Problem:
Let O(0,0),A(21,0), and B(0,23) be points in the coordinate plane. Let F be the family of segments PQ of unit length lying in the first quadrant with P on the x-axis and Q on the y-axis. There is a unique point C on AB, distinct from A and B, that does not belong to any segment from F other than AB. Then OC2=qp, where p and q are relatively prime positive integers. Find p+q.
Solution:
Note that A=(cos3π,0) and B=(0,sin3π). For any PQ in F, there is a real number θ in the interval [0,2π] such that P=(cosθ,0) and Q=(0,sinθ). The equations of the lines AB and PQ are, respectively,
cos3πx+sin3πy=1andcosθx+sinθy=1.
For θ=3π, the x-coordinate of the intersection of the lines is
is a monotonically decreasing function of x. Therefore x∗ is a monotonically decreasing function of θ for θ∈(0,2π).
As θ varies in the interval (0,2π), the value of x∗ varies between 0 and 21. The only value that x∗ cannot attain occurs when θ=3π, because in that case segments AB and PQ coincide. As θ approaches 3π, x∗ approaches (cos3π)3, so this is the unattainable value for x∗. This gives the point C=((cos3π)3,(sin3π)3), so OC2=641+6427=167. The requested sum is 7+16=23.