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

Revision as of 20:16, 12 June 2011

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

With Vieta's formula, 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 the other 2 $\pm \sqrt{2011}$ .

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

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

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

$a^2 = 2011 + bc$

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:

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: $098$

@@ Line 22: / Line 22: @@
 We know that <math>b</math>, <math>c</math> have the same sign. So <math>|a| \ge 45</math>. (<math>44^2<2011</math> and <math>45^2 = 2025</math>)
-Also, <math>bc</math> maximize when <math>b = c</math> if we fixed <math>a</math>. Hence, <math>2011 = a^2 - bc < \frac{3}{4}a^2</math>.
+Also, <math>bc</math> maximize when <math>b = c</math> if we fixed <math>a</math>. Hence, <math>2011 = a^2 - bc > \frac{3}{4}a^2</math>.
 So <math>a ^2 < \frac{(4)2011}{3} = 2681+\frac{1}{3}</math>.

2011 AIME I (Problems • Answer Key • Resources)
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