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

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==Solution==
 
==Solution==
  
With Vieta's formula, we know that <math>a+b+c = 0</math>, and <math>ab+bc+ac = -2011</math>.  
+
With Vieta's formulas, we know that <math>a+b+c = 0</math>, and <math>ab+bc+ac = -2011</math>.  
  
  
 
<br />
 
<br />
<math>a,b,c\neq 0</math> since any one being zero will make the the other 2 <math>\pm \sqrt{2011}</math>.  
+
<math>a,b,c\neq 0</math> since any one being zero will make the other two <math> \pm \sqrt{2011}</math>.  
  
 
<math>a = -(b+c)</math>. WLOG, let <math>|a| \ge |b| \ge |c|</math>.  
 
<math>a = -(b+c)</math>. WLOG, let <math>|a| \ge |b| \ge |c|</math>.  
  
Then if <math>a > 0</math>, then <math>b,c < 0</math> and if <math>a < 0</math>, <math>b,c > 0</math>.
+
Then if <math>a > 0</math>, then <math>b,c < 0</math> and if <math>a < 0</math>, then <math>b,c > 0</math>.
  
 
<br />
 
<br />
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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>)
 
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, if we fix <math>a</math>, <math>bc</math> is maximized when <math>b = c</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>.  
 
So <math>a ^2 < \frac{(4)2011}{3} = 2681+\frac{1}{3}</math>.  
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<br />
 
<br />
Now we have limited a to <math>45\le |a| \le 51</math>.
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Now we have limited <math>a</math> to <math>45\le |a| \le 51</math>.
  
Let's us analyze <math>a^2 = 2011 + bc</math>.
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Let's analyze <math>a^2 = 2011 + bc</math>.
  
 
<br />
 
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<math>105 = (3)(5)(7)</math>, So <math>46</math> won't work. <math>198/47 > 4</math>,  
 
<math>105 = (3)(5)(7)</math>, So <math>46</math> won't work. <math>198/47 > 4</math>,  
  
<math>198</math> is not divisible by <math>5</math>, <math>198/6 = 33</math>, which is too small to get <math>47</math>
+
<math>198</math> is not divisible by <math>5</math>, <math>198/6 = 33</math>, which is too small to get <math>47</math>.
  
<math>293/48 > 6</math>, <math>293</math> is not divisible by <math>7</math> or <math>8</math> or <math>9</math>, we can clearly tell that <math>10</math> is too much
+
<math>293/48 > 6</math>, <math>293</math> is not divisible by <math>7</math> or <math>8</math> or <math>9</math>, we can clearly tell that <math>10</math> is too much.
  
  
 
Hence, <math>|a| = 49</math>, <math>a^2 -2011 = 390</math>. <math>b = 39</math>, <math>c = 10</math>.
 
Hence, <math>|a| = 49</math>, <math>a^2 -2011 = 390</math>. <math>b = 39</math>, <math>c = 10</math>.
  
Answer: <math>089</math>
+
Answer: <math>\boxed{098}</math>
  
 
== Solution 2==
 
== Solution 2==
  
  
Starting off like the previous solution, we know that a + b + c = 0, and ab + bc + ac = -2011
+
Starting off like the previous solution, we know that <math>a + b + c = 0</math>, and <math>ab + bc + ac = -2011</math>.
  
Therefore, <math>c = -b-a</math>
+
Therefore, <math>c = -b-a</math>.
  
Substituting, <math>ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011</math>
+
Substituting, <math>ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011</math>.
  
Factoring the perfect square, we get: <math>ab-(b+a)^2=-2011</math> or <math>(b+a)^2-ab=2011</math>
+
Factoring the perfect square, we get: <math>ab-(b+a)^2=-2011</math> or <math>(b+a)^2-ab=2011</math>.
  
Therefore, a sum (<math>a+b</math>) squared minus a product (<math>ab</math>) gives <math>2011</math>.
+
Therefore, a sum (<math>a+b</math>) squared minus a product (<math>ab</math>) gives <math>2011</math>..
  
 
<br/>
 
<br/>
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We can guess and check different <math>a+b</math>’s starting with <math>45</math> since <math>44^2 < 2011</math>.
 
We can guess and check different <math>a+b</math>’s starting with <math>45</math> since <math>44^2 < 2011</math>.
  
<math>45^2 = 2025</math> therefore <math>ab = 2025-2011 = 14</math>  
+
<math>45^2 = 2025</math> therefore <math>ab = 2025-2011 = 14</math>.
  
 
Since no factors of <math>14</math> can sum to <math>45</math> (<math>1+14</math> being the largest sum), a + b cannot equal <math>45</math>.
 
Since no factors of <math>14</math> can sum to <math>45</math> (<math>1+14</math> being the largest sum), a + b cannot equal <math>45</math>.
  
<math>46^2 = 2116</math> making <math>ab = 105 = 3 * 5 * 7</math>
+
<math>46^2 = 2116</math> making <math>ab = 105 = 3 * 5 * 7</math>.
  
<math>5 * 7 + 3 < 46</math> and <math>3 * 5 * 7 > 46</math> so <math>46</math> cannot work either
+
<math>5 * 7 + 3 < 46</math> and <math>3 * 5 * 7 > 46</math> so <math>46</math> cannot work either.
  
 
<br/>
 
<br/>
  
We can continue to do this until we reach <math>49</math>
+
We can continue to do this until we reach <math>49</math>.
  
<math>49^2 =  2401</math> making <math>ab = 390 = 2 * 3 * 5* 13</math>
+
<math>49^2 =  2401</math> making <math>ab = 390 = 2 * 3 * 5* 13</math>.
  
 
<math>3 * 13 + 2* 5 = 49</math>, so one root is <math>10</math> and another is <math>39</math>. The roots sum to zero, so the last root must be <math>-49</math>.
 
<math>3 * 13 + 2* 5 = 49</math>, so one root is <math>10</math> and another is <math>39</math>. The roots sum to zero, so the last root must be <math>-49</math>.
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<br/>
 
<br/>
  
<math>|-49|+10+39 = 98</math>
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<math>|-49|+10+39 = \boxed{098}</math>.
  
Answer: <math>089</math>
+
 
 +
== 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>.
 +
 
 +
==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 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
 +
 
 +
==Video Solution==
 +
 
 +
https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx
  
 
== See also ==
 
== See also ==
  
{{AIME box|year=2011|num-b=14|after=-|n=I}}
+
{{AIME box|year=2011|num-b=14|after=Last Problem|n=I}}
 +
 
  
[[Category:Intermediate Geometry Problems]]
+
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 11:53, 8 March 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

$(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc$ 

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 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

Video Solution

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

See also

2011 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Problem
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All AIME Problems and Solutions

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