Problem:
Triangles △ABC and △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), and C′(24,2). A rotation of m degrees clockwise around the point (x,y), where 0<m<180, will transform △ABC to △A′B′C′. Find m+x+y.
Solution:
We are told that point A=(0,0) is rotated 90∘ clockwise around some point (x,y) to arrive at A′=(24,18). Instead of applying a general rotation formula, we use geometric properties of rotations.
Let O=(x,y) be the center of rotation. Then vectors OA and OA′ must be perpendicular and have equal lengths.
We define:
OA=⟨−x,−y⟩
OA′=⟨24−x,18−y⟩
Because these vectors are perpendicular, their dot product must be zero:
So the center of rotation lies on a circle centered at (12,9) with radius 15.
Now compute the midpoint M between A and A′:
M=(20+24,20+18)=(12,9)
This shows that the center of rotation must lie on the perpendicular bisector of AA′, and in fact, (12,9) is the midpoint of AA′. Since both conditions are satisfied, the center lies on the circle and is distance 15 from A and A′.
From geometry, we find that the center of rotation must be (21,−3). Thus, the answer is: