2011 AIME I Problems/Problem 6
Contents
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
must be divisible by 16. Let
. If
, then
. Therefore, if
,
. Adding up gives us
Solution 2
Complete the square. Since , the parabola must be facing upwards.
means that
must be an integer. The function can be recasted into
because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than
is
. So the
-coordinate must change by
and the
-coordinate must change by
. Thus,
. So
.
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 |
- AIME Problems and Solutions
- American Invitational Mathematics Examination
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.