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Difference between revisions of "2014 AIME I Problems/Problem 6"

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The graphs <math>y=3(x-h)^2+j</math> and <math>y=2(x-h)^2+k</math> have y-intercepts of <math>2013</math> and <math>2014</math>, respectively, and each graph has two positive integer x-intercepts. Find <math>h</math>.
 
The graphs <math>y=3(x-h)^2+j</math> and <math>y=2(x-h)^2+k</math> have y-intercepts of <math>2013</math> and <math>2014</math>, respectively, and each graph has two positive integer x-intercepts. Find <math>h</math>.
  
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== Solution ==
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== Solution 1 ==
 
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\ge32</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\ge32</math>.
  
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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>.
 
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>.
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== Solution 2 ==
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Let <math>x=0</math> and <math>y=2013</math> for the first equation, resulting in <math>j=2013-3h^2</math>.  Substituting back in to the original equation, we get <math>y=3(x-h)^2+2013-3h^2</math>.
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Now we set <math>y</math> equal to zero, since there are two distinct positive integer roots.  Rearranging, we get <math>2013=3h^2-3(x-h)^2</math>, which simplifies to <math>671=h^2-(x-h)^2</math>.  Applying difference of squares, we get <math>671=(2h-x)(x)</math>.
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Now, we know that <math>x</math> and <math>h</math> are both integers, so we can use the fact that <math>671=61\times11</math>, and set <math>2h-x=11</math> and <math>x=61</math> (note that letting <math>x=11</math> gets the same result).  Therefore, <math>h=\boxed{036}</math>.
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Note that we did not use the second equation since we took advantage of the fact that AIME answers must be integers.  However, one can enter <math>h=36</math> into the second equation to verify the validity of the answer.
  
 
== 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 17:50, 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 1

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, $h\ge32$.

You'll know that $h^2=\frac{2014-k}{2}$, so you now need to find a positive integer $h$ which has positive integer 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 $\boxed{036}$.


Solution 2

Let $x=0$ and $y=2013$ for the first equation, resulting in $j=2013-3h^2$. Substituting back in to the original equation, we get $y=3(x-h)^2+2013-3h^2$.

Now we set $y$ equal to zero, since there are two distinct positive integer roots. Rearranging, we get $2013=3h^2-3(x-h)^2$, which simplifies to $671=h^2-(x-h)^2$. Applying difference of squares, we get $671=(2h-x)(x)$.

Now, we know that $x$ and $h$ are both integers, so we can use the fact that $671=61\times11$, and set $2h-x=11$ and $x=61$ (note that letting $x=11$ gets the same result). Therefore, $h=\boxed{036}$.

Note that we did not use the second equation since we took advantage of the fact that AIME answers must be integers. However, one can enter $h=36$ into the second equation to verify the validity of the answer.

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

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