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2016 AMC 10A Problem 16

2016 AMC 10A Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Transformations & Symmetry · difficulty 2 of 5 · about 3 min · position P16-20

A triangle with vertices $A(0, 2)$ , $B(-3, 2)$ , and $C(-3, 0)$ is reflected about the $x$ -axis, then the image $\triangle A'B'C'$ is rotated counterclockwise about the origin by $90^{\circ}$ to produce $\triangle A''B''C''$ . Which of the following transformations will return $\triangle A''B''C''$ to $\triangle ABC$ ?

A)
counterclockwise rotation about the origin by $90^{\circ}$ .
B)
clockwise rotation about the origin by $90^{\circ}$ .
C)
reflection about the $x$ -axis
D)
reflection about the line $y = x
E)
reflection about the $y$ -axis.

Answer

Key insight

Track (x, y): the reflection gives (x, -y), the rotation then gives (y, x), so the combined map is reflection over y = x, its own inverse.

Solution

Follow a general point $(x, y)$ through both steps. Reflecting over the $x$ -axis gives $(x, -y)$ . A $90^\circ$ counterclockwise rotation about the origin sends $(u, v)$ to $(-v, u)$ ; applied to $(x, -y)$ it gives $(y, x)$ .

So the combined transformation swaps the coordinates: $(x, y) \mapsto (y, x)$ , which is reflection over the line $y = x$ . Concretely, $A(0,2) \to A''(2,0)$ , $B(-3,2) \to B''(2,-3)$ , $C(-3,0) \to C''(0,-3)$ .

Reflecting twice over the same line returns every point to where it started, so reflecting $\triangle A''B''C''$ over $y = x$ swaps the coordinates back and recovers $\triangle ABC$ .

The answer is $\boxed{\textbf{(D)}\ \text{reflection about the line } y = x}$ .

Why this works

Compositions of rigid motions are easiest to identify by their effect on coordinates. A reflection followed by a rotation is again a reflection (orientation is reversed once), and here the axis turns out to be $y = x$ . Any reflection is an involution, so the inverse of the whole process is the same reflection.

Alternative approach

Eliminate: a rotation preserves orientation, but $\triangle A''B''C''$ has the opposite orientation from $\triangle ABC$ (one reflection was applied), so (A) and (B) fail. Reflecting $A''(2,0)$ over the $x$ -axis or $y$ -axis gives $(2,0)$ or $(-2,0)$ , not $A(0,2)$ , so (C) and (E) fail, leaving (D).

The trap

Using the clockwise rotation formula (x, y) -> (y, -x) for a counterclockwise turn, or assuming the inverse must be some rotation.

Common mistakes

Using the clockwise rotation formula (x, y) -> (y, -x) for a counterclockwise turn, or assuming the inverse must be some rotation.
Checking only one vertex such as $C$ and picking a transformation that happens to fix that point but not the others.

Techniques

Set up the equation/formula and compute; no special trick needed · Test small/specific values or special cases to find or verify the answer