Difference between revisions of "2013 AIME I Problems/Problem 10"
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== Solution == | == 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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| + | <math>(x-q)(x-r-si)(x-r+si) = x^3 -ax^2 + bx -65</math> | ||
| + | |||
| + | Applying difference of squares, and regrouping, we have | ||
| + | |||
| + | <math>(x-q)(x^2 - 2rx + (r^2 + s^2) = x^3 -ax^2 + bx -65</math> | ||
| + | |||
| + | So matching coefficients, we obtain | ||
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| + | <math>q(r^2 + s^2) = 65</math> | ||
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| + | <math>b = r^2 + s^2 + 2rq</math> | ||
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| + | <math>a = q + 2r</math> | ||
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| + | By Vieta's each <math> {p}_{a,b} = a </math> so we just need to find the values of <math> a </math> in each pair. | ||
| + | We proceed by determining possible values for <math>q</math>, <math>r</math>, and <math>s</math> and using these to determine <math>a</math> and <math>b</math>. | ||
| + | If <math>q = 1</math> : <math>r^2 + s^2 = 65</math> | ||
| + | so (r, s) = <math>(\pm1, \pm 8), (\pm8, \pm 1), (\pm4, \pm 7), (\pm7, \pm 4)</math> | ||
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| + | Similarly, for <math>q = 5</math>, <math>r^2 + s^2 = 13</math> so the pairs <math>(r,s)</math> are | ||
| + | <math>(\pm2, \pm 3), (\pm3, \pm 2)</math> | ||
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| + | For <math>q = 13</math> ,<math> r^2 + s^2 = 5</math> so the pairs <math> (r,s)</math> are | ||
| + | <math>(\pm2, \pm 1), (\pm1, \pm 2)</math> | ||
| + | |||
| + | 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>. | ||
| + | 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 r generates a pair <math>(a,b)</math>). | ||
| + | Our answer is then <math>(1)(8) + (5)(4) + (13)(4) = 80</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}} | ||
Revision as of 22:11, 16 March 2013
Problem 10
There are nonzero integers 
, 
, 
, and 
such that the complex number 
is a zero of the polynomial 
. For each possible combination of and , let 
be the sum of the zeros of 
. Find the sum of the 's for all possible combinations of and .
Solution
Since is a root, by the Complex Conjugate Root Theorem, 
must be the other imaginary root. Using 
to represent the rational root, we have
Applying difference of squares, and regrouping, we have
So matching coefficients, we obtain
By Vieta's each 
so we just need to find the values of in each pair.
We proceed by determining possible values for , , and and using these to determine and .
If 
: 
so (r, s) = 
Similarly, for 
, 
so the pairs 
are
For 
,
so the pairs are
Then, since only 
but not appears in the equations for and , we can ignore the plus minus sign for .
The positive and negative values of r will cancel, so the sum of the for is times the number of distinct values (as each value of r generates a pair 
).
Our answer is then 
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 | ||
