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 "2014 AIME I Problems/Problem 6"
(→‎Solution)
Line 4: Line 4:
  
 
== Solution ==
 
== Solution ==
−
We use the fact that 2013 and 2014 are y-intercepts by plugging them into the equations to get <math>2013=3h^2+j</math> and <math>2014=2h^2+k.</math>
+
Begin by setting <math>x</math> to 0, then set both equations to <math>h^2=\frac{2013-j}{3}</math> and <math>h^2=\frac{2014-k}{2}</math>, respectively. You'll notice that because the two parabolas have to have x-intercepts, <math>h\le32</math>.
 +
 
 +
You'll know that <math>h^2=\frac{2014-k}{2}</math>, so you now need to find a positive integer <math>h</math> which has [b]positive integer[/b] x-intercepts for both equations.
 +
 
 +
Notice that if <math>k=2014-2h^2</math> is -2 times a square number, then you have found a value of <math>h</math> for which the second equation has positive x-intercepts. We guess and check <math>h=36</math> to obtain <math>k=-578=-2(17^2)</math>.
 +
 
 +
Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is <math>\boxed{036}</math>.
 +
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=5|num-a=7}}
 
{{AIME box|year=2014|n=I|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:44, 15 March 2014

Problem 6

The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$, respectively, and each graph has two positive integer x-intercepts. Find $h$.

Solution

Begin by setting $x$ to 0, then set both equations to $h^2=\frac{2013-j}{3}$ and $h^2=\frac{2014-k}{2}$, respectively. You'll notice that because the two parabolas have to have x-intercepts, $h\le32$.

You'll know that $h^2=\frac{2014-k}{2}$, so you now need to find a positive integer $h$ which has [b]positive integer[/b] x-intercepts for both equations.

Notice that if $k=2014-2h^2$ is -2 times a square number, then you have found a value of $h$ for which the second equation has positive x-intercepts. We guess and check $h=36$ to obtain $k=-578=-2(17^2)$.

Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is $\boxed{036}$.

See also

2014 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=2014_AIME_I_Problems/Problem_6&oldid=61055"