Difference between revisions of "2023 AMC 10B Problems/Problem 20"
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+ | ==Problem== | ||
+ | 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. | ||
+ | The length of the curve is <math>\pi\sqrt{n}</math>. What is <math>n</math>? | ||
+ | |||
+ | [[Image:202310bQ20.jpeg|center]] | ||
+ | |||
+ | <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== | ||
Line 28: | Line 37: | ||
</asy> | </asy> | ||
− | This is a cross-section of the sphere seen from the side. We know that <math> | + | 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 | ||
+ | ~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 == | ||
+ | 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 | ||
+ | |||
+ | ==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>. | ||
+ | We have <math>M = \left( 1, 0, 1 \right)</math>. | ||
+ | Thus, <math>OM = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}</math>. | ||
+ | |||
+ | In the right triangle <math>\triangle OAM</math>, we have <math>MA = \sqrt{OA^2 - OM^2} = \sqrt{2}</math>. | ||
+ | |||
+ | Therefore, the length of the curve is | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 4 \cdot \frac{1}{2} 2 \pi \cdot MA | ||
+ | = \pi \sqrt{32} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | 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 | ||
+ | |||
+ | ==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
Contents
Problem
Four congruent semicircles are drawn on the surface of a sphere with radius , as
shown, creating a close curve that divides the surface into two congruent regions.
The length of the curve is
. What is
?
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 Theorem, length of
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 ~minor edits by JiuruAops
Note:
TLDR:
The radius of gives us a line segment connecting diagonal vertices of the semi-circles with a measure of
, giving us through
relations and Pythagorean theorem a diameter for each semi-circle of
, which we can use to bash out the circumference of a full circle, multiply by
, and move inside and under the root to get
.
~Aryan Mukherjee
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
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: ,
,
,
.
Now, we compute the radius of each semicircle.
Denote by the midpoint of
and
. Thus,
is the center of the semicircle that ends at
and
.
We have
.
Thus,
.
In the right triangle , we have
.
Therefore, the length of the curve is
Therefore, the answer is .
~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 . Thus, the entire curve is
. Therefore, the answer is
.
~andliu766
Solution 5 (Cheese! Narrow it down to 2 choices!) and actual way
Cheese: You can immediately say that the answer choice is either or
because there are four semicircles in that curve; there are
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
or
.
Actual way: Take a cross-section of the sphere to get four different points equidistant from the center
of the sphere,
,
,
,
such that
, and so
is a square with side length
, and proceed as in Solution 1 to get
. ~get-rickrolled ~LaTeX errors fixed by get-rickrolled
Video Solution 1 by OmegaLearn
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.