2014 AIME I Problems/Problem 6

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 $\boxed{036}$ .

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

Solution 4 (unnecessarily long)

First, we expand both equations to get $y=3x^2-6hx+3h^2+j$ and $y=2x^2-4hx+2h^2+k$ . The $y$ -intercept for the first equation can be expressed as $3h^2+j$ . From this, the x-intercepts for the first equation can be written as

$x=h \pm \sqrt{(-6h)^2-4*3(3h^2+j)}=h \pm \sqrt{36h^2-12(2013)}=h \pm \sqrt{36h^2-24156}$ .

Since the $x$ -intercepts must be integers, $\sqrt{36h^2-24156}$ must also be an integer. From solution 1, we know $h$ must be greater than or equal to 32. Solving for $h$ , we find that $h=36$ .

We can test this using the second equation, whose $x$ -intercepts are $x=h \pm \sqrt{(-4h)^2-4*2(2h^2+k)}=h \pm \sqrt{16h^2-8(2014)}=h \pm \sqrt{16h^2-16112}$ . Substituting 36 in for $h$ , we get $h=36 \pm 68$ , which satisfies the requirement that all x-intercepts must be (positive) integers.

Thus, $h=\boxed{036}$ .

2014 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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