Difference between revisions of "2011 AIME I Problems/Problem 15"
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Factoring the perfect square, we get: <math>ab-(b+a)^2=-2011</math> or <math>(b+a)^2-ab=2011</math> | Factoring the perfect square, we get: <math>ab-(b+a)^2=-2011</math> or <math>(b+a)^2-ab=2011</math> | ||
− | Therefore, a sum (a+b) squared minus a product (ab) gives 2011. | + | Therefore, a sum (<math>a+b</math>) squared minus a product (<math>ab</math>) gives <math>2011</math>. |
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Revision as of 16:23, 20 June 2011
Contents
Problem
For some integer , the polynomial
has the three integer roots
,
, and
. Find
.
Solution
With Vieta's formula, we know that , and
.
since any one being zero will make the the other 2
.
. 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 a to .
Let's us analyze .
Here is a table:
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---|---|
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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: 089
Solution 2
Starting off like the previous solution, we know that a + b + c = 0, and ab + bc + ac = -2011
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
.
|-49|+10+39 = 98
Answer: 089
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 |