2016 AMC 10A Problem 16

The safest way to handle two moves in a row is to follow actual points through them.

1. Reflect over the $x$-axis. This keeps $x$ and flips the sign of $y$:
- $A(0, 2) \to A'(0, -2)$
- $B(-3, 2) \to B'(-3, -2)$
- $C(-3, 0) \to C'(-3, 0)$
2. Rotate $90^\circ$ counterclockwise about the origin. This sends a point $(x, y)$ to $(-y, x)$. (Check with an easy point: $(1, 0)$ on the right goes to $(0, 1)$ at the top, which is a quarter turn counterclockwise.)
- $A'(0, -2) \to A''(2, 0)$
- $B'(-3, -2) \to B''(2, -3)$
- $C'(-3, 0) \to C''(0, -3)$
3. Compare the start and the end: $A(0,2)$ became $(2,0)$, $B(-3,2)$ became $(2,-3)$, $C(-3,0)$ became $(0,-3)$. Every time, the two coordinates just swapped places.
4. Swapping the coordinates is exactly what a reflection over the line $y = x$ does. So the two moves together are one reflection over $y = x$.
5. Reflecting over the same line a second time swaps the coordinates back, returning every point to where it started.

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

To figure out what a chain of moves does, track a few points and look for a pattern in their coordinates. A reflection followed by a rotation always acts like a single reflection, because the triangle gets flipped over exactly once. Any reflection is undone by doing the same reflection again.

Rule out the other choices. The triangle was flipped once, so it now faces the "mirror" way round; a rotation can never flip it back, so (A) and (B) are out. For (C) and (E), look at where $A''(2, 0)$ goes: reflecting over the $x$-axis leaves it at $(2, 0)$, and reflecting over the $y$-axis sends it to $(-2, 0)$. Neither is $A(0, 2)$. Only (D) is left.

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