Question

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.)

$12$

$15$

$17$

$20$

$25$

Answer: A

Concepts:transformation casework

Difficulty rating: 1910

Small Hint:

Since $T$ is scalene, the only isometry fixing $T$ is the identity, so the three must compose to the identity

Big Hint:

The third transformation is forced by the first two; it is allowed exactly when the product of the first two is again one of the five maps

Solution:

Because $T$ is a scalene right triangle, the only isometry carrying $T$ to itself is the identity, so a sequence works exactly when the three transformations compose to the identity.

Let $r$ be the $90^\circ$ rotation and $s$ reflection across the $x$ -axis. The five allowed maps are $r,r^2,r^3,s,r^2s;$ the missing nonidentity maps are the diagonal reflections $rs,r^3s.$ In an ordered triple, the third map is forced by the first two and is allowed precisely when their product is one of the five.

Exactly $5$ ordered pairs have product identity: each first map is paired with its inverse. A diagonal reflection requires one rotation and one axis reflection; $r$ or $r^3$ may be paired on either side with $s$ or $r^2s,$ giving $2\cdot2\cdot2=8$ pairs. The remaining $25-5-8=12$ ordered pairs give valid sequences.

Thus, A is the correct answer.

2020 AMC 12A Problem 20

Answer: A

Difficulty rating: 1910

Small Hint:

Big Hint:

Solution:

Problem 20 in Other Years