Difference between revisions of "Mock AIME I 2015 Problems/Problem 11"
(2 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Suppose <math>\alpha</math>, <math>\beta</math>, and <math>\gamma</math> are complex numbers that satisfy the system of equations \begin{align*}\alpha+\beta+\gamma&=6,\\\alpha^3+\beta^3+\gamma^3&=87,\\(\alpha+1)(\beta+1)(\gamma+1)&=33.\end{align*}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>. | + | 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== | ==Solution 1== | ||
Line 32: | Line 32: | ||
Our solutions are <math>x = \frac{113}{7}</math> and <math>y = \frac{69}{7}</math>. | 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> | <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>