Posted on June 12, 2026
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.?
Put the island in coordinates as
\[ x^2+y^2\le 1,\qquad y\ge 0. \]
For a fixed wave direction, the wave is a moving line. At 10:05, halfway between first and last contact, that line is halfway between the two supporting lines of the semidisk perpendicular to the wave's motion.
By symmetry, let the wave's normal vector be
\[ n=(\sin\phi,\cos\phi),\qquad 0\le \phi\le \frac{\pi}{2}. \]
A line perpendicular to \(n\) can be written as
\[ n\cdot(x,y)=t. \]
As the wave moves, \(n\) stays fixed and only \(t\) changes. So first contact and last contact happen at the smallest and largest possible values of \(n\cdot(x,y)\) among all points \((x,y)\) on the island.
The largest possible value is \(1\). This is because \(n\) is a unit vector, so the farthest point of the unit circle in direction \(n\) is exactly \(n\) itself, and
\[ n\cdot n=1. \]
That point is on the upper circular arc, so it belongs to the semicircle island.
The smallest possible value occurs on the flat side \(y=0\). On that side,
\[ n\cdot(x,0)=x\sin\phi. \]
Since the flat side runs from \(x=-1\) to \(x=1\), the smallest value is reached at \((-1,0)\), giving
\[ n\cdot(-1,0)=-\sin\phi. \]
At 10:05, the wave is halfway between these two projection values, so the midpoint line has equation
\[ n\cdot (x,y)=c,\qquad c=\frac{1-\sin\phi}{2}. \]
Now the problem has become a chord-length problem: for each possible direction \(\phi\), compute the length of the chord cut out by this midpoint line, then choose the direction that makes that length as large as possible.
The full chord of the unit disk at distance \(c\) from the origin has length
\[ 2\sqrt{1-c^2}=\sqrt{3+2\sin\phi-\sin^2\phi}. \]
But the island is only the upper half of the disk. This full chord lies entirely in the semidisk exactly when it does not dip below the diameter \(y=0\). Writing \(s=\sin\phi\), that condition reduces to
\[ s\le \frac{1}{3}. \]
On that interval, the chord length is
\[ L(s)=\sqrt{3+2s-s^2}, \]
which is increasing for \(0\le s\le 1/3\). Therefore the best full-chord case occurs at \(s=1/3\), giving
\[ L=\sqrt{3+\frac{2}{3}-\frac{1}{9}}=\sqrt{\frac{32}{9}}=\frac{4\sqrt{2}}{3}. \]
For \(s>1/3\), the line is clipped by the flat side of the island, and the resulting segment is shorter. So this is the global maximum.
Answer: \( \boxed{\frac{4\sqrt{2}}{3}} \) miles, about \(1.886\) miles.
Another wave is approaching the island, but no one knows which direction it's coming from; for the moment, all directions are equally likely.
On average, what is the length of the stretch of land directly under the wave halfway between when the wave first and last makes contact with the island?
The same setup gives the length as a function of \(\phi\). Let \(s=\sin\phi\), \(c=(1-s)/2\), and
\[ a=\sqrt{1-c^2}. \]
Here \(a\) is half the corresponding chord length in the full unit disk. If \(0\le s\le 1/3\), the whole chord is inside the semidisk, so
\[ L(\phi)=2a. \]
If \(1/3<s\le 1\), the flat side \(y=0\) cuts the chord, and the length becomes
\[ L(\phi)=a+c\cot\phi. \]
Because directions are uniform and the semidisk is symmetric left-to-right, it is enough to average over \(0\le\phi\le\pi/2\) with density \(2/\pi\). Thus
\[ \mathbb{E}[L]=\frac{2}{\pi}\left( \int_0^{\arcsin(1/3)} 2a\,d\phi+ \int_{\arcsin(1/3)}^{\pi/2} (a+c\cot\phi)\,d\phi \right), \]
where \(c=(1-\sin\phi)/2\) and \(a=\sqrt{1-c^2}\). Numerically, this is
\[ \mathbb{E}[L]\approx 1.304439455. \]
Extra credit answer: \( \boxed{1.304439455\ldots} \) miles.
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