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2023 AMC 10B Problems/Problem 20: Difference between revisions

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The answers will be uploaded soon
==Solution 1==
- MRP
 
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 $A$ and $B$. Similarly, let the bottom two dots be $C$ and $D$, 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 $\overline{AO}=\overline{BO}=\overline{CO}=\overline{DO}=2$, and by Pythagorean therorem, $\overline{AB}=2\sqrt2.$

Each of the four congruent semicircles has the length $AB$ as a diameter (since $AB$ is congruent to $BC,CD,$ and $DA$), so its radius is $\dfrac{2\sqrt2}2=\sqrt2.$ Each one's arc length is thus $\pi\cdot\sqrt2=\sqrt2\pi.$

We have $4$ of these, so the total length is $4\sqrt2\pi=\sqrt{32}\pi$, so thus our answer is $\boxed{\textbf{(A) }32.}$

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