Problem:
Suppose that a parabola has vertex (41,−89) and equation y=ax2+bx+c, where a>0 and a+b+c is an integer. The minimum possible value of a can be written in the form qp, where p and q are relatively prime positive integers. Find p+q.
Solution:
Because the vertex of the parabola is (41,−89), the equation of the parabola can be written as y=a(x−41)2−89. Note that a+b+c is equal to the value of y at x=1. Hence a+b+c=a(1−41)2−89=169(a−2), which is given to be an integer. If 169(a−2)≤−2, then a≤−914<0. Therefore 169(a−2)≥−1, which is equivalent to a≥92, and thus p+q=11.
The problems on this page are the property of the MAA's American Mathematics Competitions