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2020 AMC 10A Problem 23

2020 AMC 10A Problem 23AMC 10 Step by Step

Name: Date: September 13, 2026

Transformations & Symmetry · difficulty 4 of 5 · about 8 min · position P21-25

Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$ -axis, and reflection across the $y$ -axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by a reflection across the $y$ -axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$ -axis, followed by another reflection across the $x$ -axis will not return $T$ to its original position.)

A)
$12$
B)
$15$
C)
$17$
D)
$20$
E)
$25$

Answer

Key insight

T has no symmetry, so the composite must be the identity; reflections flip orientation, so there are 0 or 2 of them, and each case gives 6 sequences.

Solution

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 $\boxed{\textbf{(A)}\ 12}$ .

Why this works

These five transformations live inside the symmetry group of a square (order $8$ ), where orientation is an invariant that immediately kills all odd-reflection cases. Within the even cases, the only non-obvious rule is that a reflection conjugates a rotation to its inverse, which is why the middle rotation must be the self-inverse $180^\circ$ . Tracking the image of one asymmetric triangle is equivalent to tracking the group element.

The trap

Forgetting that a rotation sandwiched between two reflections is reversed, or allowing sequences with one or three reflections.

Common mistakes

Forgetting that a rotation sandwiched between two reflections is reversed, or allowing sequences with one or three reflections.
Counting $\{90, 270\}$ pairs as cancelling and then thinking a third transformation can be "nothing" (the identity is not among the five choices).

Techniques

Split into exhaustive cases and handle each · Use an invariant, parity, or coloring argument