2011 AIME I Problems/Problem 6


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 $p + q$.


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 $2+9=\boxed{011}$.

Solution 3

To do this, we caN use the formula for the minimum (or maximum) value of the $x$ coordinate at a vertex of a parabola, $-\frac{b}{2a}$ and equate this to $\frac{1}{4}$. Solving, we get $-\frac{a}{2}=b$. Enter $x=\frac{1}{4}$ to get $-\frac{9}{8}=\frac{a}{16}+\frac{b}{4}+c=-\frac{a}{16}+c$ so $c=\frac{a-18}{16}$. This means that $\frac{9a-18}{16}\in Z$ so the minimum of $a>0$ is when the fraction equals -1, so $a=\frac{2}{9}$. Therefore, $p+q=2+9=\boxed{011}$. -Gideontz

Video Solution


See also

2011 AIME I (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS