Difference between revisions of "2011 AIME I Problems/Problem 15"
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==Solution 4== | ==Solution 4== | ||
− | We have <math>(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc</math> | + | We have |
+ | |||
+ | <math>(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc</math> | ||
+ | |||
As a result, we have | As a result, we have | ||
+ | |||
<math>a+b+c=0</math> | <math>a+b+c=0</math> | ||
+ | |||
<math>ab+bc+ac=-2011</math> | <math>ab+bc+ac=-2011</math> | ||
+ | |||
<math>abc=-m</math> | <math>abc=-m</math> | ||
+ | |||
So, <math>a=-b-c</math> | So, <math>a=-b-c</math> | ||
+ | |||
As a result, <math>ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011</math> | As a result, <math>ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011</math> | ||
+ | |||
Solve <math>b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}</math> and | Solve <math>b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}</math> and | ||
<math>\Delta =8044-3c^2=k^2</math>, where <math>k</math> is an integer | <math>\Delta =8044-3c^2=k^2</math>, where <math>k</math> is an integer | ||
+ | |||
Cause <math>89<\sqrt{8044}<90</math> | Cause <math>89<\sqrt{8044}<90</math> | ||
+ | |||
So, after we tried for <math>2</math> times, we get <math>k=88</math> and <math>c=10</math> | So, after we tried for <math>2</math> times, we get <math>k=88</math> and <math>c=10</math> | ||
+ | |||
then <math>b=39</math>, <math>a=-b-c=-49</math> | then <math>b=39</math>, <math>a=-b-c=-49</math> | ||
+ | |||
As a result, <math>|a|+|b|+|c|=10+39+49=\boxed{098}</math> | As a result, <math>|a|+|b|+|c|=10+39+49=\boxed{098}</math> | ||
+ | |||
Solution By FYC | Solution By FYC | ||
Revision as of 07:57, 13 August 2018
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 two
.
. WLOG, let
.
Then if , then
and if
, then
.
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
.
.
Solution 3
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
. -Gideontz
Solution 4
We have
As a result, we have
So,
As a result,
Solve and
, where
is an integer
Cause
So, after we tried for times, we get
and
then ,
As a result,
Solution By FYC
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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.