2016 AMC 10A Problem 16

P16-20~3 minPrint

Transformations & Symmetry Coordinate Geometry Set up the equation/formula and compute; no special trick needed Test small/specific values or special cases to find or verify the answer

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 $\triangle ABC$ ?

Show solution without answering

Key insight

Follow the corners: every point (x, y) ends at (y, x). That is a reflection over y = x, and repeating it undoes it.

Solution

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

Why this works

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.

Alternative approach

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.

Concise solution

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

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

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
Introduction to Algebra, Chapter 8: Graphing Lines