# Difference between revisions of "Mock AIME I 2015 Problems/Problem 11"

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− | ==Solution== | + | ==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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From the given information, we know that the coefficient of the <math>x^2</math> term is <math>6</math>, and we also know that <math>P(-1) = -33</math>, or in other words, <math>-j + k = -26</math>. By Newton's Sums (since we are given <math>a^3 + b^3 + c^3</math>), we also find that <math>6j + k = 43</math>. Solving this system, we find that <math>(j, k) \in (\frac{69}{7}, -\frac{113}{7})</math>. Thus, <math>\frac{j}{-k} = \frac{69}{113}</math>, so our final answer is <math>69 + 113 = \boxed{182}</math>. | From the given information, we know that the coefficient of the <math>x^2</math> term is <math>6</math>, and we also know that <math>P(-1) = -33</math>, or in other words, <math>-j + k = -26</math>. By Newton's Sums (since we are given <math>a^3 + b^3 + c^3</math>), we also find that <math>6j + k = 43</math>. Solving this system, we find that <math>(j, k) \in (\frac{69}{7}, -\frac{113}{7})</math>. Thus, <math>\frac{j}{-k} = \frac{69}{113}</math>, so our final answer is <math>69 + 113 = \boxed{182}</math>. | ||

+ | |||

+ | ==Solution 2== | ||

+ | |||

+ | Let <math>\alpha = a</math>, <math>\beta = b</math>, and <math>\beta = c</math>. Then our system becomes | ||

+ | <cmath>a + b + c</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>. | ||

+ | |||

+ | <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>. | ||

+ | |||

+ | 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>. | ||

+ | Our solutions are <math>x = \frac{113}{7}</math> and <math>y = \frac{69}{7}</math>. | ||

+ | |||

+ | Then <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> |

## Revision as of 11:19, 11 October 2019

## 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 .

Then . So, .

<baker77>