Difference between revisions of "2011 AIME I Problems/Problem 15"
Tempaccount (talk | contribs) (Remove extra problem section) |
(→Solution 3) |
||
Line 103: | Line 103: | ||
== Solution 3 == | == Solution 3 == | ||
Let us first note the obvious that is derived from Vieta's formulas: <math>a+b+c=0, ab+bc+ac=-2011</math>. Now, due to the first equation, let us say that <math>a+b=-c</math>, meaning that <math>a,b>0</math> and <math>c<0</math>. Now, since both <math>a</math> and <math>b</math> are greater than 0, their absolute values are both equal to <math>a</math> and <math>b</math>, respectively. Since <math>c</math> is less than 0, it equals <math>-a-b</math>. Therefore, <math>|c|=|-a-b|=a+b</math>, meaning <math>|a|+|b|+|c|=2(a+b)</math>. We now apply Newton's sums to get that <math>a^2+b^2+ab=2011</math>,or <math>(a+b)^2-ab=2011</math>. Solving, we find that <math>49^2-390</math> satisfies this, meaning <math>a+b=49</math>, so <math>2(a+b)=\boxed{098}</math>. -Gideontz | Let us first note the obvious that is derived from Vieta's formulas: <math>a+b+c=0, ab+bc+ac=-2011</math>. Now, due to the first equation, let us say that <math>a+b=-c</math>, meaning that <math>a,b>0</math> and <math>c<0</math>. Now, since both <math>a</math> and <math>b</math> are greater than 0, their absolute values are both equal to <math>a</math> and <math>b</math>, respectively. Since <math>c</math> is less than 0, it equals <math>-a-b</math>. Therefore, <math>|c|=|-a-b|=a+b</math>, meaning <math>|a|+|b|+|c|=2(a+b)</math>. We now apply Newton's sums to get that <math>a^2+b^2+ab=2011</math>,or <math>(a+b)^2-ab=2011</math>. Solving, we find that <math>49^2-390</math> satisfies this, meaning <math>a+b=49</math>, so <math>2(a+b)=\boxed{098}</math>. -Gideontz | ||
+ | |||
+ | ==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> | ||
+ | As a result, we have | ||
+ | <math>a+b+c=0</math> | ||
+ | <math>ab+bc+ac=-2011</math> | ||
+ | <math>abc=-m</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> | ||
+ | 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 | ||
+ | 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> | ||
+ | then <math>b=39</math>, <math>a=-b-c=-49</math> | ||
+ | As a result, <math>|a|+|b|+|c|=10+39+49=\boxed{098}</math> | ||
+ | Solution By FYC | ||
== See also == | == See also == |
Revision as of 23:52, 12 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.