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

Revision as of 00:18, 11 December 2023

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

Video Solution

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

@@ Line 3: / Line 3: @@
 ==Solution==
-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 <cmath>y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}.</cmath>
-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>
+Expanding, we find that <cmath>y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8},</cmath> and <cmath>y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}.</cmath> 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 <cmath>\frac{9a-18}{16}=a+b+c.</cmath> We know that <math>a+b+c</math> is an integer, so <math>9a-18</math> must be divisible by <math>16</math>.  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>
+==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>.
+==Solution 3==
+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 \mathbb{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
+==Solution 5 (You don't remember conic section formulae)==
+Take the derivative to get that the vertex is at <math>2ax+b=0</math> and note that this implies <math>\frac{1}{2} \cdot a = -b</math> and proceed with any of the solutions above.
+~Dhillonr25
+==Video Solution==
+https://www.youtube.com/watch?v=vkniYGN45F4
 == See also ==
-{{AIME box|year=2011|before=Problem 5|num-a=7}}
+{{AIME box|year=2011|n=I|num-b=5|num-a=7}}
+[[Category:Intermediate Algebra Problems]]
 * [[AIME Problems and Solutions]]
 * [[American Invitational Mathematics Examination]]
 * [[Mathematics competition resources]]
+{{MAA Notice}}

2011 AIME I (Problems • Answer Key • Resources)
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

Page

Toolbox

Search