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Difference between revisions of "2023 AMC 10B Problems/Problem 20"
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==Problem 20==
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==Problem==
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Four congruent semicircles are drawn on the surface of a sphere with radius 2, as
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Four congruent semicircles are drawn on the surface of a sphere with radius <math>2</math>, as
 
shown, creating a close curve that divides the surface into two congruent regions.
 
shown, creating a close curve that divides the surface into two congruent regions.
−
The length of the curve is <math>\pi\sqrt{n}</math>. What is 𝑛?
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The length of the curve is <math>\pi\sqrt{n}</math>. What is <math>n</math>?
 +
 
 +
[[Image:202310bQ20.jpeg|center]]
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<math>\textbf{(A) } 32 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 48 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 27</math>
  
 
==Solution 1==
 
==Solution 1==
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</asy>
 
</asy>
  
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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>  
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This is a cross-section of the sphere seen from the side. We know that <math>{AO}={BO}={CO}={DO}=2</math>, and by Pythagorean Theorem, length of  <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>  
+
Each of the four congruent semicircles has the length <math>AB</math> as a diameter (since <math>\overline{AB}</math> is congruent to <math>\overline{BC},\overline{CD},</math> and <math>\overline{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>
 
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>
  
 
~Technodoggo
 
~Technodoggo
 +
~minor edits by JiuruAops
 +
 +
 +
 +
Note:
 +
 +
TLDR:
 +
 +
The radius of <math>2</math> gives us a line segment connecting diagonal vertices of the semi-circles with a measure of <math>4</math>, giving us through <math>45^{\circ}-45^{\circ}-90^{\circ}</math> relations and Pythagorean theorem a diameter for each semi-circle of <math>2\sqrt{2}</math>, which we can use to bash out the circumference of a full circle, multiply by <math>2</math>, and move inside and under the root to get <math>32</math>.
 +
 +
~Aryan Mukherjee
  
 
== Solution 2 ==
 
== Solution 2 ==
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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>.
+
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} = 2\sqrt2.</math>, and the rest is the same as the second half of solution <math>1</math>.
  
 
~jonathanzhou18
 
~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: <math>A = \left( 0, 0, 2 \right)</math>, <math>B = \left( 2, 0, 0 \right)</math>, <math>C = \left( 0, 0, -2 \right)</math>, <math>D = \left( -2, 0, 0 \right)</math>.
 +
 +
Now, we compute the radius of each semicircle.
 +
Denote by <math>M</math> the midpoint of <math>A</math> and <math>B</math>. Thus, <math>M</math> is the center of the semicircle that ends at <math>A</math> and <math>B</math>.
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We have <math>M = \left( 1, 0, 1 \right)</math>.
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Thus, <math>OM = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}</math>.
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In the right triangle <math>\triangle OAM</math>, we have <math>MA = \sqrt{OA^2 - OM^2} = \sqrt{2}</math>.
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 +
Therefore, the length of the curve is
 +
<cmath>
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\begin{align*}
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4 \cdot \frac{1}{2} 2 \pi \cdot MA
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= \pi \sqrt{32} .
 +
\end{align*}
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</cmath>
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Therefore, the answer is <math>\boxed{\textbf{(A) 32}}</math>.
 +
 +
~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 <math>2\sqrt{2}</math>. Thus, the entire curve is <math>2\sqrt{2} \cdot \pi \cdot \frac{1}{2} \cdot 4 = 4\sqrt{2} \pi = \sqrt{32} \pi</math>. Therefore, the answer is <math>\boxed{\textbf{(A) 32}}</math>.
 +
~andliu766
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 +
==Solution 5 (Cheese! Narrow it down to 2 choices!) and actual way==
 +
Cheese: You can immediately say that the answer choice is either <math>{\text{(A) }32}</math> or <math>{\text{(C) }48}</math> because there are four semicircles in that curve; there are <math>4 = \sqrt{16}</math> semicircles in the curve, so n has to be a multiple of 16, and if you don't know how to do this problem, just guess one of <math>{\text{(A)}}</math> or <math>{\text{(C)}}</math>.
 +
Actual way: Take a cross-section of the sphere to get four different points equidistant from the center <math>O</math> of the sphere, <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> such that <math>AO = BO = CO = DO = 2</math>, and so <math>ABCD</math> is a square with side length <math>2\sqrt{2}</math>, and proceed as in Solution 1 to get <math>\boxed{\textbf{(A) 32}}</math>.      ~get-rickrolled ~LaTeX errors fixed by get-rickrolled
 +
 +
==Video Solution 1 by OmegaLearn==
 +
https://youtu.be/bQfD2S1HS4c
 +
 +
==Video Solution==
 +
 +
https://youtu.be/nkwCDGYAkiw
 +
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 +
 +
==See also==
 +
{{AMC10 box|year=2023|ab=B|num-b=19|num-a=21}}
 +
{{MAA Notice}}

Latest revision as of 11:53, 29 February 2024

Problem

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 $\pi\sqrt{n}$. What is $n$?

202310bQ20.jpeg

$\textbf{(A) } 32 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 48 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 27$

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 ${AO}={BO}={CO}={DO}=2$, and by Pythagorean Theorem, length of $\overline{AB}=2\sqrt2.$

Each of the four congruent semicircles has the length $AB$ as a diameter (since $\overline{AB}$ is congruent to $\overline{BC},\overline{CD},$ and $\overline{DA}$), 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 $\boxed{\textbf{(A) }32.}$

~Technodoggo ~minor edits by JiuruAops


Note:

TLDR:

The radius of $2$ gives us a line segment connecting diagonal vertices of the semi-circles with a measure of $4$, giving us through $45^{\circ}-45^{\circ}-90^{\circ}$ relations and Pythagorean theorem a diameter for each semi-circle of $2\sqrt{2}$, which we can use to bash out the circumference of a full circle, multiply by $2$, and move inside and under the root to get $32$.

~Aryan Mukherjee

Solution 2

Assume $A$, $B$, $C$, and $D$ are the four points connecting the semicircles. By law of symmetry, we can pretty confidently assume that $ABCD$ is a square. Then, $\overline{AB} = 2\sqrt2.$, and the rest is the same as the second half of solution $1$.

~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: $A = \left( 0, 0, 2 \right)$, $B = \left( 2, 0, 0 \right)$, $C = \left( 0, 0, -2 \right)$, $D = \left( -2, 0, 0 \right)$.

Now, we compute the radius of each semicircle. Denote by $M$ the midpoint of $A$ and $B$. Thus, $M$ is the center of the semicircle that ends at $A$ and $B$. We have $M = \left( 1, 0, 1 \right)$. Thus, $OM = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$.

In the right triangle $\triangle OAM$, we have $MA = \sqrt{OA^2 - OM^2} = \sqrt{2}$.

Therefore, the length of the curve is \begin{align*} 4 \cdot \frac{1}{2} 2 \pi \cdot MA = \pi \sqrt{32} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(A) 32}}$.

~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 $2\sqrt{2}$. Thus, the entire curve is $2\sqrt{2} \cdot \pi \cdot \frac{1}{2} \cdot 4 = 4\sqrt{2} \pi = \sqrt{32} \pi$. Therefore, the answer is $\boxed{\textbf{(A) 32}}$. ~andliu766

Solution 5 (Cheese! Narrow it down to 2 choices!) and actual way

Cheese: You can immediately say that the answer choice is either ${\text{(A) }32}$ or ${\text{(C) }48}$ because there are four semicircles in that curve; there are $4 = \sqrt{16}$ semicircles in the curve, so n has to be a multiple of 16, and if you don't know how to do this problem, just guess one of ${\text{(A)}}$ or ${\text{(C)}}$. Actual way: Take a cross-section of the sphere to get four different points equidistant from the center $O$ of the sphere, $A$, $B$, $C$, $D$ such that $AO = BO = CO = DO = 2$, and so $ABCD$ is a square with side length $2\sqrt{2}$, and proceed as in Solution 1 to get $\boxed{\textbf{(A) 32}}$. ~get-rickrolled ~LaTeX errors fixed by get-rickrolled

Video Solution 1 by OmegaLearn

https://youtu.be/bQfD2S1HS4c

Video Solution

https://youtu.be/nkwCDGYAkiw

~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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