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Difference between revisions of "2011 AIME I Problems/Problem 6"

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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>.
 
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 <math>9a-18</math> 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\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 <math>9a-18</math> 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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 +
==Solution 2==
 +
Complete the square. Since <math>a>0</math>, the parabola must be facing upwards. <math>a+b+c=\text{integer}</math> means that <math>f(1)</math> must be an integer. The function can be recasted into <math>a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}</math> because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than <math>-\frac{9}{8}</math> is <math>-1</math>. So the <math>y</math>-coordinate must change by <math>\frac{1}{8}</math> and the <math>x</math>-coordinate must change by <math>1-\frac{1}{4}=\frac{3}{4}</math>. Thus, <math>a\left(\frac{3}{4}\right)^2=\frac{1}{8}\implies \frac{9a}{16}=\frac{1}{8}\implies a=\frac{2}{9}</math>. So <math>2+9=\boxed{011}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:40, 2 March 2016

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 $2+9=\boxed{011}$

Solution 2

Complete the square. Since $a>0$, the parabola must be facing upwards. $a+b+c=\text{integer}$ means that $f(1)$ must be an integer. The function can be recasted into $a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}$ because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than $-\frac{9}{8}$ is $-1$. So the $y$-coordinate must change by $\frac{1}{8}$ and the $x$-coordinate must change by $1-\frac{1}{4}=\frac{3}{4}$. Thus, $a\left(\frac{3}{4}\right)^2=\frac{1}{8}\implies \frac{9a}{16}=\frac{1}{8}\implies a=\frac{2}{9}$. So $2+9=\boxed{011}$.

See also

2011 AIME I (Problems • Answer Key • Resources)
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Problem 5 		Followed by
Problem 7
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