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

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==Solution 3==
 
==Solution 3==
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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>.
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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
 
-Gideontz
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==Solution 4==
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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>.
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-bobthegod78, krwang, Simplest14
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==Solution 5 (You don't remember conic section formulae)==
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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.
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~Dhillonr25
  
 
==Video Solution==
 
==Video Solution==
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== See also ==
 
== See also ==
 
{{AIME box|year=2011|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2011|n=I|num-b=5|num-a=7}}
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[[Category:Intermediate Algebra Problems]]
 +
 
* [[AIME Problems and Solutions]]
 
* [[AIME Problems and Solutions]]
 
* [[American Invitational Mathematics Examination]]
 
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest 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

See also

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

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