Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2011 AIME I Problems/Problem 15"

(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 22:52, 12 August 2018

Problem

For some integer $m$, the polynomial $x^3 - 2011x + m$ has the three integer roots $a$, $b$, and $c$. Find $|a| + |b| + |c|$.

Solution

With Vieta's formulas, we know that $a+b+c = 0$, and $ab+bc+ac = -2011$.



$a,b,c\neq 0$ since any one being zero will make the other two $\pm \sqrt{2011}$.

$a = -(b+c)$. WLOG, let $|a| \ge |b| \ge |c|$.

Then if $a > 0$, then $b,c < 0$ and if $a < 0$, then $b,c > 0$.


$ab+bc+ac = -2011 = a(b+c)+bc = -a^2+bc$


$a^2 = 2011 + bc$

We know that $b$, $c$ have the same sign. So $|a| \ge 45$. ($44^2<2011$ and $45^2 = 2025$)

Also, $bc$ maximize when $b = c$ if we fixed $a$. Hence, $2011 = a^2 - bc > \frac{3}{4}a^2$.

So $a ^2 < \frac{(4)2011}{3} = 2681+\frac{1}{3}$.

$52^2 = 2704$ so $|a| \le 51$.



Now we have limited $a$ to $45\le |a| \le 51$.

Let's us analyze $a^2 = 2011 + bc$.


Here is a table:

$|a|$$a^2 = 2011 + bc$
$45$$14$
$46$$14 + 91 =105$
$47$$105 + 93 = 198$
$48$$198 + 95 = 293$
$49$$293 + 97 = 390$


We can tell we don't need to bother with $45$,

$105 = (3)(5)(7)$, So $46$ won't work. $198/47 > 4$,

$198$ is not divisible by $5$, $198/6 = 33$, which is too small to get $47$.

$293/48 > 6$, $293$ is not divisible by $7$ or $8$ or $9$, we can clearly tell that $10$ is too much.


Hence, $|a| = 49$, $a^2 -2011 = 390$. $b = 39$, $c = 10$.

Answer: $\boxed{098}$

Solution 2

Starting off like the previous solution, we know that $a + b + c = 0$, and $ab + bc + ac = -2011$.

Therefore, $c = -b-a$.

Substituting, $ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011$.

Factoring the perfect square, we get: $ab-(b+a)^2=-2011$ or $(b+a)^2-ab=2011$.

Therefore, a sum ($a+b$) squared minus a product ($ab$) gives $2011$..


We can guess and check different $a+b$’s starting with $45$ since $44^2 < 2011$.

$45^2 = 2025$ therefore $ab = 2025-2011 = 14$.

Since no factors of $14$ can sum to $45$ ($1+14$ being the largest sum), a + b cannot equal $45$.

$46^2 = 2116$ making $ab = 105 = 3 * 5 * 7$.

$5 * 7 + 3 < 46$ and $3 * 5 * 7 > 46$ so $46$ cannot work either.


We can continue to do this until we reach $49$.

$49^2 = 2401$ making $ab = 390 = 2 * 3 * 5* 13$.

$3 * 13 + 2* 5 = 49$, so one root is $10$ and another is $39$. The roots sum to zero, so the last root must be $-49$.


$|-49|+10+39 = \boxed{098}$.


Solution 3

Let us first note the obvious that is derived from Vieta's formulas: $a+b+c=0, ab+bc+ac=-2011$. Now, due to the first equation, let us say that $a+b=-c$, meaning that $a,b>0$ and $c<0$. Now, since both $a$ and $b$ are greater than 0, their absolute values are both equal to $a$ and $b$, respectively. Since $c$ is less than 0, it equals $-a-b$. Therefore, $|c|=|-a-b|=a+b$, meaning $|a|+|b|+|c|=2(a+b)$. We now apply Newton's sums to get that $a^2+b^2+ab=2011$,or $(a+b)^2-ab=2011$. Solving, we find that $49^2-390$ satisfies this, meaning $a+b=49$, so $2(a+b)=\boxed{098}$. -Gideontz

Solution 4

We have $(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc$ As a result, we have $a+b+c=0$ $ab+bc+ac=-2011$ $abc=-m$ So, $a=-b-c$ As a result, $ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011$ Solve $b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}$ and $\Delta =8044-3c^2=k^2$, where $k$ is an integer Cause $89<\sqrt{8044}<90$ So, after we tried for $2$ times, we get $k=88$ and $c=10$ then $b=39$, $a=-b-c=-49$ As a result, $|a|+|b|+|c|=10+39+49=\boxed{098}$ 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_15&oldid=97205"