Difference between revisions of "2011 AIME I Problems/Problem 15"
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− | <math>a,b,c\neq 0</math> since any one being zero will make | + | <math>a,b,c\neq 0</math> since any one being zero will make the other <math>2 \pm \sqrt{2011}</math>. |
<math>a = -(b+c)</math>. WLOG, let <math>|a| \ge |b| \ge |c|</math>. | <math>a = -(b+c)</math>. WLOG, let <math>|a| \ge |b| \ge |c|</math>. |
Revision as of 20:45, 28 February 2016
Contents
Problem
For some integer , the polynomial has the three integer roots , , and . Find .
Solution
With Vieta's formulas, we know that , and .
since any one being zero will make the other .
. WLOG, let .
Then if , then and if , .
We know that , have the same sign. So . ( and )
Also, maximize when if we fixed . Hence, .
So .
so .
Now we have limited to .
Let's us analyze .
Here is a table:
We can tell we don't need to bother with ,
, So won't work. ,
is not divisible by , , which is too small to get .
, is not divisible by or or , we can clearly tell that is too much.
Hence, , . , .
Answer:
Solution 2
Starting off like the previous solution, we know that , and .
Therefore, .
Substituting, .
Factoring the perfect square, we get: or .
Therefore, a sum () squared minus a product () gives ..
We can guess and check different ’s starting with since .
therefore .
Since no factors of can sum to ( being the largest sum), a + b cannot equal .
making .
and so cannot work either.
We can continue to do this until we reach .
making .
, so one root is and another is . The roots sum to zero, so the last root must be .
.
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by - | |
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.