Difference between revisions of "2023 AMC 10B Problems/Problem 20"
m (latex) |
|||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Four congruent semicircles are drawn on the surface of a sphere with radius 2, as | + | Four congruent semicircles are drawn on the surface of a sphere with radius <math>2</math>, as |
shown, creating a close curve that divides the surface into two congruent regions. | shown, creating a close curve that divides the surface into two congruent regions. | ||
− | The length of the curve is <math>\pi\sqrt{n}</math>. What is | + | The length of the curve is <math>\pi\sqrt{n}</math>. What is <math>n</math>? |
<math>\textbf{(A) } 32 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 48 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 27</math> | <math>\textbf{(A) } 32 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 48 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 27</math> |
Revision as of 20:12, 16 November 2023
Contents
Problem
Four congruent semicircles are drawn on the surface of a sphere with radius , as shown, creating a close curve that divides the surface into two congruent regions. The length of the curve is . What is ?
Solution 1
There are four marked points on the diagram; let us examine the top two points and call them and . Similarly, let the bottom two dots be and , as shown:
This is a cross-section of the sphere seen from the side. We know that , and by Pythagorean therorem,
Each of the four congruent semicircles has the length as a diameter (since is congruent to and ), so its radius is Each one's arc length is thus
We have of these, so the total length is , so thus our answer is
~Technodoggo
Solution 2
Assume , , , and are the four points connecting the semicircles. By law of symmetry, we can pretty confidently assume that is a square. Then, , and the rest is the same as the second half of solution .
~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: , , , .
Now, we compute the radius of each semicircle. Denote by the midpoint of and . Thus, is the center of the semicircle that ends at and . We have . Thus, .
In the right triangle , we have .
Therefore, the length of the curve is
Therefore, the answer is .
~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 . Thus, the entire curve is . Therefore, the answer is . ~andliu766
Video Solution 1 by OmegaLearn
See also
2023 AMC 10B (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.