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Difference between revisions of "2024 AMC 10A Problems/Problem 13"
(→‎Solution 1 (Generalized): Added in counterexamples to speed up.)
 
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<math>\textbf{(A)}~1\qquad\textbf{(B)}~2\qquad\textbf{(C)}~3\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5</math>
 
<math>\textbf{(A)}~1\qquad\textbf{(B)}~2\qquad\textbf{(C)}~3\qquad\textbf{(D)}~4\qquad\textbf{(E)}~5</math>
  
−
== Solution 1 (General) ==
+
== Solution 1 (Generalized) ==
 
Label the given transformations <math>T_1, T_2, T_3,</math> and <math>T_4,</math> respectively. The rules of transformations are:
 
Label the given transformations <math>T_1, T_2, T_3,</math> and <math>T_4,</math> respectively. The rules of transformations are:
  
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   <li>Applying <math>T_1</math> and then <math>T_2</math> gives <math>(x,y)\to(x+2,y)\to(-y,x+2).</math> <p>
 
   <li>Applying <math>T_1</math> and then <math>T_2</math> gives <math>(x,y)\to(x+2,y)\to(-y,x+2).</math> <p>
 
Applying <math>T_2</math> and then <math>T_1</math> gives <math>(x,y)\to(-y,x)\to(-y+2,x).</math> <p>
 
Applying <math>T_2</math> and then <math>T_1</math> gives <math>(x,y)\to(-y,x)\to(-y+2,x).</math> <p>
−
Therefore, <math>T_1</math> and <math>T_2</math> do not commute.
+
Therefore, <math>T_1</math> and <math>T_2</math> do not commute. One counterexample is the preimage <math>(0,0).</math>
 
</li><p>
 
</li><p>
 
<li>Applying <math>T_1</math> and then <math>T_3</math> gives <math>(x,y)\to(x+2,y)\to(x+2,-y).</math> <p>
 
<li>Applying <math>T_1</math> and then <math>T_3</math> gives <math>(x,y)\to(x+2,y)\to(x+2,-y).</math> <p>
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<li>Applying <math>T_1</math> and then <math>T_4</math> gives <math>(x,y)\to(x+2,y)\to(2x+4,2y).</math> <p>
 
<li>Applying <math>T_1</math> and then <math>T_4</math> gives <math>(x,y)\to(x+2,y)\to(2x+4,2y).</math> <p>
 
Applying <math>T_4</math> and then <math>T_1</math> gives <math>(x,y)\to(2x,2y)\to(2x+2,2y).</math> <p>
 
Applying <math>T_4</math> and then <math>T_1</math> gives <math>(x,y)\to(2x,2y)\to(2x+2,2y).</math> <p>
−
Therefore, <math>T_1</math> and <math>T_4</math> do not commute.
+
Therefore, <math>T_1</math> and <math>T_4</math> do not commute. One counterexample is the preimage <math>(0,0).</math>
 
</li><p>
 
</li><p>
 
<li>Applying <math>T_2</math> and then <math>T_3</math> gives <math>(x,y)\to(-y,x)\to(-y,-x).</math> <p>
 
<li>Applying <math>T_2</math> and then <math>T_3</math> gives <math>(x,y)\to(-y,x)\to(-y,-x).</math> <p>
 
Applying <math>T_3</math> and then <math>T_2</math> gives <math>(x,y)\to(x,-y)\to(y,x).</math> <p>
 
Applying <math>T_3</math> and then <math>T_2</math> gives <math>(x,y)\to(x,-y)\to(y,x).</math> <p>
−
Therefore, <math>T_2</math> and <math>T_3</math> do not commute.
+
Therefore, <math>T_2</math> and <math>T_3</math> do not commute. One counterexample is the preimage <math>(1,0).</math>
 
</li><p>
 
</li><p>
 
<li>Applying <math>T_2</math> and then <math>T_4</math> gives <math>(x,y)\to(-y,x)\to(-2y,2x).</math> <p>
 
<li>Applying <math>T_2</math> and then <math>T_4</math> gives <math>(x,y)\to(-y,x)\to(-2y,2x).</math> <p>
 
Applying <math>T_4</math> and then <math>T_2</math> gives <math>(x,y)\to(2x,2y)\to(-2y,2x).</math> <p>
 
Applying <math>T_4</math> and then <math>T_2</math> gives <math>(x,y)\to(2x,2y)\to(-2y,2x).</math> <p>
 
Therefore, <math>T_2</math> and <math>T_4</math> commute.
 
Therefore, <math>T_2</math> and <math>T_4</math> commute.
 +
</li><p>
 +
<li>Applying <math>T_3</math> and then <math>T_4</math> gives <math>(x,y)\to(x,-y)\to(2x,-2y).</math> <p>
 +
Applying <math>T_4</math> and then <math>T_3</math> gives <math>(x,y)\to(2x,2y)\to(2x,-2y).</math> <p>
 +
Therefore, <math>T_3</math> and <math>T_4</math> commute.
 
</li><p>
 
</li><p>
 
</ul>
 
</ul>
 +
Together, <math>\boxed{\textbf{(C)}~3}</math> pairs of transformations commute: <math>T_1</math> and <math>T_3, T_2</math> and <math>T_4,</math> and <math>T_3</math> and <math>T_4.</math>
 +
 +
<u><b>Remark</b></u>
 +
 +
To show that two transformations do not commute, we only need one counterexample.
 +
 +
~MRENTHUSIASM
  
 
== Solution 2 (Specific) ==
 
== Solution 2 (Specific) ==
 
Label the transformations as follows:
 
Label the transformations as follows:
  
−
• a translation 2 units to the right <math>(W)</math>
+
• a translation <math>2</math> units to the right <math>(W)</math>
  
−
• a 90°-rotation counterclockwise about the origin <math>(X)</math>
+
• a <math>90^{\circ}</math>-rotation counterclockwise about the origin <math>(X)</math>
  
−
• a reflection across the 𝑥-axis <math>(Y)</math>
+
• a reflection across the <math>x</math>-axis <math>(Y)</math>
  
−
• a dilation centered at the origin with scale factor 2 <math>(Z)</math>
+
• a dilation centered at the origin with scale factor <math>2</math> <math>(Z)</math>
  
 
Now, examine each possible pair of transformations with the point <math>(1,0)</math>:
 
Now, examine each possible pair of transformations with the point <math>(1,0)</math>:
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Label the transformations as follows:
 
Label the transformations as follows:
  
−
• a translation 2 units to the right <math>(A)</math>
+
• a translation <math>2</math> units to the right <math>(A)</math>
  
−
• a 90°-rotation counterclockwise about the origin <math>(B)</math>
+
• a <math>90^{\circ}</math>-rotation counterclockwise about the origin <math>(B)</math>
  
−
• a reflection across the 𝑥-axis <math>(C)</math>
+
• a reflection across the <math>x</math>-axis <math>(C)</math>
  
−
• a dilation centered at the origin with scale factor 2 <math>(D)</math>
+
• a dilation centered at the origin with scale factor <math>2</math> <math>(D)</math>
  
 
Now, we count each transformation individually. It is not hard to see that <math>AC, BD,</math> and <math>CD</math> are commutative (an easy way to test commutativity for some cases would be to have the original point on the <math>x</math>-axis).  
 
Now, we count each transformation individually. It is not hard to see that <math>AC, BD,</math> and <math>CD</math> are commutative (an easy way to test commutativity for some cases would be to have the original point on the <math>x</math>-axis).  
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~xHypotenuse
 
~xHypotenuse
 +
==Solution 4 (quick with reasoning)==
 +
Alright so realize that obviously rotation and dilation are commute, if you ever looked at the polar plane. Also, reflecting around the x-axis means that moving to the right two units and dilating are commute. You can easily see this by taking an arbitrary point and applying the transformations. The rest don’t work since you can easily test out using random points on the plane, which means the answer is <math>\boxed{C}</math>.
 +
 +
~EaZ_Shadow
 +
== Video Solution by Pi Academy ==
 +
 +
https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
 +
 +
==Video Solution 1 by Power Solve ==
 +
https://youtu.be/QVDbm5sDxxU
 +
 +
==Video Solution 2 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=6SQ74nt3ynw
  
 
==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}}

Latest revision as of 23:41, 18 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 $90^{\circ}$-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 (Generalized)

Label the given transformations $T_1, T_2, T_3,$ and $T_4,$ respectively. The rules of transformations are:

  • $T_1: \ (x,y)\to(x+2,y)$
  • $T_2: \ (x,y)\to(-y,x)$
  • $T_3: \ (x,y)\to(x,-y)$
  • $T_4: \ (x,y)\to(2x,2y)$

Note that:

  • Applying $T_1$ and then $T_2$ gives $(x,y)\to(x+2,y)\to(-y,x+2).$

    Applying $T_2$ and then $T_1$ gives $(x,y)\to(-y,x)\to(-y+2,x).$

    Therefore, $T_1$ and $T_2$ do not commute. One counterexample is the preimage $(0,0).$

  • Applying $T_1$ and then $T_3$ gives $(x,y)\to(x+2,y)\to(x+2,-y).$

    Applying $T_3$ and then $T_1$ gives $(x,y)\to(x,-y)\to(x+2,-y).$

    Therefore, $T_1$ and $T_3$ commute. They form a glide reflection.

  • Applying $T_1$ and then $T_4$ gives $(x,y)\to(x+2,y)\to(2x+4,2y).$

    Applying $T_4$ and then $T_1$ gives $(x,y)\to(2x,2y)\to(2x+2,2y).$

    Therefore, $T_1$ and $T_4$ do not commute. One counterexample is the preimage $(0,0).$

  • Applying $T_2$ and then $T_3$ gives $(x,y)\to(-y,x)\to(-y,-x).$

    Applying $T_3$ and then $T_2$ gives $(x,y)\to(x,-y)\to(y,x).$

    Therefore, $T_2$ and $T_3$ do not commute. One counterexample is the preimage $(1,0).$

  • Applying $T_2$ and then $T_4$ gives $(x,y)\to(-y,x)\to(-2y,2x).$

    Applying $T_4$ and then $T_2$ gives $(x,y)\to(2x,2y)\to(-2y,2x).$

    Therefore, $T_2$ and $T_4$ commute.

  • Applying $T_3$ and then $T_4$ gives $(x,y)\to(x,-y)\to(2x,-2y).$

    Applying $T_4$ and then $T_3$ gives $(x,y)\to(2x,2y)\to(2x,-2y).$

    Therefore, $T_3$ and $T_4$ commute.

Together, $\boxed{\textbf{(C)}~3}$ pairs of transformations commute: $T_1$ and $T_3, T_2$ and $T_4,$ and $T_3$ and $T_4.$

Remark

To show that two transformations do not commute, we only need one counterexample.

~MRENTHUSIASM

Solution 2 (Specific)

Label the transformations as follows:

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

• a $90^{\circ}$-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 $\boxed{\textbf{(C) }3}$.

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

~Tacos_are_yummy_1

Solution 3 (Specific)

Label the transformations as follows:

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

• a $90^{\circ}$-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).

In total, $\boxed{\textbf{(C) }3}$ transformation pairs commute.

~xHypotenuse

Solution 4 (quick with reasoning)

Alright so realize that obviously rotation and dilation are commute, if you ever looked at the polar plane. Also, reflecting around the x-axis means that moving to the right two units and dilating are commute. You can easily see this by taking an arbitrary point and applying the transformations. The rest don’t work since you can easily test out using random points on the plane, which means the answer is $\boxed{C}$.

~EaZ_Shadow

Video Solution by Pi Academy

https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM

Video Solution 1 by Power Solve

https://youtu.be/QVDbm5sDxxU

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

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
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