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

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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.
 
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.
 +
 +
==Solution 3==
 +
Similar to the first two solutions, we deduce that <math>\text{(-)}j</math> and <math>\text{(-)}k</math> are of the form <math>3a^2</math> and <math>2b^2</math>, respectively, because the roots are integers and so is the <math>y</math>-intercept of both equations. So the <math>x</math>-intercepts should be integers also.
 +
 +
The first parabola gives
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<cmath>3h^2+j=3\left(h^2-a^2\right)=2013</cmath>
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<cmath>h^2-a^2=671</cmath>
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And the second parabola gives
 +
<cmath>2h^2+k=2\left(h^2-b^2\right)=2014</cmath>
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<cmath>h^2-b^2=1007</cmath>
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 +
We know that <math>671=11\cdot 61</math> and that <math>1007=19\cdot 53</math>. It is just a fitting coincidence that the average of <math>11</math> and <math>61</math> is the same as the average of <math>19</math> and <math>53</math>. That is <math>\boxed{036}</math>.
 +
 +
To check, we have
 +
<cmath>(h-a)(h+a)=671=11\cdot 61</cmath>
 +
<cmath>(h-b)(h+b)=1007=19\cdot 53</cmath>
 +
Those are the only two prime factors of <math>671</math> and <math>1007</math>, respectively. So we don't need any new factorizations for those numbers.
 +
 +
<math>h+a=61,h-a=11\implies (h,a)=\{36,25\}</math>
 +
 +
<math>h+b=53,h-b=19\implies (h,b)=\{36,17\}</math>
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 +
Thus the common integer value for <math>h</math> 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 12:09, 14 February 2016

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. Notice that because the two parabolas have to have positive x-intercepts, $h\ge32$.

We see that $h^2=\frac{2014-k}{2}$, so we 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.

Solution 3

Similar to the first two solutions, we deduce that $\text{(-)}j$ and $\text{(-)}k$ are of the form $3a^2$ and $2b^2$, respectively, because the roots are integers and so is the $y$-intercept of both equations. So the $x$-intercepts should be integers also.

The first parabola gives \[3h^2+j=3\left(h^2-a^2\right)=2013\] \[h^2-a^2=671\] And the second parabola gives \[2h^2+k=2\left(h^2-b^2\right)=2014\] \[h^2-b^2=1007\]

We know that $671=11\cdot 61$ and that $1007=19\cdot 53$. It is just a fitting coincidence that the average of $11$ and $61$ is the same as the average of $19$ and $53$. That is $\boxed{036}$.

To check, we have \[(h-a)(h+a)=671=11\cdot 61\] \[(h-b)(h+b)=1007=19\cdot 53\] Those are the only two prime factors of $671$ and $1007$, respectively. So we don't need any new factorizations for those numbers.

$h+a=61,h-a=11\implies (h,a)=\{36,25\}$

$h+b=53,h-b=19\implies (h,b)=\{36,17\}$

Thus the common integer value for $h$ 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

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