2023 AMC 10B Problems/Problem 20
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 Theorem, length of
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 ~minor edits by JiuruAops
Note:
TLDR:
The radius of gives us a line segment connecting diagonal vertices of the semi-circles with a measure of , giving us through relations and Pythagorean theorem a diameter for each semi-circle of , which we can use to bash out the circumference of a full circle, multiply by , and move inside and under the root to get .
~Aryan Mukherjee
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
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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 |
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