Difference between revisions of "2014 AIME I Problems/Problem 6"
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== Problem 6 == | == Problem 6 == | ||
Revision as of 14:41, 9 August 2018
Problem
Problem 6
The graphs 
and 
have y-intercepts of 
and 
, respectively, and each graph has two positive integer x-intercepts. Find 
.
Solution 1
Begin by setting 
to 0, then set both equations to 
and 
, respectively. Notice that because the two parabolas have to have positive x-intercepts, 
.
We see that , so we now need to find a positive integer which has positive integer x-intercepts for both equations.
Notice that if 
is -2 times a square number, then you have found a value of for which the second equation has positive x-intercepts. We guess and check 
to obtain 
.
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 
.
Solution 2
Let 
and 
for the first equation, resulting in 
. Substituting back in to the original equation, we get 
.
Now we set 
equal to zero, since there are two distinct positive integer roots. Rearranging, we get 
, which simplifies to 
. Applying difference of squares, we get 
.
Now, we know that and are both integers, so we can use the fact that 
, and set 
and 
(note that letting 
gets the same result). Therefore, 
.
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 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 
, yielding 
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 
and 
are of the form 
and 
, respectively, because the roots are integers and so is the -intercept of both equations. So the -intercepts should be integers also.
The first parabola gives
![\[3h^2+j=3\left(h^2-a^2\right)=2013\]](http://latex.artofproblemsolving.com/0/a/8/0a874c62d4bdaaf0a5837dfeb8fbe45f0e8ea71e.png)
![\[h^2-a^2=671\]](http://latex.artofproblemsolving.com/8/d/d/8dd7bd718c60b439b2e3dede8667e2f1b698dc6c.png)
And the second parabola gives
![\[2h^2+k=2\left(h^2-b^2\right)=2014\]](http://latex.artofproblemsolving.com/1/e/b/1eb7e43ba232fc9ac018912a97466419fd1b11cc.png)
![\[h^2-b^2=1007\]](http://latex.artofproblemsolving.com/4/5/0/45016c7c10cafbacd9682f88e1714471a13ecd0c.png)
We know that 
and that 
. It is just a fitting coincidence that the average of 
and 
is the same as the average of 
and 
. That is .
To check, we have
![\[(h-a)(h+a)=671=11\cdot 61\]](http://latex.artofproblemsolving.com/3/a/e/3ae3d42dc525bf72b754ea062ee6fc36b2edcba8.png)
![\[(h-b)(h+b)=1007=19\cdot 53\]](http://latex.artofproblemsolving.com/4/2/f/42fcb33e218f8471c1948692c73ae5f9341bd9f8.png)
Those are the only two prime factors of 
and 
, respectively. So we don't need any new factorizations for those numbers.
Thus the common integer value for is .
![\[y=3(x-h)^2+j\rightarrow y=3(x-11)(x-61)=3x^2-216x+2013\]](http://latex.artofproblemsolving.com/8/9/5/895e8d8d9108133044e5c467c0cd9752b1e5680d.png)
![\[y=2(x-h)^2+k\rightarrow y=2(x-19)(x-53)=2x^2-144x+2014\]](http://latex.artofproblemsolving.com/6/5/b/65bcb7f7ba7f33075896aeea632838eedc3263c2.png)
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. 
