2019 AMC 10A Problem 8

First describe the picture carefully.

• Above the line: squares sit on $\ell$, each with a short slanted tail going up and to the right from its top-right corner.
• Below the line: squares hang from $\ell$, each with a tail going down and to the left from its bottom-left corner.
• The lower squares are not directly under the upper ones. Each sits halfway between two upper squares.
• The whole pattern repeats as you slide along the line.

Now test the four motions one at a time. The tails are the smallest detail that can go wrong, so watch them.

1. Slide along $\ell$. Slide by one full repeat (from one upper square to the next). Every square and tail lands on a matching one. Works.
2. Half-turn about a point on $\ell$. Pick the point on $\ell$ halfway between an upper square and the next lower square. A half-turn swaps "above" and "below" and "left" and "right". An upper square with a tail going up-right from its top-right corner becomes a lower square with a tail going down-left from its bottom-left corner. That is exactly what the lower squares look like, and the half-period shift lines up too. Works.
3. Flip across $\ell$. An upper square would land directly under itself, but no lower square is there (they are shifted). Also its tail would point down-right, but the real lower tails point down-left. Fails.
4. Flip across a line at right angles to $\ell$. A flip left-to-right turns a tail slanting up-right into one slanting up-left. No tail in the picture slants up-left. Fails.

Two of the four motions keep the figure the same.

The answer is $\boxed{\textbf{(C)}\ 2}$.

A mirror flip reverses the slant of every tail, while a half-turn reverses it twice, which puts it back. So in a pattern where all tails on one side slant the same way, flips tend to fail and half-turns can succeed. When checking a symmetry, always follow the smallest lopsided detail, not just the big shapes.

Describe the pattern precisely. Above $\ell$, unit squares sit on the line with a short diagonal tail leaving the top-right corner toward the upper right. Below $\ell$, congruent squares hang from the line, shifted half a period to the right, with a tail leaving the bottom-left corner toward the lower left. The whole picture repeats every $4$ units along $\ell$.

• Translation parallel to $\ell$ by $4$ units: the pattern repeats, so this works.
• Rotation by $180^\circ$ about a point of $\ell$ midway between an upper square and the next lower square: a half-turn sends "square above, tail up-right from the top-right corner" to "square below, tail down-left from the bottom-left corner," which is exactly how the lower squares look, and the offset matches. This works.
• Reflection across $\ell$: an upper square would land directly below itself, but the lower squares are shifted, and the tail would point down-right instead of down-left. Fails.
• Reflection across a line perpendicular to $\ell$: every tail slanting up-right becomes one slanting up-left, and no tail in the figure does that. Fails.

Two of the four motions preserve the figure.

The answer is $\boxed{\textbf{(C)}\ 2}$.