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2024 AMC 10A Problem 13

2024 AMC 10A Problem 13AMC 10 Step by Step

Name: Date: September 13, 2026

Transformations & Symmetry · difficulty 3 of 5 · about 4 min · position P11-15

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?

A)
$1$
B)
$2$
C)
$3$
D)
$4$
E)
$5$

Answer

Key insight

Write each map as a formula on (x, y): the dilation commutes with the rotation and the reflection, the translation commutes only with the reflection; the other three pairs fail.

Solution

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

Why this works

Explicit formulas make "commute" a one-line check. The structural reasons are worth remembering: scaling about the origin commutes with every linear map (rotations, reflections through the origin); a translation commutes with a reflection whose axis is parallel to the translation, but not with rotations or dilations, which move the direction or the size of the shift; and a rotation and a reflection generally do not commute.

Alternative approach

Test one point, say $(1, 0)$ , in both orders for each pair; a single disagreement proves non-commuting, and for the three commuting pairs the formulas above confirm agreement for every point.

The trap

Assuming the rotation and reflection commute because both fix the origin; applying them to (1, 0) in the two orders gives (0, −1) versus (0, 1).

Common mistakes

Assuming the rotation and reflection commute because both fix the origin; applying them to (1, 0) in the two orders gives (0, −1) versus (0, 1).
Concluding two maps commute after testing only one point that happens to agree (e.g. the origin for $R$ and $F$ ).

Techniques

Organized listing / direct enumeration · Test small/specific values or special cases to find or verify the answer