Problem

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 $0<m<180$ , will transform $\triangle ABC$ to $\triangle A'B'C'$ . Find $m+x+y$ .

Solution

After sketching, it is clear a $90^{\circ}$ rotation is done about $(x,y)$ . Looking between $A$ and $A'$ , $x+y=18$ and $x-y=24$ . Solving gives $(x,y)\implies(21,-3)$ . Thus $90+21-3=\boxed{108}$ . ~mn28407

Solution 2 (Official MAA)

Because the rotation sends the vertical segment $\overline{AB}$ to the horizontal segment $\overline{A'B'}$ , the angle of rotation is $90^\circ$ degrees clockwise. For any point $(x,y)$ not at the origin, the line segments from $(0,0)$ to $(x,y)$ and from $(x,y)$ to $(x-y,y+x)$ are perpendicular and are the same length. Thus a $90^\circ$ clockwise rotation around the point $(x,y)$ sends the point $A(0,0)$ to the point $(x-y,y+x) = A'(24,18)$ . This has the solution $(x,y) = (21,-3)$ . The requested sum is $90+21-3=108$ .

Video Solution

https://youtu.be/atqPgGG0Ekk

~IceMatrix

See Also=

2020 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2020 AIME II Problems/Problem 4

Contents

Problem

Solution

Solution 2 (Official MAA)

Video Solution

See Also=