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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>.
 
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>.
 
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>
 
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>
 +
 +
== See also ==
 +
{{AIME box|year=2011|before=Problem 5|num-a=7}}
 +
* [[AIME Problems and Solutions]]
 +
* [[American Invitational Mathematics Examination]]
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* [[Mathematics competition resources]]

Revision as of 00:28, 19 March 2011

If the vertex is at $(\frac{1}{4}, -\frac{9}{8})$, the equation of the parabola can be expressed in the form $y=a(x-\frac{1}{4})^2-\frac{9}{8}$. Expanding, we find that $y=a(x^2-\frac{x}{2}+\frac{1}{16})-\frac{9}{8}$ , and $y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}$. From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$, where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$, $-\frac{a}{2}=b$, and $\frac{a}{16}-\frac{9}{8}=c$. Adding up all of these gives us $\frac{9a-18}{16}=a+b+c$. We know that $a+b+c$ is an integer, so 9a-18 must be divisible by 16. Let $9a=z$. If ${z-18}\equiv {0} \pmod{16}$, then ${z}\equiv {2} \pmod{16}$. Therefore, if $9a=2$, $a=\frac{2}{9}$. Adding up gives us $2+9=\boxed{011}$

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
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