Question

Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0, 0),$ $B(0, 12),$ $C(16, 0),$ $A'(24, 18),$ $B'(36, 18),$ $C'(24, 2).$ A rotation of $m$ degrees clockwise around the point $(x, y),$ where $0 \lt m \lt 180,$ will transform $\triangle ABC$ to $\triangle A'B'C'.$ Find $m + x + y.$

Answer

Answer: 108

Concepts:transformation coordinate geometry

Difficulty rating: 2300

Solution:

The vector $\overrightarrow{AB} = (0, 12)$ is vertical, while $\overrightarrow{A'B'} = (12, 0)$ is horizontal and of the same length, so the rotation turns directions by $90^\circ$ clockwise, and $m = 90.$

A $90^\circ$ clockwise rotation about $(a, b)$ sends $(p, q)$ to $(a + q - b,\ b - p + a).$ Applying this to $A = (0, 0)$ and setting the image equal to $A' = (24, 18)$ gives $a - b = 24$ and $a + b = 18,$ so $a = 21$ and $b = -3.$ Checking the other vertices: $B = (0, 12)$ maps to $(21 + 12 + 3,\ -3 + 21)$ $= (36, 18) = B',$ and $C = (16, 0)$ maps to $(21 + 3,\ -3 - 16 + 21)$ $= (24, 2) = C'.$

Therefore $m + x + y = 90$ $+ 21$ $+ (-3) = 108.$

2020 AIME II Problem 4

Answer: 108

Difficulty rating: 2300

Solution:

Problem 4 in Other Years