Difference between revisions of "2022 AMC 10A Problems/Problem 18"
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− | {{duplicate|[[2022 AMC 10A Problems/Problem | + | {{duplicate|[[2022 AMC 10A Problems/Problem 18|2022 AMC 10A #18]] and [[2022 AMC 12A Problems/Problem 18|2022 AMC 12A #18]]}} |
==Problem== | ==Problem== | ||
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<math>\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721 </math> | <math>\textbf{(A) } 359 \qquad \textbf{(B) } 360 \qquad \textbf{(C) } 719 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 721 </math> | ||
+ | ==Solution 1== | ||
+ | Let <math>P=(r,\theta)</math> be a point in polar coordinates, where <math>\theta</math> is in degrees. | ||
+ | |||
+ | Rotating <math>P</math> by <math>k^{\circ}</math> counterclockwise around the origin gives the transformation <math>(r,\theta)\rightarrow(r,\theta+k^{\circ}).</math> Reflecting <math>P</math> across the <math>y</math>-axis gives the transformation <math>(r,\theta)\rightarrow(r,180^{\circ}-\theta).</math> Note that | ||
+ | <cmath>\begin{align*} | ||
+ | T_k(P)&=(r,180^{\circ}-\theta-k^{\circ}), \\ | ||
+ | T_{k+1}(T_k(P)) &= (r,\theta -1^{\circ}). | ||
+ | \end{align*}</cmath> | ||
+ | We start with <math>(1,0^{\circ})</math> in polar coordinates. For the sequence of transformations <math>T_1, T_2, T_3, \cdots, T_k,</math> it follows that | ||
+ | |||
+ | * After <math>T_1,</math> we have <math>(1,179^{\circ}).</math> | ||
+ | |||
+ | * After <math>T_2,</math> we have <math>(1,-1^{\circ}).</math> | ||
+ | |||
+ | * After <math>T_3,</math> we have <math>(1,178^{\circ}).</math> | ||
+ | |||
+ | * After <math>T_4,</math> we have <math>(1,-2^{\circ}).</math> | ||
+ | |||
+ | * After <math>T_5,</math> we have <math>(1,177^{\circ}).</math> | ||
+ | |||
+ | * After <math>T_6,</math> we have <math>(1,-3^{\circ}).</math> | ||
+ | |||
+ | * ... | ||
+ | |||
+ | * After <math>T_{2k-1},</math> we have <math>(1,180^{\circ}-k^{\circ}).</math> | ||
+ | |||
+ | * After <math>T_{2k},</math> we have <math>(1,-k^{\circ}).</math> | ||
+ | |||
+ | The least such positive integer <math>k</math> is <math>180.</math> Therefore, the least such positive integer <math>n</math> is <math>2k-1=\boxed{\textbf{(A) } 359}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
==Solution 2== | ==Solution 2== | ||
− | Note that since we're reflecting across the <math>y</math>-axis, if the point ever makes it to <math>(-1,0)</math> then it will flip back to the original point. Note that after <math>T_1</math> the point will be <math>1</math> degree clockwise from the negative <math>x</math>-axis. Applying <math>T_2</math> will rotate it to be <math>1</math> degree counterclockwise from the negative <math>x</math>-axis, and then flip it so that it is <math>1</math> degree clockwise from the positive <math>x</math>-axis. Therefore, after every <math>2</math> transformations, the point rotates <math>1</math> degree clockwise. To rotate it so that it will rotate <math>179</math> degrees clockwise will require <math>179 \cdot 2 = 358</math> transformations. Then finally on the last transformation, it will rotate on to <math>(-1,0)</math> and then flip back to | + | Note that since we're reflecting across the <math>y</math>-axis, if the point ever makes it to <math>(-1,0)</math> then it will flip back to the original point. Note that after <math>T_1</math> the point will be <math>1</math> degree clockwise from the negative <math>x</math>-axis. Applying <math>T_2</math> will rotate it to be <math>1</math> degree counterclockwise from the negative <math>x</math>-axis, and then flip it so that it is <math>1</math> degree clockwise from the positive <math>x</math>-axis. Therefore, after every <math>2</math> transformations, the point rotates <math>1</math> degree clockwise. To rotate it so that it will rotate <math>179</math> degrees clockwise will require <math>179 \cdot 2 = 358</math> transformations. Then finally on the last transformation, it will rotate on to <math>(-1,0)</math> and then flip back to its original position. Therefore, the answer is <math>358+1 = 359 = \boxed{\textbf{(A) } 359}</math>. |
~KingRavi | ~KingRavi | ||
==Solution 3== | ==Solution 3== | ||
− | + | In degrees: | |
+ | |||
+ | Starting with <math>n=0</math>, the sequence goes <math>{0}\rightarrow {179}\rightarrow {359}\rightarrow {178}\rightarrow {358}\rightarrow {177}\rightarrow {357}\rightarrow\cdots.</math> | ||
+ | |||
+ | We see that it takes <math>2</math> steps to downgrade the point by <math>1^{\circ}</math>. Since the <math>1</math>st point in the sequence is <math>{179}</math>, the answer is <math>1+2(179)=\boxed{\textbf{(A) } 359}.</math> | ||
− | + | ==Solution 4 (Simple)== | |
+ | We recognize that the reflection actually just makes it turn clockwise. So, each time it turns the step degree. Looking at the answer choices, the closest number to make a multiple of 360 is 359, which leads us to our answer D. | ||
− | + | This is because (1+359)359/2 = something with 360, so we get the reflection back to the spot. | |
==Video Solution== | ==Video Solution== | ||
Line 26: | Line 62: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution (Simple and Fun!!!)== | ||
+ | https://youtu.be/7yAh4MtJ8a8?si=2UC_9X7DjkL8UW5C&t=4968 | ||
+ | |||
+ | ~Math-X | ||
==See also== | ==See also== |
Latest revision as of 11:13, 26 August 2024
- The following problem is from both the 2022 AMC 10A #18 and 2022 AMC 12A #18, so both problems redirect to this page.
Contents
Problem
Let be the transformation of the coordinate plane that first rotates the plane
degrees counterclockwise around the origin and then reflects the plane across the
-axis. What is the least positive
integer
such that performing the sequence of transformations
returns the point
back to itself?
Solution 1
Let be a point in polar coordinates, where
is in degrees.
Rotating by
counterclockwise around the origin gives the transformation
Reflecting
across the
-axis gives the transformation
Note that
We start with
in polar coordinates. For the sequence of transformations
it follows that
- After
we have
- After
we have
- After
we have
- After
we have
- After
we have
- After
we have
- ...
- After
we have
- After
we have
The least such positive integer is
Therefore, the least such positive integer
is
~MRENTHUSIASM
Solution 2
Note that since we're reflecting across the -axis, if the point ever makes it to
then it will flip back to the original point. Note that after
the point will be
degree clockwise from the negative
-axis. Applying
will rotate it to be
degree counterclockwise from the negative
-axis, and then flip it so that it is
degree clockwise from the positive
-axis. Therefore, after every
transformations, the point rotates
degree clockwise. To rotate it so that it will rotate
degrees clockwise will require
transformations. Then finally on the last transformation, it will rotate on to
and then flip back to its original position. Therefore, the answer is
.
~KingRavi
Solution 3
In degrees:
Starting with , the sequence goes
We see that it takes steps to downgrade the point by
. Since the
st point in the sequence is
, the answer is
Solution 4 (Simple)
We recognize that the reflection actually just makes it turn clockwise. So, each time it turns the step degree. Looking at the answer choices, the closest number to make a multiple of 360 is 359, which leads us to our answer D.
This is because (1+359)359/2 = something with 360, so we get the reflection back to the spot.
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (Simple and Fun!!!)
https://youtu.be/7yAh4MtJ8a8?si=2UC_9X7DjkL8UW5C&t=4968
~Math-X
See also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.