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 "1991 AIME Problems/Problem 7"

(→‎Solution)
m (→‎Solution 1)
 
(15 intermediate revisions by 12 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Find <math>A^2_{}</math>, where <math>A^{}_{}</math> is the sum of the absolute values of all roots of the following equation:
+
<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Find <math>A^2_{}</math>, where <math>A^{}_{}</math> is the sum of the [[absolute value]]s of all roots of the following equation:
−
<div style="text-align:center"><math>x = \sqrt{19} + \frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{{\displaystyle \sqrt{19}+\frac{91}{x}}}}}}}}}
+
<div style="text-align:center"><math>x = \sqrt{19} + \frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{x}}}}}}}}}
−
</math></div>
+
</math></div><!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
−
== Solution ==
+
== Solution 1 ==
−
The given finite expansion can be easily seen that reduces to solve the quadratic equation <math>x_{}^{2}-\sqrt{19}x-91=0</math>. The solutions are <math>x_{\pm}^{}=\frac{\sqrt{19}\pm\sqrt{383}}{2}</math>. Therefore, <math>A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}</math>. Therefore, <math>A_{}^{2}=383</math>.
+
<math>x=\sqrt{19}+\underbrace{\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}}_{x}</math>
 +
 
 +
<math>x=\sqrt{19}+\frac{91}{x}</math>
 +
 
 +
<math>x^2=x\sqrt{19}+91</math>
 +
 
 +
<math>x^2-x\sqrt{19}-91 = 0</math>
 +
 
 +
<math>\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|+|x_2|\Rightarrow\sqrt{383}</math>
 +
 
 +
<math>A^2=\boxed{383}</math>
 +
 
 +
== Solution 2 ==
 +
Let <math>f(x) = \sqrt{19} + \frac{91}{x}</math>. Then <math>x = f(f(f(f(f(x)))))</math>, from which we realize that <math>f(x) = x</math>. This is because if we expand the entire expression, we will get a fraction of the form <math>\frac{ax + b}{cx + d}</math> on the right hand side, which makes the equation simplify to a quadratic. As this quadratic will have two roots, they must be the same roots as the quadratic <math>f(x)=x</math>.
 +
 
 +
The given finite expansion can then be easily seen to reduce to the [[quadratic equation]] <math>x_{}^{2}-\sqrt{19}x-91=0</math>. The solutions are <math>x_{\pm}^{}=</math><math>\frac{\sqrt{19}\pm\sqrt{383}}{2}</math>. Therefore, <math>A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}</math>. We conclude that <math>A_{}^{2}=\boxed{383}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1991|num-b=6|num-a=8}}
 
{{AIME box|year=1991|num-b=6|num-a=8}}
 +
 +
[[Category:Intermediate Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 04:15, 13 June 2022

Problem

Find $A^2_{}$, where $A^{}_{}$ is the sum of the absolute values of all roots of the following equation:

$x = \sqrt{19} + \frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{x}}}}}}}}}$

Solution 1

$x=\sqrt{19}+\underbrace{\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}}_{x}$

$x=\sqrt{19}+\frac{91}{x}$

$x^2=x\sqrt{19}+91$

$x^2-x\sqrt{19}-91 = 0$

$\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|+|x_2|\Rightarrow\sqrt{383}$

$A^2=\boxed{383}$

Solution 2

Let $f(x) = \sqrt{19} + \frac{91}{x}$. Then $x = f(f(f(f(f(x)))))$, from which we realize that $f(x) = x$. This is because if we expand the entire expression, we will get a fraction of the form $\frac{ax + b}{cx + d}$ on the right hand side, which makes the equation simplify to a quadratic. As this quadratic will have two roots, they must be the same roots as the quadratic $f(x)=x$.

The given finite expansion can then be easily seen to reduce to the quadratic equation $x_{}^{2}-\sqrt{19}x-91=0$. The solutions are $x_{\pm}^{}=$$\frac{\sqrt{19}\pm\sqrt{383}}{2}$. Therefore, $A_{}^{}=\vert x_{+}\vert+\vert x_{-}\vert=\sqrt{383}$. We conclude that $A_{}^{2}=\boxed{383}$.

See also

1991 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_7&oldid=174789"