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

Revision as of 21:48, 16 February 2021

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, if we fix $a$ , $bc$ is maximized when $b = c$ . 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 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 $2(a+b)=\boxed{098}$ .

Solution 4

We have

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 5 (mod to help bash)

First, derive the equations $a=-b-c$ and $ab+bc+ca=-2011\implies b^2+bc+c^2=2011$ . Since the product is negative, $a$ is negative, and $b$ and $c$ positive. Now, a simple mod 3 testing of all cases shows that $b\equiv \{1,2\} \pmod{3}$ , and $c$ has the repective value. We can choose $b$ not congruent to 0, make sure you see why. Now, we bash on values of $b$ , testing the quadratic function to see if $c$ is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for $b=10$ , $c=39, -49$ . Choosing $c$ positive we get $a=-49$ , so $|a|+|b|+|c|=10+29+39=\boxed{098}$ ~firebolt360

Note

This is a misplaced number theory problem.Which requires a bit of trial and error strategy.

Video Solution

https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx

@@ Line 134: / Line 134: @@
 ==Solution 5 (mod to help bash)==
 First, derive the equations <math>a=-b-c</math> and <math>ab+bc+ca=-2011\implies b^2+bc+c^2=2011</math>. Since the product is negative, <math>a</math> is negative, and <math>b</math> and <math>c</math> positive. Now, a simple mod 3 testing of all cases shows that <math>b\equiv \{1,2\} \pmod{3}</math>, and <math>c</math> has the repective value. We can choose <math>b</math> not congruent to 0, make sure you see why. Now, we bash on values of <math>b</math>, testing the quadratic function to see if <math>c</math> is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for <math>b=10</math>, <math>c=39, -49</math>. Choosing <math>c</math> positive we get <math>a=-49</math>, so <math>|a|+|b|+|c|=10+29+39=\boxed{098}</math>
+~firebolt360
 ==Note==

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

Page

Toolbox

Search