Difference between revisions of "2014 AIME I Problems/Problem 6"
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| − | Begin by setting <math>x</math> to 0, then set both equations to <math>h^2=\frac{2013-j}{3}</math> and <math>h^2=\frac{2014-k}{2}</math>, respectively. You'll notice that because the two parabolas have to have x-intercepts, <math>h\ | + | Begin by setting <math>x</math> to 0, then set both equations to <math>h^2=\frac{2013-j}{3}</math> and <math>h^2=\frac{2014-k}{2}</math>, respectively. You'll notice that because the two parabolas have to have x-intercepts, <math>h\ge32</math>. |
You'll know that <math>h^2=\frac{2014-k}{2}</math>, so you now need to find a positive integer <math>h</math> which has positive integer x-intercepts for both equations. | You'll know that <math>h^2=\frac{2014-k}{2}</math>, so you now need to find a positive integer <math>h</math> which has positive integer x-intercepts for both equations. | ||
Revision as of 13:48, 15 March 2014
Problem 6
The graphs 
and 
have y-intercepts of 
and 
, respectively, and each graph has two positive integer x-intercepts. Find 
.
Solution
Begin by setting 
to 0, then set both equations to 
and 
, respectively. You'll notice that because the two parabolas have to have x-intercepts, 
.
You'll know that , so you 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 
.
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. 
