2023 AMC 10B Problems/Problem 20

Problem

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

$\textbf{(A) } 32 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 48 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 27$

Solution 1

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:

[asy] import graph; import geometry;  unitsize(1cm);  pair A = (-1.41, 1.41); pair B = (1.41, 1.41); pair C = (1.41, -1.41); pair D = (-1.41, -1.41); pair O = (0, 0);  draw(circle(O,2)); draw(A--O--B,black+dashed); draw(C--O--D,black+dashed);  dot(A);dot(B);dot(C);dot(D);dot(O);  label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, SW); label("$O$", (0,0.1), N); [/asy]

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

Solution 2

Assume $A$, $B$, $C$, and $D$ are the four points connecting the semicircles. By law of symmetry, we can pretty confidently assume that $ABCD$ is a square. Then, $\overline{AB} = 2\sqrt2.$, and the rest is the same as the second half of solution $1$.

~jonathanzhou18

Solution 3

We put the sphere to a coordinate space by putting the center at the origin. The four connecting points of the curve have the following coordinates: $A = \left( 0, 0, 2 \right)$, $B = \left( 2, 0, 0 \right)$, $C = \left( 0, 0, -2 \right)$, $D = \left( -2, 0, 0 \right)$.

Now, we compute the radius of each semicircle. Denote by $M$ the midpoint of $A$ and $B$. Thus, $M$ is the center of the semicircle that ends at $A$ and $B$. We have $M = \left( 1, 0, 1 \right)$. Thus, $OM = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$.

In the right triangle $\triangle OAM$, we have $MA = \sqrt{OA^2 - OM^2} = \sqrt{2}$.

Therefore, the length of the curve is \begin{align*} 4 \cdot \frac{1}{2} 2 \pi \cdot MA = \pi \sqrt{32} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(A) 32}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 4

Note that each of the diameters are the chord of the sphere of a quarter arc. Thus, the semicircles diameter's length is $2\sqrt{2}$. Thus, the entire curve is $2\sqrt{2} \cdot \pi \cdot \frac{1}{2} \cdot 4 = 4\sqrt{2} \pi = \sqrt{32} \pi$. Therefore, the answer is $\boxed{\textbf{(A) 32}}$. ~andliu766

Video Solution 1 by OmegaLearn

https://youtu.be/bQfD2S1HS4c

Video Solution

https://youtu.be/nkwCDGYAkiw

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)


See also

2023 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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