Difference between revisions of "2014 AIME I Problems/Problem 6"
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Thus the common integer value for <math>h</math> is <math>\boxed{036}</math>. | Thus the common integer value for <math>h</math> is <math>\boxed{036}</math>. | ||
+ | |||
+ | <cmath>y=3(x-h)^2+j\rightarrow y=3(x-11)(x-61)=3x^2-72x+2013</cmath> | ||
+ | <cmath>y=2(x-h)^2+k\rightarrow y=2(x-19)(x-53)=2x^2-72x+2014</cmath> | ||
== 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 13:11, 14 February 2016
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.
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
.
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.