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Algebra Difficulty 4.0 AIME Find the answer

The graphs y=3(xh)2+jy=3(x-h)^2+j and y=2(xh)2+ky=2(x-h)^2+k have y-intercepts of 20132013 and 20142014, respectively, and each graph has two positive integer x-intercepts. Find hh.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Begin by setting xx to 0, then set both equations to h2=2013j3h^2=\frac{2013-j}{3} and h2=2014k2h^2=\frac{2014-k}{2}, respectively. Notice that because the two parabolas have to have positive x-intercepts, h32h\ge32.
We see that h2=2014k2h^2=\frac{2014-k}{2}, so we now need to find a positive integer hh which has positive integer x-intercepts for both equations.
Notice that if k=20142h2k=2014-2h^2 is -2 times a square number, then you have found a value of hh for which the second equation has positive x-intercepts. We guess and check h=36h=36 to obtain k=578=2(172)k=-578=-2(17^2).
Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is 036\boxed{036}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.