The longest wave

Semicircle Island is shaped like a perfect semicircle (or semidisk, technically), with a radius of $1$ mile. It doesn’t have any permanent residents, but it’s a very popular destination for surfers.

Rumor has it that a big wave is headed toward the island. This thin, straight wall of water never changes speed or direction. It will first make contact with the island at 10 a.m. and it will last be in contact with the island at 10:10 a.m. What is the longest possible stretch of land that is directly under the wave at 10:05 a.m.?

With my deepest apologies for my bad Paint skills, let's use a modified version of picture from the Extra Credit prompt and note that I inserted a ray that is perpendicular to the direction of travel of the wave that happens to make an angle of $\theta$ with the positive $x$-axis.

Let's assume that the center of the semi-disk is the point $(0,0),$ the place where the wave first hits the island is at the point $(\cos \theta, \sin \theta)$ and the formula for the ray connecting the origin and this point is $y = x \tan \theta$ as long as $\theta \ne \frac{\pi}{2}.$ Since the wave is perpendicular to this ray, the wave's slope must be the negative reciprocal of that of the ray, so we have the fomula for the wave at 10:00 a.m. is $y= -x \cot \theta + \csc \theta,$ which works as long as $x \not\in \{ 0, \pi\}.$ We similarly see that since the wave at each future point will be parallel to this line that the equation that models the position of the wave at 10:10 a.m. is either $y= -(x+1) \cot \theta,$ if $\theta \in (0, \frac{\pi}{2}),$ or $y = -(x-1) \cot \theta$ if $\theta \in ( \frac{\pi}{2}, \pi ).$ For simplicity, and by symmetry let's just assume going forward that $\theta \in (0, \frac{\pi}{2}).$ In this case, we see that the formula that models the position of the wave at 10:05 a.m., based on the uniform motion of the wave, is then $$y = -x \cot \theta + \frac{\left(\csc \theta + (-\cot \theta)\right)}{2} = -x \cot \theta + \frac{1 - \cos \theta}{2 \sin \theta}.$$

Let's pause here and think about what the length of a chord of a full circle would be. For simplicity let's assume that the chord is parallel to the $x$-axis, at say $y = h.$ In this case, we see that the endpoints of the chord are at $(\pm \sqrt{1-h^2}, h),$ so the length of the curve is $2\sqrt{1-h^2}.$ We further notice that if we rotate the entire $xy$-plane clockwise by angle $\frac{\pi}{2} - \theta,$ then the positive $y$-axis would be mapped to the ray $y= x \tan \theta,$ as we had before. In the case of the circle case, the chord length is obviously preserved in this orthogonal rotation, but now we want to only include the portion of the rotated chord that is now above the $x$-axis. We see that the portion of the chord that was in quadrant II before the rotation is still definitely in the upper half-plane post-rotation, so we should get at least $\sqrt{1-h^2}.$ The only thing to calculate is how much of the half-chord that was in the quadrant I. Firstly, it is perhaps the case depending on the height $h$ and rotational angle $\theta$ that the entirety of the half-chord is in the upper half-plane post-rotation, so the maximum that we could ever get is $\sqrt{1-h^2}.$ On the other hand, if we look at the right triangle formed by the ray $y = x\tan \theta,$ the rotated half-chord and the $x$-axis. Using trigonometry we see that the side with length $h$ is adjacent to the angle whose measure is \theta, with the half-chord opposite that angle, so we have the length of the half-chord in the upper half-plane as $h \tan \theta.$ So putting this altogether, we see that in a case where the we have the line $y=h$ and a rotation of the xy-plane clockwise by $\frac{\pi}{2}-\theta$ then the length of the chord that remains in the upper half-plane is $$\ell(h, \theta) = \min \{ 2 \sqrt{1-h^2}, \sqrt{1-h^2} + h \tan \theta \}.$$

That's cool an all, but let's return to our particular wave. We see that the we certainly have $\theta,$ by design, but all we need is to determine $h$ in this case. In the case of our wave that is modeled by the line $$y = -x\cot \theta + \frac{1- \cos \theta}{\sin \theta},$$ we can use that same right triangle that we used in the generic case above and the fact that the line crosses the $x$-axis as the point $x = \frac{\sec \theta - 1}{2}$ to determine that $$h = \frac{\sec \theta - 1}{2} \cos \theta = \frac{1 - \cos \theta}{2}.$$ Therefore, we see that the length of the stretch of land covered at 10:05 a.m. if it first touches the island at a point $(\cos \theta, \sin \theta)$ is \begin{align*}\ell(\theta) = \ell\left(\frac{1- \cos\theta}{2}, \theta\right) &= \min \left\{ 2 \sqrt{ 1 - \left(\frac{1 - \cos \theta}{2} \right)^2 }, \frac{1 - \cos \theta}{2}\tan \theta + \sqrt{ 1 - \left( \frac{1 - \cos \theta}{2} \right)^2 } \right\} \\ &= \min \left\{ \sqrt{ (3 - \cos \theta) (1 + \cos \theta) }, \frac{\tan \theta ( 1 - \cos \theta) + \sqrt{(3-\cos \theta)(1 + \cos \theta)} }{2} \right\},\end{align*} for $\theta \in (0, \frac{\pi}{2}).$ Analyzing the parts we see that $\tan \theta (1 - \cos \theta)$ is always increasing on this interval, while the term within the square root is always decreasing, therefore we can reason that that maximal length occurs exactly at the cutover point when $$\tan \theta (1 - \cos \theta) = \sqrt{ (3 - \cos \theta) (1 + \cos \theta) }.$$ While I am sure there are many who may want to try to solve analytically, from a geometric intuition perspective, this cutover occurs exactly when the point where the wave crosses the $x$-axis is at the point $(1,0)$, that is, when $$\frac{\sec \theta^* - 1}{2} = 1,$$ or $\sec \theta^* = 3,$ or $\theta^* = \cos^{-1} \frac{1}{3}.$ At this critical point, the longest possible stretch of land that the wave is covering at 10:05 a.m. is $$\ell^* = \ell( \cos^{-1} \frac{1}{3} ) = \sqrt{ (3 - \frac{1}{3}) (1 + \frac{1}{3} ) } = \frac{4}{3} \sqrt{2} \approx 1.88561808316\dots$$ miles.

For absolute completeness we can cover the cases of $\theta = 0,$ in which case wave is represented by vertical lines and at 10:05 a.m., the wave would be covering the unit interval along the positive $y$-axis and have a length of one mile. For the case of $\theta = \frac{\pi}{2},$ where the wave is represented by horizontal lines and the wave would be at $y= \frac{1}{2}$ and cover a distance of $2\sqrt{1 - \frac{1}{2}} = \sqrt{3} \lt \frac{4}{3} \sqrt{2}$ at 10:05 a.m. By symmetry, we can cover the case of $\theta \in (\frac{\pi}{2} , \pi)$ and by another symmetry we can cover the case of what if instead of last hitting the point at $(1,0)$ the wave first hits the point at $(1,0)$ and then only at 10:10 a.m. arrives at the point $(\cos \theta, \sin \theta),$ to show that there is certainly not a larger possible stretch of land to be found if instead of the subset $(0, \frac{\pi}{2})$ that we spent most of our time on, the wave came with an orientation of \theta with respect to the positive $x$-axis for some $\theta \in (\frac{\pi}{2}, 2\pi)$... but more on this later.