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

Revision as of 23:10, 6 November 2022

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

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

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

Video Solution

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

@@ Line 12: / Line 12: @@
 To do this, we caN use the formula for the minimum (or maximum) value of the <math>x</math> coordinate at a vertex of a parabola, <math>-\frac{b}{2a}</math> and equate this to <math>\frac{1}{4}</math>. Solving, we get <math>-\frac{a}{2}=b</math>. Enter <math>x=\frac{1}{4}</math> to get <math>-\frac{9}{8}=\frac{a}{16}+\frac{b}{4}+c=-\frac{a}{16}+c</math> so <math>c=\frac{a-18}{16}</math>. This means that <math>\frac{9a-18}{16}\in Z</math> so the minimum of <math>a>0</math> is when the fraction equals -1, so <math>a=\frac{2}{9}</math>. Therefore, <math>p+q=2+9=\boxed{011}</math>.
 -Gideontz
+==Solution 4==
+Write this as <math>a\left( x- \frac 14 \right)^2 - \frac 98</math>. Since <math>a+b+c</math> is equal to the value of this expression when you plug <math>x=1</math> in, we just need <math>\frac{9a}{16}- \frac 98</math> to be an integer. Since <math>a>0</math>, we also have <math>\frac{9a}{16}>0</math> which means <math>\frac{9a}{16}- \frac 98 > -\frac{9}{8}</math>. The least possible value of <math>a</math> is when this is equal to <math>-1</math>, or <math>a=\frac 29</math>, which gives answer <math>11</math>.
+-bobthegod78, krwang, Simplest14
 ==Video Solution==

2011 AIME I (Problems • Answer Key • Resources)
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