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

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== Problem ==
 
== Problem ==
Find <math>A^2_{}</math>, where <math>A^{}_{}</math> is the sum of the [[absolute value]]s of all roots of the following equation:
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<!-- 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}}}}}}}}}
 
<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>
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</math></div><!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
== Solution ==
 
== Solution ==

Revision as of 17:50, 27 March 2015

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

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 $A_{}^{2}=383$.

See also

1991 AIME (ProblemsAnswer KeyResources)
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

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