AIME 2011 I · 第 6 题

Solution 1

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 $\frac{9a-18}{16}\in \mathbb{Z}$ 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

Solution 4

Write this as $a\left( x- \frac 14 \right)^2 - \frac 98$. Since $a+b+c$ is equal to the value of this expression when you plug $x=1$ in, we just need $\frac{9a}{16}- \frac 98$ to be an integer. Since $a>0$, we also have $\frac{9a}{16}>0$ which means $\frac{9a}{16}- \frac 98 > -\frac{9}{8}$. The least possible value of $a$ is when this is equal to $-1$, or $a=\frac 29$, which gives answer $11$.

-bobthegod78, krwang, Simplest14

Solution 5 (You don't remember conic section formulae)

Take the derivative to get that the vertex is at $2ax+b=0$ and note that this implies $\frac{1}{2} \cdot a = -b$ and proceed with any of the solutions above.

~Dhillonr25

$\textbf{Appendix to Solution 5 (You don't remember differential calculus)}$

Note that the quadratic formula for finding roots of parabolas is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, so if you average the two x-values of the roots the $\pm\sqrt{b^2-4ac}$ part will cancel out and leave you with $x=-\frac{b}{2a}$. Then proceed as above.

~WhatdoHumanitariansEat

Solution 6 (No Words)

$y = ax^2 + bx + c$ $y=a\left(x^2+\frac{b}{a}x\right)+c$ $y=a\left(x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2\right)-a\left(\frac{b}{2a}\right)^2+c$

$y=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c$.

$\frac{b}{2a}=-\frac{1}{4} \implies b=-\frac{1}{2}a$.

$- \frac{b^2}{4a} + c= - \frac{9}{8}$

$c=\frac{b^2}{4a}- \frac{9}{8} =\frac{b}{2} \cdot \frac{b}{2a} - \frac{9}{8} = -\frac{b}{8}-\frac{9}{8} = \frac{a}{16}-\frac{9}{8}$.

$a+b+c=a-\frac{1}{2}a+\frac{a}{16}-\frac{9}{8} = \frac{16a-8a+a}{16}-\frac{9}{8} = \frac{9a-18}{16} \in \mathbb{Z}$.

$16 \mid 9(a-2) \implies a-2=b \cdot \frac{16}{9}, b \in \mathbb{Z}$.

$a=\frac{16b+18}{9} \implies a = \ldots, -\frac{14}{9}, \frac{2}{9}, \frac{18}{9}, \frac{34}{9}, \ldots$.

$a>0 \implies a_{min} = \frac{2}{9} \implies 2+9 = \boxed{11}$.

Video Solution

https://www.youtube.com/watch?v=vkniYGN45F4