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>.
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==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>
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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>.
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==Solution==
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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 $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$, where $a > 0$ and $a + b + c$ is an integer. The minimum possible value of $a$ can be written in the form $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$.

Solution

If the vertex is at $\left(\frac{1}{4}, -\frac{9}{8}\right)$, the equation of the parabola can be expressed in the form $y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}$. Expanding, we find that $y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8}$ , 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions
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