(h,0)$ for $h\approx 0.854102\dots.$ So there is some value of $\phi$ for which Anita would cross the $1$-inch segment on the $n=4$ segment.

Let's look at another case where $\phi=\frac{3\pi}{4}.$ Here we see the $X_3=(1-\sqrt{2}, \sqrt{2}-3)$ and $X_4=(1+\sqrt{2},3\sqrt{2}-3).$ The line segment between $X_3$ and $X_4$ narrowly missed the $1$-inch segment, crossing at $(4-2\sqrt{2},0)$. We can show (or at least vigorously wave our hands) that there are no angle $\phi \leq \frac{3\pi}{4}$ that there are similarly no instances where Anita would cross the $1$-inch segment.

Therefore, we see that the length of the segment that Anita crossed the initial path with was four.