2020 AMC 10A Problem 23

The legs of $T$ have different lengths ($4$ along the $x$-axis, $3$ along the $y$-axis), so no rotation or reflection can move $T$ onto itself except by putting every point back where it was. So we need the three moves together to do nothing at all.

It helps to know what each move does to a point $(x, y)$:

1. Count the flips. A flip turns the triangle into its mirror image, and a turn does not. Two flips undo the mirror effect; one or three leave a mirror image, which can never be the original. So the sequence has $0$ or $2$ flips.
2. No flips: three turns. The angles must add to a full number of turns ($360^\circ$ or $720^\circ$; three turns add to at most $3 \times 270^\circ = 810^\circ$, so $1080^\circ$ is out of reach).
- Total $360^\circ$: $90 + 90 + 180$, in $3$ orders.
- Total $720^\circ$: $270 + 270 + 180$, in $3$ orders.

That is $6$ sequences.
3. Two flips and one turn. Use the table to see how flips combine.
- The same flip twice in a row undoes itself, leaving just a turn of $90^\circ$, $180^\circ$ or $270^\circ$. That never works.
- The same flip with a turn between, like $x$-flip, turn, $x$-flip, is the turn seen in a mirror: a turn the other way by the same angle. That never works either.
- Two different flips in a row: $(x, y) \to (x, -y) \to (-x, -y)$. That is the same as a $180^\circ$ turn (in either order). So with the turn first or last, the turn must also be $180^\circ$. Sequences: turn $180$, $x$-flip, $y$-flip; turn $180$, $y$-flip, $x$-flip; $x$-flip, $y$-flip, turn $180$; $y$-flip, $x$-flip, turn $180$. That is $4$.
- Two different flips with the turn between. Track $(x,y)$ through $x$-flip, turn, $y$-flip. With $180^\circ$: $(x, y) \to (x, -y) \to (-x, y) \to (x, y)$. Back home. With $90^\circ$: $(x,y) \to (x,-y) \to (y, x) \to (-y, x)$, which is a $90^\circ$ turn, not home. With $270^\circ$ you get a $270^\circ$ turn. So only $180^\circ$ works, and the flips can come in either order: $2$ sequences.

That is $4 + 2 = 6$ sequences.
4. Total: $6 + 6 = 12$.

The answer is $\boxed{\textbf{(A)}\ 12}$.

Thinking about mirror images removes most sequences at once: you need an even number of flips. After that, a few facts do the rest: two different axis flips equal a $180^\circ$ turn, and a turn viewed through a mirror goes the other way. Tracking a single point $(x, y)$ through the moves is a reliable way to check any of these facts.

The triangle $T$ has no symmetries, so it returns to its position only if the net transformation is the identity. Write $R_\theta$ for the rotations and $F_x$, $F_y$ for the reflections.

Reflections reverse orientation and rotations preserve it, so an identity composite needs an even number of reflections: $0$ or $2$.

No reflections. Three rotation angles from $\{90, 180, 270\}$ must sum to a multiple of $360$. Sum $360$: $\{90, 90, 180\}$ in $3$ orders. Sum $720$: $\{270, 270, 180\}$ in $3$ orders. Sum $1080$ is impossible. That is $6$ sequences.

Two reflections and one rotation. Key facts: $F_x F_x = F_y F_y = I$, $F_x F_y = F_y F_x = R_{180}$, and $R_{180}$ commutes with everything, while $F R_{\theta} = R_{-\theta} F$ for any reflection $F$.

• Same reflection twice: if adjacent, they cancel and a lone rotation remains; if the rotation sits between them, the result is $R_{-\theta}$. Never the identity.
• Different reflections, rotation first or last: the reflections combine to $R_{180}$, so the rotation must be $R_{180}$. Sequences: $(R_{180}, F_x, F_y)$, $(R_{180}, F_y, F_x)$, $(F_x, F_y, R_{180})$, $(F_y, F_x, R_{180})$: $4$.
• Different reflections with the rotation between: $F_x R_\theta F_y = R_{-\theta} F_x F_y = R_{-\theta} R_{180}$, identity only for $\theta = 180$. Sequences $(F_x, R_{180}, F_y)$ and $(F_y, R_{180}, F_x)$: $2$.

That is $6$ more. Total $6 + 6 = 12$.

The answer is $\boxed{\textbf{(A)}\ 12}$.