Difference between revisions of "Mock AIME I 2015 Problems/Problem 11"
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− | ==Solution 1= | + | ==Problem== |
+ | |||
+ | Suppose <math>\alpha</math>, <math>\beta</math>, and <math>\gamma</math> are complex numbers that satisfy the system of equations <cmath>\begin{align*}\alpha+\beta+\gamma&=6,\\\alpha^3+\beta^3+\gamma^3&=87,\\(\alpha+1)(\beta+1)(\gamma+1)&=33.\end{align*}</cmath> If <math>\frac1\alpha+\frac1\beta+\frac1\gamma=\tfrac mn</math> for positive relatively prime integers <math>m</math> and <math>n</math>, find <math>m+n</math>. | ||
+ | |||
+ | ==Solution 1== | ||
For convenience, let's use <math>a, b, c</math> instead of <math>\alpha, \beta, \gamma</math>. Define a polynomial <math>P(x)</math> such that <math>P(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc</math>. Let <math>j = ab + ac + bc</math> and <math>k = -abc</math>. Then, our polynomial becomes <math>P(x) = x^3 - (a+b+c)x^2 + jx + k</math>. | For convenience, let's use <math>a, b, c</math> instead of <math>\alpha, \beta, \gamma</math>. Define a polynomial <math>P(x)</math> such that <math>P(x) = (x-a)(x-b)(x-c) = x^3 - (a+b+c)x^2 + (ab+ac+bc)x - abc</math>. Let <math>j = ab + ac + bc</math> and <math>k = -abc</math>. Then, our polynomial becomes <math>P(x) = x^3 - (a+b+c)x^2 + jx + k</math>. | ||
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==Solution 2== | ==Solution 2== | ||
− | Let <math>\alpha = a</math>, <math>\beta = b</math>, and <math>\ | + | Let <math>\alpha = a</math>, <math>\beta = b</math>, and <math>\gamma = c</math>. Then our system becomes |
− | + | <cmath>a + b + c = 6</cmath> | |
− | + | <cmath>a^3 + b^3 + c^3 = 87</cmath> | |
− | + | <cmath>(a + 1)(b + 1)(c + 1) = 33</cmath>. | |
− | + | <cmath>a^3 + b^3 + c^3 = 87</cmath> | |
− | < | + | <cmath>a^3 + b^3 + c^3 - 3abc = 87 - 3abc</cmath> |
− | < | + | <cmath>(a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = 87 - 3abc</cmath> |
− | < | + | <cmath>(a + b + c)((a + b + c)^2 - 3(ab + ac + bc)) = 87 - 3abc</cmath> |
Since <math>a + b + c = 6</math>, this equation becomes <math>6(6^2 - 3(ab + ac + bc)) = 87 - 3abc</math>. | Since <math>a + b + c = 6</math>, this equation becomes <math>6(6^2 - 3(ab + ac + bc)) = 87 - 3abc</math>. | ||
− | + | ||
+ | <cmath>33 = (a + 1)(b + 1)(c + 1) = abc + ab + ac + bc + a + b + c + 1</cmath>. | ||
Since <math>a + b + c = 6</math>, this equation becomes <math>26 = abc + ab + ac + bc</math>. | Since <math>a + b + c = 6</math>, this equation becomes <math>26 = abc + ab + ac + bc</math>. | ||
We will now use these <math>2</math> equations to solve the problem. Let <math>x = abc</math>, and <math>y = ab + ac + bc</math>. Then we have | We will now use these <math>2</math> equations to solve the problem. Let <math>x = abc</math>, and <math>y = ab + ac + bc</math>. Then we have | ||
− | <cmath>6(6^2 - 3y) = 87 - 3x</cmath><cmath>26 = x + y</cmath>. | + | <cmath>6(6^2 - 3y) = 87 - 3x</cmath> |
− | + | <cmath>26 = x + y</cmath>. | |
+ | Our solutions are <math>x = \frac{113}{7}</math> and <math>y = \frac{69}{7}</math>. | ||
+ | |||
+ | Therefore, <math>\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + ac + bc}{abc} = \frac{69/7}{113/7} = \frac{69}{113}</math>. So, <math>m + n = 69 + 113 = \boxed{182}</math>. | ||
− | + | <baker77> |
Latest revision as of 11:29, 29 October 2019
Problem
Suppose ,
, and
are complex numbers that satisfy the system of equations
If
for positive relatively prime integers
and
, find
.
Solution 1
For convenience, let's use instead of
. Define a polynomial
such that
. Let
and
. Then, our polynomial becomes
.
Note that we want to compute
.
From the given information, we know that the coefficient of the term is
, and we also know that
, or in other words,
. By Newton's Sums (since we are given
), we also find that
. Solving this system, we find that
. Thus,
, so our final answer is
.
Solution 2
Let ,
, and
. Then our system becomes
.
Since
, this equation becomes
.
.
Since
, this equation becomes
.
We will now use these equations to solve the problem. Let
, and
. Then we have
.
Our solutions are
and
.
Therefore, . So,
.
<baker77>