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Difference between revisions of "2023 AMC 10B Problems/Problem 20"

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~Technodoggo
 
~Technodoggo
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== Solution 2 ==
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Assume <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math> are the four points connecting the semicircles. By law of symmetry, we can pretty confidently assume that <math>ABCD</math> is a square. Then, <math>/overline{AB} = 2sqrt2.</math>, and the rest is the same as the second half of solution <math>1</math>.
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~jonathanzhou18

Revision as of 15:19, 15 November 2023

Problem 20

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?

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 $\overline{AO}=\overline{BO}=\overline{CO}=\overline{DO}=2$, and by Pythagorean therorem, $\overline{AB}=2\sqrt2.$

Each of the four congruent semicircles has the length $AB$ as a diameter (since $AB$ is congruent to $BC,CD,$ and $DA$), 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

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, $/overline{AB} = 2sqrt2.$, and the rest is the same as the second half of solution $1$.

~jonathanzhou18

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