Difference between revisions of "2014 AIME I Problems/Problem 6"
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<cmath>2h=671/x_1+x_1^2/x^1</cmath> <cmath>2h=671/x_1+x_1</cmath> <cmath>x_1+x_2=671/x_1+x_1</cmath> <cmath>x_2=671/x_1</cmath> <cmath>x_1x_2=671.</cmath> Prime factorization of 671 gives 11 and 61. So now we know <math>x_1=11</math> and <math>x_2=61</math>. Lastly, we plug in the numbers,11 and 61, into <math>x_1+x_2=2h</math>, so <math>\boxed{h=36}</math>. | <cmath>2h=671/x_1+x_1^2/x^1</cmath> <cmath>2h=671/x_1+x_1</cmath> <cmath>x_1+x_2=671/x_1+x_1</cmath> <cmath>x_2=671/x_1</cmath> <cmath>x_1x_2=671.</cmath> Prime factorization of 671 gives 11 and 61. So now we know <math>x_1=11</math> and <math>x_2=61</math>. Lastly, we plug in the numbers,11 and 61, into <math>x_1+x_2=2h</math>, so <math>\boxed{h=36}</math>. | ||
+ | ==Solution 6 (Vieta's solution)== | ||
+ | First, we start of exactly like solutions above and we find out that <math>j=2013-3h^2</math> and <math>k=2014-2h^2</math> We then plug j and k into <math>3(x-h)^2+j</math> and <math>y=2(x-h)^2+k</math> respectively. After that, we get two equations, <math>y=3x^2-6xh+2013</math> and <math>y=2x^2-4xh+2014</math>. We can apply vieta's. Let the roots of the first equation be <math>a, b</math> and the roots of the second equation be <math>x, y</math>. Thus, we have that <math>a*b=1007</math> <math>a+b=2h</math> and <math>x*y=671</math> <math>x+y=2h</math> Simple evaluations finds that <math>\boxed{h=36}</math> | ||
+ | |||
+ | ~Jske25 | ||
== 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}} |
Latest revision as of 20:58, 9 April 2021
Contents
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 And the second parabola gives
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 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 .
Solution 4
First, we expand both equations to get and . The -intercept for the first equation can be expressed as . From this, the x-intercepts for the first equation can be written as
Since the -intercepts must be integers, must also be an integer. From solution 1, we know must be greater than or equal to 32. We can substitute increasing integer values for starting from 32; we find that .
We can test this result using the second equation, whose -intercepts are Substituting 36 in for , we get , which satisfies the requirement that all x-intercepts must be (positive) integers.
Thus, .
Solution 5
We have the equation
We know: , so after plugging in the values and isolating . Therefore, .
Lets call the x-intercepts , . Since both and are positive there is a relationship between , and . Namely, . The is because: ,
Similarly, we know: , so . Combining the two equations gives us
Now since we have this relationship, , we can just multiply the last equation by 2(so that we get on the left side) which gives us Prime factorization of 671 gives 11 and 61. So now we know and . Lastly, we plug in the numbers,11 and 61, into , so .
Solution 6 (Vieta's solution)
First, we start of exactly like solutions above and we find out that and We then plug j and k into and respectively. After that, we get two equations, and . We can apply vieta's. Let the roots of the first equation be and the roots of the second equation be . Thus, we have that and Simple evaluations finds that
~Jske25
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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