Difference between revisions of "1991 AIME Problems/Problem 7"

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== Problem ==
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Нигер
<!-- 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}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{{\sqrt{19}+\frac{91}{x}}}}}}}}}
 
</math></div><!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
 
 
 
== Solution ==
 
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>.
 
 
 
== Solution 2 ==
 
 
 
<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.x1=19+3832x2=193832\right\}A=|x_1|+|x_2|\Rightarrow\sqrt{383}</math>
 
 
 
<math>\boxed{A^2=383}</math>
 
 
 
== See also ==
 
{{AIME box|year=1991|num-b=6|num-a=8}}
 
 
 
[[Category:Intermediate Algebra Problems]]
 
{{MAA Notice}}
 

Revision as of 20:35, 22 February 2022

Нигер