Difference between revisions of "2011 AIME I Problems/Problem 15"
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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== |
− | + | 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: | ||
+ | <table border = 1 cellspacing = 0 cellpadding = 5 style = "text-align:center;"> | ||
+ | |||
+ | <tr><th><math>|a|</math></th><th><math>bc=a^2-2011</math></th></tr> | ||
+ | |||
+ | <tr><td><math>45</math></td><td><math>14</math></td></tr> | ||
+ | <tr><td><math>46</math></td><td><math>105</math></td></tr> | ||
+ | <tr><td><math>47</math></td><td><math>198</math></td></tr> | ||
+ | <tr><td><math>48</math></td><td><math>293</math></td></tr> | ||
+ | <tr><td><math>49</math></td><td><math>390</math></td></tr> | ||
+ | </table> | ||
<br /> | <br /> | ||
− | |||
− | <math> | + | We can tell we don't need to bother with <math>45</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>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>. | ||
+ | |||
+ | Answer: <math>\boxed{098}</math> | ||
+ | |||
+ | == Solution 2== | ||
+ | |||
+ | |||
+ | 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>. | |
− | < | + | Substituting, <math>ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011</math>. |
− | <math>ab+ | + | |
+ | 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>.. | ||
+ | |||
+ | <br/> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>5 * 7 + 3 < 46</math> and <math>3 * 5 * 7 > 46</math> so <math>46</math> cannot work either. | ||
+ | |||
+ | <br/> | ||
+ | |||
+ | 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>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>. | ||
− | <br / | + | <br/> |
− | |||
− | + | <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> | + | <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> | |
− | <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> | |
− | <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 | ||
+ | ==Solution 6== | ||
+ | Note that <math>-c=b+a</math>, so <math>c^2=a^2+2ab+b^2</math>, or <math>-c^2+ab=-a^2-ab-b^2</math>. Also, <math>ab+bc+ca=-2011</math>, so <math>(a+b)c+ab=-c^2+ab=-2011</math>. Substituting <math>-c^2+ab=-a^2-ab-b^2</math>, we can obtain <math>a^2+ab+b^2=2011</math>, or <math>\frac{a^3-b^3}{a-b}=2011</math>. If it is not known that <math>2011</math> is prime, it may be proved in <math>5</math> minutes or so by checking all primes up to <math>43</math>. If <math>2011</math> divided either of <math>a, b</math>, then in order for <math>a^3-b^3</math> to contain an extra copy of <math>2011</math>, both <math>a, b</math> would need to be divisible by <math>2021</math>. But then <math>c</math> would also be divisible by <math>2011</math>, and the sum <math>ab+bc+ca</math> would clearly be divisible by <math>2011^2</math>. | ||
− | + | By LTE, <math>v_{2011}(a^3-b^3)=v_{2011}(a-b)</math> if <math>a-b</math> is divisible by <math>2011</math> and neither <math>a,b</math> are divisible by <math>2011</math>. Thus, the only possibility remaining is if <math>a-b</math> did not divide <math>2011</math>. Let <math>a=k+b</math>. Then, we have <math>(b+k)^3-b^3=2011k</math>. Rearranging gives <math>3b(b+k)=2011-k^2</math>. As in the above solutions, we may eliminate certain values of <math>k</math> by using mods. Then, we may test values until we obtain <math>k=29</math>, and <math>a=10</math>. Thus, <math>b=39</math>, <math>c=-49</math>, and our answer is <math>49+39+10=098</math>. | |
+ | ==Video Solution== | ||
− | + | https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx | |
== See also == | == See also == | ||
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{{AIME box|year=2011|num-b=14|after=Last Problem|n=I}} | {{AIME box|year=2011|num-b=14|after=Last Problem|n=I}} | ||
− | [[Category:Intermediate | + | |
+ | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 21:12, 30 September 2024
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
Solution 6
Note that , so , or . Also, , so . Substituting , we can obtain , or . If it is not known that is prime, it may be proved in minutes or so by checking all primes up to . If divided either of , then in order for to contain an extra copy of , both would need to be divisible by . But then would also be divisible by , and the sum would clearly be divisible by .
By LTE, if is divisible by and neither are divisible by . Thus, the only possibility remaining is if did not divide . Let . Then, we have . Rearranging gives . As in the above solutions, we may eliminate certain values of by using mods. Then, we may test values until we obtain , and . Thus, , , and our answer is .
Video Solution
https://www.youtube.com/watch?v=QNbfAu5rdJI&t=26s ~ MathEx
See also
2011 AIME I (Problems • Answer Key • Resources) | ||
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