2011 AIME I Problems/Problem 15
For some integer , the polynomial has the three integer roots , , and . Find .
With Vieta's formulas, we know that , and .
since any one being zero will make the other two .
. WLOG, let .
Then if , then and if , then .
We know that , have the same sign. So . ( and )
Also, if we fix , is maximized when . Hence, .
Now we have limited to .
Let's 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, , . , .
Starting off like the previous solution, we know that , and .
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 .
Since no factors of can sum to ( being the largest sum), a + b cannot equal .
and so cannot work either.
We can continue to do this until we reach .
, so one root is and another is . The roots sum to zero, so the last root must be .
Let us first note the obvious that is derived from Vieta's formulas: . Now, due to the first equation, let us say that , meaning that and . Now, since both and are greater than 0, their absolute values are both equal to and , respectively. Since is less than 0, it equals . Therefore, , meaning . We now apply Newton's sums to get that ,or . Solving, we find that satisfies this, meaning , so .
As a result, we have
As a result,
Solve and , where is an integer
So, after we tried for times, we get and
As a result,
This is a misplaced number theory problem.
|2011 AIME I (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|