Difference between revisions of "2011 AIME I Problems/Problem 15"
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== Problem == | == Problem == | ||
For some integer <math>m</math>, the polynomial <math>x^3 - 2011x + m</math> has the three integer roots <math>a</math>, <math>b</math>, and <math>c</math>. Find <math>|a| + |b| + |c|</math>. | For some integer <math>m</math>, the polynomial <math>x^3 - 2011x + m</math> has the three integer roots <math>a</math>, <math>b</math>, and <math>c</math>. Find <math>|a| + |b| + |c|</math>. | ||
− | ==Solution== | + | ==Solution 1== |
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− | + | From Vieta's formulas, we know that <math>a+b+c = 0</math>, and <math>ab+bc+ac = -2011</math>. Thus <math>a = -(b+c)</math>. All three of <math>a</math>, <math>b</math>, and <math>c</math> are non-zero: say, if <math>a=0</math>, then <math>b=-c=\pm\sqrt{2011}</math> (which is not an integer). <math>\textsc{wlog}</math>, let <math>|a| \ge |b| \ge |c|</math>. If <math>a > 0</math>, then <math>b,c < 0</math> and if <math>a < 0</math>, then <math>b,c > 0</math>. We have <cmath>-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc</cmath> | |
+ | Thus <math>a^2 = 2011 + bc</math>. We know that <math>b</math>, <math>c</math> have the same sign. So <math>|a| \ge 45 = \lceil \sqrt{2011} \rceil</math>. | ||
− | + | Also, if we fix <math>a</math>, <math>b+c</math> is fixed, so <math>bc</math> is maximized when <math>b = c</math> . Hence, <cmath>2011 = a^2 - bc > \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 < \tfrac{4}{3}\cdot 2011 = 2681+\tfrac{1}{3}</cmath> | |
− | + | So <math>|a| \le 51</math>. Thus we have bounded <math>a</math> as <math>45\le |a| \le 51</math>, i.e. <math>45\le |b+c| \le 51</math> since <math>a=-(b+c)</math>. Let's analyze <math>bc=(b+c)^2-2011</math>. Here is a table: | |
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− | Here is a table: | ||
<table border = 1 cellspacing = 0 cellpadding = 5 style = "text-align:center;"> | <table border = 1 cellspacing = 0 cellpadding = 5 style = "text-align:center;"> | ||
− | <tr><th><math>|a|</math></th><th><math>a^2 | + | <tr><th><math>|a|</math></th><th><math>bc=2011-a^2</math></th></tr> |
<tr><td><math>45</math></td><td><math>14</math></td></tr> | <tr><td><math>45</math></td><td><math>14</math></td></tr> | ||
− | <tr><td><math>46</math></td><td><math> | + | <tr><td><math>46</math></td><td><math>105</math></td></tr> |
− | <tr><td><math>47</math></td><td><math> | + | <tr><td><math>47</math></td><td><math>198</math></td></tr> |
− | <tr><td><math>48</math></td><td><math> | + | <tr><td><math>48</math></td><td><math>293</math></td></tr> |
− | <tr><td><math>49</math></td><td><math> | + | <tr><td><math>49</math></td><td><math>390</math></td></tr> |
</table> | </table> | ||
<br /> | <br /> | ||
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== 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>. - | + | 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>. |
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+ | ==Solution 4== | ||
+ | We have | ||
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+ | <math>(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc</math> | ||
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+ | As a result, we have | ||
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+ | <math>a+b+c=0</math> | ||
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+ | <math>ab+bc+ac=-2011</math> | ||
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+ | <math>abc=-m</math> | ||
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+ | So, <math>a=-b-c</math> | ||
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+ | As a result, <math>ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011</math> | ||
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+ | 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 | ||
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+ | Cause <math>89<\sqrt{8044}<90</math> | ||
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+ | So, after we tried for <math>2</math> times, we get <math>k=88</math> and <math>c=10</math> | ||
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+ | then <math>b=39</math>, <math>a=-b-c=-49</math> | ||
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+ | As a result, <math>|a|+|b|+|c|=10+39+49=\boxed{098}</math> | ||
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+ | ==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 | ||
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+ | ==Video Solution== | ||
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+ | https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx | ||
== See also == | == See also == | ||
{{AIME box|year=2011|num-b=14|after=Last Problem|n=I}} | {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}} | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:31, 27 January 2022
Contents
Problem
For some integer , the polynomial has the three integer roots , , and . Find .
Solution 1
From Vieta's formulas, we know that , and . Thus . All three of , , and are non-zero: say, if , then (which is not an integer). , let . If , then and if , then . We have Thus . We know that , have the same sign. So .
Also, if we fix , is fixed, so is maximized when . Hence, So . Thus we have bounded as , i.e. since . 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 |
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