2024 AMC 10A Problem 13

P11-15~4 minPrint

Transformations & Symmetry Coordinate Geometry Organized listing / direct enumeration Test small/specific values or special cases to find or verify the answer

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?

Show solution without answering

Key insight

Write each move as a rule for where (x, y) goes, then do each pair in both orders. Only three pairs give the same point both ways.

Solution

"Commute" means the order does not matter. The surest way to check is to write each move as a rule saying where a point $(x, y)$ goes.

1. The four rules.
- Translation $T$ (2 units right): $(x, y) \to (x + 2,\ y)$ .
- Rotation $R$ ( $90^\circ$ counterclockwise): $(x, y) \to (-y,\ x)$ . For example $(1, 0) \to (0, 1)$ .
- Reflection $F$ (across the $x$ -axis): $(x, y) \to (x,\ -y)$ .
- Dilation $D$ (scale by $2$ ): $(x, y) \to (2x,\ 2y)$ .
2. Check each pair both ways. "First $T$ , then $R$ " means apply $T$ 's rule, then apply $R$ 's rule to the result.

pair	first one, then the other	other order	same?
$T$ , $R$	$T$ then $R$ : $(-y,\ x + 2)$	$R$ then $T$ : $(-y + 2,\ x)$	no
$T$ , $F$	$T$ then $F$ : $(x + 2,\ -y)$	$F$ then $T$ : $(x + 2,\ -y)$	yes
$T$ , $D$	$T$ then $D$ : $(2x + 4,\ 2y)$	$D$ then $T$ : $(2x + 2,\ 2y)$	no
$R$ , $F$	$R$ then $F$ : $(-y,\ -x)$	$F$ then $R$ : $(y,\ x)$	no
$R$ , $D$	$R$ then $D$ : $(-2y,\ 2x)$	$D$ then $R$ : $(-2y,\ 2x)$	yes
$F$ , $D$	$F$ then $D$ : $(2x,\ -2y)$	$D$ then $F$ : $(2x,\ -2y)$	yes

Three of the six pairs commute: translation with reflection, rotation with dilation, and reflection with dilation.

The answer is $\boxed{\textbf{(C)}\ 3}$ .

Why this works

Writing a move as a rule for $(x, y)$ turns "does the order matter?" into a quick algebra check. Some reasons are worth remembering. Scaling from the origin just makes everything bigger, so it works with any rotation or reflection about the origin. Sliding right does not care about flipping up and down, so those two commute. But a slide gets turned by a rotation and stretched by a dilation, so it does not commute with those.

Alternative approach

To show two moves do not commute, one point is enough. Take $(1, 0)$ :

$T$ then $R$ : $(3, 0) \to (0, 3)$ ; $R$ then $T$ : $(0, 1) \to (2, 1)$ . Different.
$T$ then $D$ : $(3, 0) \to (6, 0)$ ; $D$ then $T$ : $(2, 0) \to (4, 0)$ . Different.
$R$ then $F$ : $(0, 1) \to (0, -1)$ ; $F$ then $R$ : $(1, 0) \to (0, 1)$ . Different.

The other three pairs agree on every point (see the table above), so the count is $3$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Write the four maps as formulas:
$T(x,y) = (x+2,\, y),\quad R(x,y) = (-y,\, x),\quad F(x,y) = (x,\, -y),\quad D(x,y) = (2x,\, 2y).$
Now compose each pair both ways.

$T, R$ : $R(T(x,y)) = (-y,\, x+2)$ but $T(R(x,y)) = (-y+2,\, x)$ . Fail.
$T, F$ : $F(T(x,y)) = (x+2,\, -y) = T(F(x,y))$ . Commute.
$T, D$ : $D(T(x,y)) = (2x+4,\, 2y)$ but $T(D(x,y)) = (2x+2,\, 2y)$ . Fail.
$R, F$ : $F(R(x,y)) = (-y,\, -x)$ but $R(F(x,y)) = (y,\, x)$ . Fail.
$R, D$ : $D(R(x,y)) = (-2y,\, 2x) = R(D(x,y))$ . Commute.
$F, D$ : $D(F(x,y)) = (2x,\, -2y) = F(D(x,y))$ . Commute.

Exactly three pairs commute.

The answer is $\boxed{\textbf{(C)}\ 3}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 16: The More Things Change. . . (§16.1, §16.2, §16.3, §16.4)
Introduction to Geometry, Chapter 17: Analytic Geometry
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.2)
Introduction to Algebra, Chapter 8: Graphing Lines