Problem:
The graphs of y=3(x−h)2+j and y=2(x−h)2+k have y-intercepts of 2013 and 2014, respectively, and each graph has two positive integer x-intercepts. Find h.
Solution:
The first equation can be rewritten as y=3x2−6xh+2013. If its x-intercepts are p and q, then pq=32013=671=11⋅61, so {p,q}={1,671} or {11,61}. The value of h is the average of p and q, so h=336 or 36. The second equation can be rewritten as y=2x2−4xh+2014. If its x-intercepts are r and s, then rs=22014=1007=19⋅53, so {r,s}={1,1007} or {19,53}, and h=504 or 36. The only value of h that is possible for both equations is 36. Note that y=3(x−36)2−1875 and y=2(x−36)2−578 satisfy the conditions in the problem.
The problems on this page are the property of the MAA's American Mathematics Competitions