2023 AMC 10B Problem 20

The curve is like the seam on a tennis ball. The main job is to find the size of one semicircle.

1. Where the semicircles meet. Call the four meeting points $P_1, P_2, P_3, P_4$. The four semicircles are the same size and the curve splits the sphere into two matching halves, so the picture is evenly balanced: the four meeting points sit on the sphere's "equator" (a circle of radius $2$ around the center), a quarter turn apart. The semicircles bulge up and down in turn.
2. Distance between neighboring points. Take the center $O$ and two neighboring points $P_1$, $P_2$. Then $OP_1 = OP_2 = 2$ (radii) and the angle at $O$ is $90^\circ$ (a quarter turn). By Pythagoras,
$$P_1P_2 = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}.$$
3. Size of one semicircle. Each semicircle has two neighboring meeting points as the ends of its diameter. So its diameter is $2\sqrt{2}$ and its radius is $\sqrt{2}$.
4. Length of one semicircle. Half of a full circle's length $2\pi r$:
$$\frac{1}{2} \cdot 2\pi \cdot \sqrt{2} = \pi\sqrt{2}.$$
5. Total. Four of them:
$$4\pi\sqrt{2} = \pi\sqrt{16 \cdot 2} = \pi\sqrt{32}.$$
So $n = 32$.

(Check that such a semicircle fits on the sphere. The middle of $P_1P_2$ is $\sqrt{2}$ from $O$. A flat slice of the sphere at distance $\sqrt{2}$ from the center is a circle of radius $\sqrt{2^2 - 2} = \sqrt{2}$, the same size as our semicircle. So it lies on the sphere.)

The answer is $\boxed{\textbf{(A)}\ 32}$.

A circle drawn on a sphere is still an ordinary flat circle, so all you need is the distance between its endpoints. Turning the words "congruent" and "divides into two congruent regions" into a balanced picture with the meeting points a quarter turn apart is the key step; after that, one right triangle gives everything.

Let the four points where consecutive semicircles meet be $P_1, P_2, P_3, P_4$. Since the semicircles are congruent and the curve splits the sphere into two congruent pieces (like the seam of a tennis ball), the picture has fourfold symmetry about an axis through the center: the four junction points lie on a great circle, spaced $90^\circ$ apart, and the semicircles bulge alternately to one side and the other.

Each semicircle joins two adjacent junction points and, being a semicircle, has that segment as its diameter. The chord between adjacent points on a great circle of radius $2$ subtending $90^\circ$ has length $2\sqrt{2}$: it is the hypotenuse of the right isosceles triangle formed with the center, whose legs are radii of length $2$. So each semicircle has radius $\sqrt{2}$ and length $\pi\sqrt{2}$.

Check that such a semicircle really lies on the sphere: the plane circle with that diameter has center at distance $\sqrt{2}$ from the sphere's center, and the sphere's section by any plane at distance $\sqrt{2}$ is a circle of radius $\sqrt{4 - 2} = \sqrt{2}$, matching.

Total length: $4 \cdot \pi\sqrt{2} = 4\sqrt{2}\,\pi = \pi\sqrt{32}$, so $n = 32$.

The answer is $\boxed{\textbf{(A)}\ 32}$.