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Difference between revisions of "2024 AMC 10A Problems/Problem 13"

(→‎Solution 1)
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Note: It is easier to just visualize this problem instead of actually calculating points on paper.
 
Note: It is easier to just visualize this problem instead of actually calculating points on paper.
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== Solution 2 ==
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Label the transformations as follows:
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• a translation 2 units to the right <math>(A)</math>
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 +
• a 90°- rotation counterclockwise about the origin <math>(B)</math>
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• a reflection across the 𝑥-axis <math>(C)</math>
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• a dilation centered at the origin with scale factor 2 <math>(D)</math>
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Now, we count each transformation individually. It is not hard to see that <math>AC, BD, and CD</math> are commutative (an easy way to test commutativity for some cases would be to have the original point on the x-axis).
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Therefore, only <math>\boxed{\text{(C) }3}</math> transformation pairs commute.
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~xHypotenuse
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=12|num-a=14}}
 
{{AMC10 box|year=2024|ab=A|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:52, 9 November 2024

Problem

Two transformations are said to commute if applying the first followed by the second gives the same result as applying the second followed by the first. Consider these four transformations of the coordinate plane:

• a translation 2 units to the right,

• a 90°- rotation counterclockwise about the origin,

• a reflection across the 𝑥-axis, and

• a dilation centered at the origin with scale factor 2 .

Of the 6 pairs of distinct transformations from this list, how many commute?


$\textbf{(A)1}\qquad\textbf{(B)2}\qquad\textbf{(C)3}\qquad\textbf{(D)4}\qquad\textbf{(E)5}$

Solution 1

Label the transformations as follows:

• a translation 2 units to the right $(W)$

• a 90°- rotation counterclockwise about the origin $(X)$

• a reflection across the 𝑥-axis $(Y)$

• a dilation centered at the origin with scale factor 2 $(Z)$


Now, examine each possible pair of transformations with the point $(1,0)$

$1.$ $W$ and $X$. $W\rightarrow X$ ends with the point $(0,3)$. Going $X\rightarrow W$ ends in the point $(1,2)$, so this pair does not work

$2.$ $W$ and $Y$. $W\rightarrow Y$ gives the point $(3,0)$, and going $Y\rightarrow W$ ends in the same point. This pair is valid.

$3.$ $W$ and $Z$. $W\rightarrow Z$ ends in the point $(6,0)$, while going the other way gives $(4,0)$. This pair isn't commute.

$4.$ $X$ and $Y$. $X\rightarrow Y$. gives the point $(0,-1)$, while the other way gives $(0,1)$. Not a valid pair

$5.$ $X$ and $Z$. $X\rightarrow Z$ ends in the point $(0,2)$, and $Z\rightarrow X$ also ends in $(0,2)$. This pair works.

$6.$ $Y$ and $Z$. $Y\rightarrow Z$ gives the point $(2,0)$, and going the other way also ends in $(2,0)$. This pair is valid.

Therefore, the answer is $\boxed{\text{(C) }3}$ ~Tacos_are_yummy_1

Note: It is easier to just visualize this problem instead of actually calculating points on paper.


Solution 2

Label the transformations as follows:

• a translation 2 units to the right $(A)$

• a 90°- rotation counterclockwise about the origin $(B)$

• a reflection across the 𝑥-axis $(C)$

• a dilation centered at the origin with scale factor 2 $(D)$


Now, we count each transformation individually. It is not hard to see that $AC, BD, and CD$ are commutative (an easy way to test commutativity for some cases would be to have the original point on the x-axis).

Therefore, only $\boxed{\text{(C) }3}$ transformation pairs commute.


~xHypotenuse

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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