aime26-04
1.000
Proposed solution
I need to find the cosine of the acute angle θ given the rotation configuration and the distance AB'.
Let me set up a coordinate system with B at the origin and A on the positive x-axis.
Setting up coordinates:
- Let B = (0, 0)
- Let A = (1, 0) since AB = 1
First rotation: Rotate A counterclockwise around B through angle θ to get A'.
- A' = (cos θ, sin θ)
Second rotation: Rotate B clockwise around A' through angle θ to get B'.
- B' = A' + (B - A') rotated clockwise by θ
- B' = A' + (A' - B) rotated clockwise by θ
- B' = A' + (-A') rotated clockwise by θ
- B' = A' - A' rotated clockwise by θ
Let me reconsider this more carefully.
When rotating point P around center C through angle α, the new position is:
P' = C + (P - C) rotated by α
So for the second rotation:
B' = A' + (B - A') rotated clockwise by θ
B' = A' + (0, 0) - (cos θ, sin θ)) rotated clockwise by θ
B' = A' + (-cos θ, -sin θ) rotated clockwise by θ
To rotate a vector (x, y) clockwise by angle θ, I multiply by the rotation matrix:
$$\begin{pmatrix} \cos θ & \sin θ \\ -\sin θ & \cos θ \end{pmatrix}$$
So (-cos θ, -sin θ) rotated clockwise by θ becomes:
$$\begin{pmatrix} \cos θ & \sin θ \\ -\sin θ & \cos θ \end{pmatrix} \begin{pmatrix} -\cos θ \\ -\sin θ \end{pmatrix} = \begin{pmatrix} -\cos^2 θ - \sin^2 θ \\ \sin θ \cos θ - \sin θ \cos θ \end{pmatrix} = \begin{pmatrix} -1 \\ 0 \end{pmatrix}$$
Therefore:
B' = A' + (-1, 0) = (cos θ, sin θ) + (-1, 0) = (cos θ - 1, sin θ)
Now I can find AB':
AB' = |A - B'| = |(1, 0) - (cos θ - 1, sin θ)| = |(2 - cos θ, -sin θ)|
So:
AB'² = (2 - cos θ)² + sin² θ = 4 - 4cos θ + cos² θ + sin² θ = 4 - 4cos θ + 1 = 5 - 4cos θ
Given that AB' = 4/3:
AB'² = 16/9
Therefore:
5 - 4cos θ = 16/9
4cos θ = 5 - 16/9 = 45/9 - 16/9 = 29/9
cos θ = 29/36
Let me verify that 29 and 36 are relatively prime:
The factors of 29: 1, 29 (since 29 is prime)
The factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36
Since 29 is prime and doesn't divide 36, gcd(29, 36) = 1.
Therefore, m = 29 and n = 36, so m + n = 65.
$\boxed{65}$