2011 AIME I Problems/Problem 6
Revision as of 17:17, 18 March 2011 by TrainingAccount (talk | contribs) (Created page with 'If the vertex is at (1/4, -9/8), the equation of the parabola can be expressed in the form y=a(x-1/4)^2-9/8. Expanding, we find that y=a(x^2-x/2+1/16)-9/8 , and y=ax^2-ax/2+a/16-…')
If the vertex is at (1/4, -9/8), the equation of the parabola can be expressed in the form y=a(x-1/4)^2-9/8. Expanding, we find that y=a(x^2-x/2+1/16)-9/8 , and y=ax^2-ax/2+a/16-9/8. From the problem, we know that the parabola can be expressed in the form y=ax^2+bx+c, where a+b+c= an integer. From the above equation, we can conclude that a=a, -a/2=b, and a/16-9/8=c. Adding up all of these gives us that (9a-18)/16=a+b+c. We know that a+b+c is an integer, so 9a-18 must be divisible by 16. Let 9a=z. If z-18=0(mod 16), z=2(mod 16). Therefore, if 9a=2, a=2/9 Ans.
