Difference between revisions of "2023 AMC 10B Problems/Problem 20"
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We have <math>4</math> of these, so the total length is <math>4\sqrt2\pi=\sqrt{32}\pi</math>, so thus our answer is <math>\boxed{\textbf{(A) }32.}</math> | We have <math>4</math> of these, so the total length is <math>4\sqrt2\pi=\sqrt{32}\pi</math>, so thus our answer is <math>\boxed{\textbf{(A) }32.}</math> | ||
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| + | ~Technodoggo | ||
Revision as of 13:27, 15 November 2023
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:
![[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]](http://latex.artofproblemsolving.com/0/e/2/0e2436e052e566b8c2cfb41c5674fc19c3e7ece2.png)
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
