Difference between revisions of "2011 AIME I Problems/Problem 6"
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| − | If the vertex is at <math>(\frac{1}{4}, -\frac{9}{8})</math>, the equation of the parabola can be expressed in the form <math>y=a(x-\frac{1}{4})^2-\frac{9}{8}</math>. | + | ==Problem== |
| − | Expanding, we find that <math>y=a(x^2-\frac{x}{2}+\frac{1}{16})-\frac{9}{8}</math> , and <math>y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}</math>. From the problem, we know that the parabola can be expressed in the form <math>y=ax^2+bx+c</math>, where <math>a+b+c</math> is an integer. From the above equation, we can conclude that <math>a=a</math>, <math>-\frac{a}{2}=b</math>, and <math>\frac{a}{16}-\frac{9}{8}=c</math>. Adding up all of these gives us <math>\frac{9a-18}{16}=a+b+c</math>. We know that <math>a+b+c</math> is an integer, so 9a-18 must be divisible by 16. Let <math>9a=z</math>. If <math>{z-18}\equiv {0} \pmod{16}</math>, then <math>{z}\equiv {2} \pmod{16}</math>. Therefore, if <math>9a=2</math>, <math>a=\frac{2}{9}</math>. Adding up gives us <math>2+9=\boxed{011}</math> | + | Suppose that a parabola has vertex <math>\left(\frac{1}{4},-\frac{9}{8}\right)</math> and equation <math>y = ax^2 + bx + c</math>, where <math>a > 0</math> and <math>a + b + c</math> is an integer. The minimum possible value of <math>a</math> can be written in the form <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p + q</math>. |
| + | |||
| + | ==Solution== | ||
| + | If the vertex is at <math>\left(\frac{1}{4}, -\frac{9}{8}\right)</math>, the equation of the parabola can be expressed in the form <math>y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}</math>. | ||
| + | Expanding, we find that <math>y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8}</math> , and <math>y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}</math>. From the problem, we know that the parabola can be expressed in the form <math>y=ax^2+bx+c</math>, where <math>a+b+c</math> is an integer. From the above equation, we can conclude that <math>a=a</math>, <math>-\frac{a}{2}=b</math>, and <math>\frac{a}{16}-\frac{9}{8}=c</math>. Adding up all of these gives us <math>\frac{9a-18}{16}=a+b+c</math>. We know that <math>a+b+c</math> is an integer, so 9a-18 must be divisible by 16. Let <math>9a=z</math>. If <math>{z-18}\equiv {0} \pmod{16}</math>, then <math>{z}\equiv {2} \pmod{16}</math>. Therefore, if <math>9a=2</math>, <math>a=\frac{2}{9}</math>. Adding up gives us <math>2+9=\boxed{011}</math> | ||
== See also == | == See also == | ||
Revision as of 23:35, 18 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 (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 | ||
