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

Revision as of 02:57, 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 $90^{\circ}$ -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 (General)

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.

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.

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.

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.

Solution 2 (Specific)

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 $\boxed{\textbf{(C) }3}$ .

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°-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

@@ Line 6: / Line 6: @@
 * a translation <math>2</math> units to the right,
-* a <math>90^{\circ}</math>-crotation counterclockwise about the origin,
+* a <math>90^{\circ}</math>-rotation counterclockwise about the origin,
 * a reflection across the <math>x</math>-axis, and
@@ Line 36: / Line 36: @@
 Applying <math>T_3</math> and then <math>T_1</math> gives <math>(x,y)\to(x,-y)\to(x+2,-y).</math> <p>
 Therefore, <math>T_1</math> and <math>T_3</math> commute. They form a <b>glide reflection</b>.
+</li><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>
+Therefore, <math>T_1</math> and <math>T_4</math> do not commute.
+</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>
+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.
+</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>
+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.
 </li><p>
 </ul>

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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