2015 AMC 10B Problem 8

Follow a general point $(x, y)$ through the three moves.

1. Rotation $90^\circ$ clockwise about the origin: $(x, y) \to (y, -x)$.
2. Reflection in the $y$-axis: $(y, -x) \to (-y, -x)$.
3. Half turn about the origin: $(-y, -x) \to (y, x)$.

The net map is $(x, y) \to (y, x)$: a reflection across the line $y = x$.

Now apply it to the F. Its stem runs up from the $x$-axis along $x = 1$, with the two arms pointing right and the long top bar at $y = 5$. Swapping coordinates turns the stem into a segment along $y = 1$ starting at the $y$-axis, the arms now point up, and the long bar becomes vertical at $x = 5$. That is an F lying on its back in the first quadrant, which is picture (E).

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

Composite transformations are easiest to handle algebraically: write each as a coordinate rule and compose. An odd number of reflections yields a mirror image, so the answer had to be a reflected F rather than a rotated one, which alone eliminates (A), (C) and (D). The specific reflection line then picks (E) over (B).

Track two landmark points. The foot of the stem $(1, 0)$ goes $(0, -1) \to (0, -1) \to (0, 1)$, and the tip of the top bar $(0, 5)$ goes $(5, 0) \to (-5, 0) \to (5, 0)$. Only picture (E) has the stem starting on the $y$-axis at $(0,1)$ and the bar ending at $(5, 0)$.

• Turning the wrong way for the clockwise rotation or doing the steps out of order; any answer that is merely a rotated F (not a mirror image) is wrong.
• Reflecting in the $x$-axis instead of the $y$-axis, which produces the mirror image in a different position.