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2014 AIME I Problems/Problem 6

Revision as of 18:14, 11 March 2018 by Damalone (talk | contribs) (→‎Solution 2)

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.

Note on the previous note: we still must use the second equation since we could also use $671=671\times1$, yielding $h=336.$ This answer however does not check out with the second equation which is why it is invalid.

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 $\boxed{036}$.

\[y=3(x-h)^2+j\rightarrow y=3(x-11)(x-61)=3x^2-216x+2013\] \[y=2(x-h)^2+k\rightarrow y=2(x-19)(x-53)=2x^2-144x+2014\]

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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