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Difference between revisions of "2013 AIME I Problems/Problem 10"
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==Problem 10==
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==Problem ==
 
There are nonzero integers <math>a</math>, <math>b</math>, <math>r</math>, and <math>s</math> such that the complex number <math>r+si</math> is a zero of the polynomial <math>P(x)={x}^{3}-a{x}^{2}+bx-65</math>. For each possible combination of <math>a</math> and <math>b</math>, let <math>{p}_{a,b}</math> be the sum of the zeros of <math>P(x)</math>. Find the sum of the <math>{p}_{a,b}</math>'s for all possible combinations of <math>a</math> and <math>b</math>.
 
There are nonzero integers <math>a</math>, <math>b</math>, <math>r</math>, and <math>s</math> such that the complex number <math>r+si</math> is a zero of the polynomial <math>P(x)={x}^{3}-a{x}^{2}+bx-65</math>. For each possible combination of <math>a</math> and <math>b</math>, let <math>{p}_{a,b}</math> be the sum of the zeros of <math>P(x)</math>. Find the sum of the <math>{p}_{a,b}</math>'s for all possible combinations of <math>a</math> and <math>b</math>.
  
  
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== Solution ==
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== Solution==
  
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Since <math>r+si</math> is a root, by the Complex Conjugate Root Theorem, <math>r-si</math> must be the other imaginary root. Using <math>q</math> to represent the rational root, we have
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Since <math>r+si</math> is a root, by the Complex Conjugate Root Theorem, <math>r-si</math> must be the other imaginary root. Using <math>q</math> to represent the real root, we have
  
 
<math>(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65</math>
 
<math>(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65</math>
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<math>(\pm2, \pm 1), (\pm1, \pm 2)</math>
 
<math>(\pm2, \pm 1), (\pm1, \pm 2)</math>
  
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Then, since only <math>s^2</math> but not <math>s</math> appears in the equations for <math>a</math> and <math>b</math>, we can ignore the plus minus sign for <math>s</math>.
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Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases.
 
The positive and negative values of r will cancel, so the sum of the  <math> {p}_{a,b} = a </math> for <math>q = 1</math> is <math>q</math> times the number of distinct <math>r</math> values (as each value of <math>r</math> generates a pair <math>(a,b)</math>).
 
The positive and negative values of r will cancel, so the sum of the  <math> {p}_{a,b} = a </math> for <math>q = 1</math> is <math>q</math> times the number of distinct <math>r</math> values (as each value of <math>r</math> generates a pair <math>(a,b)</math>).
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Our answer is then <math>(1)(8) + (5)(4) + (13)(4) = \boxed{80}</math>.
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Our answer is then <math>(1)(8) + (5)(4) + (13)(4) = \boxed{080}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2013|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2013|n=I|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 11:05, 10 August 2021

Problem

There are nonzero integers $a$, $b$, $r$, and $s$ such that the complex number $r+si$ is a zero of the polynomial $P(x)={x}^{3}-a{x}^{2}+bx-65$. For each possible combination of $a$ and $b$, let ${p}_{a,b}$ be the sum of the zeros of $P(x)$. Find the sum of the ${p}_{a,b}$'s for all possible combinations of $a$ and $b$.


Solution

Since $r+si$ is a root, by the Complex Conjugate Root Theorem, $r-si$ must be the other imaginary root. Using $q$ to represent the real root, we have

$(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65$

Applying difference of squares, and regrouping, we have

$(x-q)(x^2 - 2rx + (r^2 + s^2)) = x^3 -ax^2 + bx -65$

So matching coefficients, we obtain

$q(r^2 + s^2) = 65$

$b = r^2 + s^2 + 2rq$

$a = q + 2r$

By Vieta's each ${p}_{a,b} = a$ so we just need to find the values of $a$ in each pair. We proceed by determining possible values for $q$, $r$, and $s$ and using these to determine $a$ and $b$.

If $q = 1$, $r^2 + s^2 = 65$ so (r, s) = $(\pm1, \pm 8), (\pm8, \pm 1), (\pm4, \pm 7), (\pm7, \pm 4)$

Similarly, for $q = 5$, $r^2 + s^2 = 13$ so the pairs $(r,s)$ are $(\pm2, \pm 3), (\pm3, \pm 2)$

For $q = 13$, $r^2 + s^2 = 5$ so the pairs $(r,s)$ are $(\pm2, \pm 1), (\pm1, \pm 2)$

Now we can disregard the plus minus signs for s because those cases are included as complex conjugates of the counted cases. The positive and negative values of r will cancel, so the sum of the ${p}_{a,b} = a$ for $q = 1$ is $q$ times the number of distinct $r$ values (as each value of $r$ generates a pair $(a,b)$). Our answer is then $(1)(8) + (5)(4) + (13)(4) = \boxed{080}$.

See also

2013 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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