Difference between revisions of "2023 AMC 10B Problems/Problem 20"
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+ | == Solution 2 == | ||
+ | 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 . What is 𝑛?
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