Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AIME I Problems/Problem 6"
(→‎Solution)
Line 4: Line 4:
 
==Solution==
 
==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 <math>y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}</math>.
−
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 <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 <math>9a-18</math> 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>
  
 
== See also ==
 
== See also ==

Revision as of 14:14, 24 June 2014

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 $2+9=\boxed{011}$

See also

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_6&oldid=62328"