2015 AMC 10B Problem 8

Instead of redrawing the F three times, follow what each move does to the coordinates of a point.

1. Turn $90^\circ$ clockwise about the origin. Try the point $(1, 0)$ on the positive $x$-axis: a quarter turn clockwise sends it down to $(0, -1)$. Try $(0, 1)$ on the positive $y$-axis: it goes to $(1, 0)$. In general, $(x, y)$ goes to $(y, -x)$.
2. Reflect in the $y$-axis. This changes the sign of the first coordinate: $(y, -x)$ goes to $(-y, -x)$.
3. Half turn about the origin. This changes the sign of both coordinates: $(-y, -x)$ goes to $(y, x)$.

So after all three moves, every point $(x, y)$ ends up at $(y, x)$. The two coordinates just swap. That is a reflection across the line $y = x$.

Now use two corners of the F to find the picture.

1. The F's upright stem sits on the $x$-axis, with its bottom right corner at about $(1, 0)$. Swapping gives $(0, 1)$: a point on the $y$-axis.
2. The top left corner of the F is at the top of the $y$-axis, about $(0, 5)$. Swapping gives $(5, 0)$: a point far out on the $x$-axis.

So the stem now lies flat along the $x$-axis, running out to about $x = 5$, with its left end against the $y$-axis, and the arms of the F now point up. That is picture (E).

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

A chain of moves is easier to handle as a coordinate rule than as a picture: write each move as a rule and apply them in order. There is also a quick check. The chain has one reflection, and one reflection always turns a shape into its mirror image, so pictures (A), (C) and (D), which are just the F turned around, cannot be right.