Difference between revisions of "2023 AMC 10B Problems/Problem 20"
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| − | + | ==Solution 1== | |
| − | - | + | |
| + | There are four marked points on the diagram; let us examine the top two points and call them <math>A</math> and <math>B</math>. Similarly, let the bottom two dots be <math>C</math> and <math>D</math>, 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 <math>\overline{AO}=\overline{BO}=\overline{CO}=\overline{DO}=2</math>, and by Pythagorean therorem, <math>\overline{AB}=2\sqrt2.</math> | ||
| + | |||
| + | Each of the four congruent semicircles has the length <math>AB</math> as a diameter (since <math>AB</math> is congruent to <math>BC,CD,</math> and <math>DA</math>), so its radius is <math>\dfrac{2\sqrt2}2=\sqrt2.</math> Each one's arc length is thus <math>\pi\cdot\sqrt2=\sqrt2\pi.</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> | ||
Revision as of 13:21, 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 
