Difference between revisions of "2011 AIME I Problems/Problem 6"
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* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] |
Revision as of 04:37, 29 March 2011
Problem
Suppose that a parabola has vertex and equation
, where
and
is an integer. The minimum possible value of
can be written in the form
, where
and
are relatively prime positive integers. Find
.
Solution
If the vertex is at , the equation of the parabola can be expressed in the form
.
Expanding, we find that
, and
. From the problem, we know that the parabola can be expressed in the form
, where
is an integer. From the above equation, we can conclude that
,
, and
. Adding up all of these gives us
. We know that
is an integer, so 9a-18 must be divisible by 16. Let
. If
, then
. Therefore, if
,
. Adding up gives us
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |