Problem:
Let O(0,0),A(21β,0), and B(0,23ββ) be points in the coordinate plane. Let F be the family of segments PQβ 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=qpβ, where p and q are relatively prime positive integers. Find p+q.
Solution:
Note that A=(cos3Οβ,0) and B=(0,sin3Οβ). For any PQβ in F, there is a real number ΞΈ in the interval [0,2Οβ] such that P=(cosΞΈ,0) and Q=(0,sinΞΈ). The equations of the lines AB and PQ are, respectively,
Note that for β2Οβ<x<2Οβ and 0β€a<2Οβ,
f(x)=cos(x)cos(x+a)β=cos(a)βtan(x)sin(a)
is a monotonically decreasing function of x. Therefore xβ is a monotonically decreasing function of ΞΈ for ΞΈβ(0,2Οβ).
As ΞΈ varies in the interval (0,2Οβ), the value of xβ varies between 0 and 21β. The only value that xβ cannot attain occurs when ΞΈ=3Οβ, because in that case segments AB and PQβ coincide. As ΞΈ approaches 3Οβ, xβ approaches (cos3Οβ)3, so this is the unattainable value for xβ. This gives the point C=((cos3Οβ)3,(sin3Οβ)3), so OC2=641β+6427β=167β. The requested sum is 7+16=23β.