Four congruent semicircles are drawn on the surface of a sphere with radius $2$, as
shown, creating a close curve that divides the surface into two congruent regions.
The length of the curve is $\pi\sqrt{n}$. What is $n$?

[202310bQ20.jpeg](https://artofproblemsolving.com/wiki/index.php/File:202310bQ20.jpeg)
$\textbf{(A) } 32 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 48 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 27$
There are four marked points on the diagram; let us examine the top two points and call them $A$ and $B$. Similarly, let the bottom two dots be $C$ and $D$, as shown: 

This is a cross-section of the sphere seen from the side. We know that ${AO}={BO}={CO}={DO}=2$, and by Pythagorean Theorem, length of  $\overline{AB}=2\sqrt2.$
Each of the four congruent semicircles has the length $AB$ as a diameter (since $\overline{AB}$ is congruent to $\overline{BC},\overline{CD},$ and $\overline{DA}$), so its radius is $\dfrac{2\sqrt2}2=\sqrt2.$ Each one's arc length is thus $\pi\cdot\sqrt2=\sqrt2\pi.$
We have $4$ of these, so the total length is $4\sqrt2\pi=\sqrt{32}\pi$, so thus our answer is $\boxed{\textbf{(A) }32.}$
~Technodoggo
~minor edits by JiuruAops

Note:
TLDR:
The radius of $2$ gives us a line segment connecting diagonal vertices of the semi-circles with a measure of $4$, giving us through $45^{\circ}-45^{\circ}-90^{\circ}$ relations and Pythagorean theorem a diameter for each semi-circle of $2\sqrt{2}$, which we can use to bash out the circumference of a full circle, multiply by $2$, and move inside and under the root to get $32$.
~Aryan Mukherjee