Difference between revisions of "2011 AIME I Problems/Problem 15"
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− | <math>a,b,c\neq 0</math> since any one being zero will make | + | <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>. |
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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> | + | 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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Now we have limited <math>a</math> to <math>45\le |a| \le 51</math>. | Now we have limited <math>a</math> to <math>45\le |a| \le 51</math>. | ||
− | Let's | + | Let's analyze <math>a^2 = 2011 + bc</math>. |
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<math>|-49|+10+39 = \boxed{098}</math>. | <math>|-49|+10+39 = \boxed{098}</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= | + | {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}} |
+ | |||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:53, 8 March 2021
Contents
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, if we fix , is maximized when . Hence, .
So .
so .
Now we have limited to .
Let's 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 .
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 5 (mod to help bash)
First, derive the equations and . Since the product is negative, is negative, and and positive. Now, a simple mod 3 testing of all cases shows that , and has the repective value. We can choose not congruent to 0, make sure you see why. Now, we bash on values of , testing the quadratic function to see if is positive. You can also use a delta argument like solution 4, but this is simpler. We get that for , . Choosing positive we get , so ~firebolt360
Video Solution
https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx
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.