Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1994 AIME Problems/Problem 13"

(solution by scorpius119)
 
(9 intermediate revisions by 7 users not shown)
Line 5: Line 5:
 
<center><math>\frac 1{r_1\overline{r_1}}+\frac 1{r_2\overline{r_2}}+\frac 1{r_3\overline{r_3}}+\frac 1{r_4\overline{r_4}}+\frac 1{r_5\overline{r_5}}.</math></center>
 
<center><math>\frac 1{r_1\overline{r_1}}+\frac 1{r_2\overline{r_2}}+\frac 1{r_3\overline{r_3}}+\frac 1{r_4\overline{r_4}}+\frac 1{r_5\overline{r_5}}.</math></center>
  
−
== Solution ==
+
== Solution 1 ==
 
Let <math>t = 1/x</math>. After multiplying the equation by <math>t^{10}</math>, <math>1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1</math>.
 
Let <math>t = 1/x</math>. After multiplying the equation by <math>t^{10}</math>, <math>1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1</math>.
  
−
Using DeMoivre, <math>13 - t = e^\frac {(2k + 1)\pi}{10}</math> where <math>k</math> is an integer between <math>0</math> and <math>9</math>.
+
Using DeMoivre, <math>13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right)</math> where <math>k</math> is an integer between <math>0</math> and <math>9</math>.
  
−
<math>t = 13 - e^\frac {(2k + 1)\pi}{10} \Rightarrow \bar{t} = 13 - e^{ - \frac {(2k + 1)\pi}{10}}</math>.
+
<math>t = 13 - \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) \Rightarrow \bar{t} = 13 - \text{cis}\left(-\frac {(2k + 1)\pi}{10}\right)</math>.
  
−
Since <math>e^{iy} + e^{ - iy} = 2\cos y</math>, <math>t\bar{t} = 170 - 2\cos \frac {(2k + 1)\pi}{10}</math> after expanding. Here <math>k</math> ranges from 0 to 4 because two angles which sum to <math>2\pi</math> are involved in the product..
+
Since <math>\text{cis}(\theta) + \text{cis}(-\theta) = 2\cos(\theta)</math>, <math>t\bar{t} = 170 - 26\cos \left(\frac {(2k + 1)\pi}{10}\right)</math> after expanding. Here <math>k</math> ranges from 0 to 4 because two angles which sum to <math>2\pi</math> are involved in the product.
  
−
The expression to find is <math>\sum t\bar{t} = 850 - 2\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}</math>.
+
The expression to find is <math>\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}</math>.
  
 
But <math>\cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0</math> so the sum is <math>\boxed{850}</math>.
 
But <math>\cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0</math> so the sum is <math>\boxed{850}</math>.
 +
 +
== Solution 2 ==
 +
Divide both sides by <math>x^{10}</math> to get <cmath>1 + \left(13-\dfrac{1}{x}\right)^{10}=0</cmath>
 +
 +
Rearranging: <cmath>\left(13-\dfrac{1}{x}\right)^{10} = -1</cmath>
 +
 +
Thus, <math>13-\dfrac{1}{x} = \omega</math> where <math>\omega = e^{i(\pi n/5+\pi/10)}</math> where <math>n</math> is an integer.
 +
 +
We see that <math>\dfrac{1}{x}=13-\omega</math>. Thus, <cmath>\dfrac{1}{x\overline{x}}=(13\, -\, \omega)(13\, -\, \overline{\omega})=169-13(\omega\, +\, \overline{\omega})\, +\, \omega\overline{\omega}=170\, -\, 13(\omega\, +\, \overline{\omega})</cmath>
 +
 +
Summing over all terms: <cmath>\dfrac{1}{r_1\overline{r_1}}+\cdots + \dfrac{1}{r_5\overline{r_5}} = 5\cdot 170 - 13(e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)})</cmath>
 +
 +
However, note that <math>e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)}=0</math> from drawing the numbers on the complex plane, our answer is just <cmath>5\cdot 170=\boxed{850}</cmath>
 +
 +
== Solution 3(Ileytyn) ==
 +
 +
Let us apply difference of squares, by writing this equation as <math>(x^5)^2-(i[13x-1]^5)^2=0</math>, where <math>i=\sqrt{-1}</math>.
 +
 +
Once applied, we have <cmath>(x^5-i[13x-1]^5)(x^5+i[13x-1]^5)=0</cmath> By factorization, we get that either
 +
<cmath>\text{equation 1:  }x^5-i[13x-1]^5=0\text{ or equation 2:    }x^5+i[13x-1]^5=0</cmath>
 +
We find the trivial solution to the first equation <math>x^5=i[13x-1]^5</math>, and since a fifth root of <math>i=i^5</math> is <math>i</math>, we can find this solution by taking the fifth root, or when <math>x=(13x-1)i</math>, which when solved, gives <math>x=\frac{i}{13i-1}=\frac{13-i}{170}</math>. By using a similar approach to the second equation, <math>x^5+i[13x-1]^5=0</math>, or <math>x=-i(13x-1)</math>, we get <math>\frac{i}{13i+1}=\frac{13+i}{170}</math>.
 +
We observe that these are conjugates. We also note that <math>z\overline{z}=|z|^2</math>. Hence, we find the magnitude of one of these two conjugates, which is <math>\frac{1}{\sqrt{170}}</math>, or as we are finding the reciprocal of the square value of this, this contributes <math>170</math> to the final answer. We claim that each of the roots of equation one is a conjugate of one of equation 2. We also notice that, by an application of DeMoivre's, each of the roots has the same magnitude, just a different angle. Hence, as each of the reciprocals of product of conjugates has the value <math>170</math>, and there are 5 of these pairs of conjugates, the answer is <cmath>170\cdot 5 = \boxed{850}</cmath>
 +
 +
== Video Solution ==
 +
https://youtu.be/3GG6tdEz0KA
  
 
== See also ==
 
== See also ==
Line 22: Line 47:
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 17:39, 22 May 2025

Problem

The equation

$x^{10}+(13x-1)^{10}=0\,$

has 10 complex roots $r_1, \overline{r_1}, r_2, \overline{r_2}, r_3, \overline{r_3}, r_4, \overline{r_4}, r_5, \overline{r_5},\,$ where the bar denotes complex conjugation. Find the value of

$\frac 1{r_1\overline{r_1}}+\frac 1{r_2\overline{r_2}}+\frac 1{r_3\overline{r_3}}+\frac 1{r_4\overline{r_4}}+\frac 1{r_5\overline{r_5}}.$

Solution 1

Let $t = 1/x$. After multiplying the equation by $t^{10}$, $1 + (13 - t)^{10} = 0\Rightarrow (13 - t)^{10} = - 1$.

Using DeMoivre, $13 - t = \text{cis}\left(\frac {(2k + 1)\pi}{10}\right)$ where $k$ is an integer between $0$ and $9$.

$t = 13 - \text{cis}\left(\frac {(2k + 1)\pi}{10}\right) \Rightarrow \bar{t} = 13 - \text{cis}\left(-\frac {(2k + 1)\pi}{10}\right)$.

Since $\text{cis}(\theta) + \text{cis}(-\theta) = 2\cos(\theta)$, $t\bar{t} = 170 - 26\cos \left(\frac {(2k + 1)\pi}{10}\right)$ after expanding. Here $k$ ranges from 0 to 4 because two angles which sum to $2\pi$ are involved in the product.

The expression to find is $\sum t\bar{t} = 850 - 26\sum_{k = 0}^4 \cos \frac {(2k + 1)\pi}{10}$.

But $\cos \frac {\pi}{10} + \cos \frac {9\pi}{10} = \cos \frac {3\pi}{10} + \cos \frac {7\pi}{10} = \cos \frac {\pi}{2} = 0$ so the sum is $\boxed{850}$.

Solution 2

Divide both sides by $x^{10}$ to get \[1 + \left(13-\dfrac{1}{x}\right)^{10}=0\]

Rearranging: \[\left(13-\dfrac{1}{x}\right)^{10} = -1\]

Thus, $13-\dfrac{1}{x} = \omega$ where $\omega = e^{i(\pi n/5+\pi/10)}$ where $n$ is an integer.

We see that $\dfrac{1}{x}=13-\omega$. Thus, \[\dfrac{1}{x\overline{x}}=(13\, -\, \omega)(13\, -\, \overline{\omega})=169-13(\omega\, +\, \overline{\omega})\, +\, \omega\overline{\omega}=170\, -\, 13(\omega\, +\, \overline{\omega})\]

Summing over all terms: \[\dfrac{1}{r_1\overline{r_1}}+\cdots + \dfrac{1}{r_5\overline{r_5}} = 5\cdot 170 - 13(e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)})\]

However, note that $e^{i\pi/10}+\cdots +e^{i(9\pi/5+\pi/10)}=0$ from drawing the numbers on the complex plane, our answer is just \[5\cdot 170=\boxed{850}\]

Solution 3(Ileytyn)

Let us apply difference of squares, by writing this equation as $(x^5)^2-(i[13x-1]^5)^2=0$, where $i=\sqrt{-1}$.

Once applied, we have \[(x^5-i[13x-1]^5)(x^5+i[13x-1]^5)=0\] By factorization, we get that either \[\text{equation 1: }x^5-i[13x-1]^5=0\text{ or equation 2: }x^5+i[13x-1]^5=0\] We find the trivial solution to the first equation $x^5=i[13x-1]^5$, and since a fifth root of $i=i^5$ is $i$, we can find this solution by taking the fifth root, or when $x=(13x-1)i$, which when solved, gives $x=\frac{i}{13i-1}=\frac{13-i}{170}$. By using a similar approach to the second equation, $x^5+i[13x-1]^5=0$, or $x=-i(13x-1)$, we get $\frac{i}{13i+1}=\frac{13+i}{170}$. We observe that these are conjugates. We also note that $z\overline{z}=|z|^2$. Hence, we find the magnitude of one of these two conjugates, which is $\frac{1}{\sqrt{170}}$, or as we are finding the reciprocal of the square value of this, this contributes $170$ to the final answer. We claim that each of the roots of equation one is a conjugate of one of equation 2. We also notice that, by an application of DeMoivre's, each of the roots has the same magnitude, just a different angle. Hence, as each of the reciprocals of product of conjugates has the value $170$, and there are 5 of these pairs of conjugates, the answer is \[170\cdot 5 = \boxed{850}\]

Video Solution

https://youtu.be/3GG6tdEz0KA

See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_13&oldid=248926"