Difference between revisions of "2011 AIME I Problems/Problem 6"
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== See also == | == See also == | ||
| − | {{AIME box|year=2011|before=Problem 5|num-a=7}} | + | {{AIME box|year=2011|n=I|before=Problem 5|num-a=7}} |
* [[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 17:51, 21 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 | ||
