AIME 1991 · 第 7 题

Solution 1

$x=\sqrt{19}+\underbrace{\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{\sqrt{19}+\frac{91}{x}}}}}}_{x}$ $x=\sqrt{19}+\frac{91}{x}$ $x^2=x\sqrt{19}+91$ $x^2-x\sqrt{19}-91 = 0$ $\left.\begin{array}{l}x_1=\frac{\sqrt{19}+\sqrt{383}}{2}\\\\x_2=\frac{\sqrt{19}-\sqrt{383}}{2}\end{array}\right\}A=|x_1|+|x_2|\Rightarrow\sqrt{383}$ $A^2=\boxed{383}$

~ NOTE: I don't actually think this solution is correct, since the original expression for x has 5 $\sqrt{19}'s$ but the one it is substituted to only has 4. This only works in infinite sums. ~eqb5000

~ NOTE 2: The solution is still correct because by substituting in the given value for $x$ on the right side, you do obtain an infinite sum, and can do this substitution. ~PojoDotCom

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