1986 AHSME Problems/Problem 30

Problem

The number of real solutions $(x,y,z,w)$ of the simultaneous equations $2y = x + \frac{17}{x}, 2z = y + \frac{17}{y}, 2w = z + \frac{17}{z}, 2x = w + \frac{17}{w}$ is

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$

Solution 1

Consider the cases $x>0$ and $x<0$ , and also note that by AM-GM, for any positive number $a$ , we have $a+\frac{17}{a} \geq 2\sqrt{17}$ , with equality only if $a = \sqrt{17}$ . Thus, if $x>0$ , considering each equation in turn, we get that $y \geq \sqrt{17}, z \geq \sqrt{17}, w \geq \sqrt{17}$ , and finally $x \geq \sqrt{17}$ .

Now suppose $x > \sqrt{17}$ . Then $y - \sqrt{17} = \frac{x^{2}+17}{2x} - \sqrt{17} = (\frac{x-\sqrt{17}}{2x})(x-\sqrt{17}) < \frac{1}{2}(x-\sqrt{17})$ , so that $x > y$ . Similarly, we can get $y > z$ , $z > w$ , and $w > x$ , and combining these gives $x > x$ , an obvious contradiction.

Thus we must have $x \geq \sqrt{17}$ , but $x \ngtr \sqrt{17}$ , so if $x > 0$ , the only possibility is $x = \sqrt{17}$ , and analogously from the other equations we get $x = y = z = w = \sqrt{17}$ ; indeed, by substituting, we verify that this works.

As for the other case, $x < 0$ , notice that $(x,y,z,w)$ is a solution if and only if $(-x,-y,-z,-w)$ is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is $x = y = z = w = -\sqrt{17}$ , so that we have $2$ solutions in total, and therefore the answer is $\boxed{B}$ .

Solution 2

Let $(a, b, c, d) = \left(\frac{x}{\sqrt{17}}, \frac{y}{\sqrt{17}}, \frac{z}{\sqrt{17}}, \frac{w}{\sqrt{17}}\right)$ . Then these equations become $2b = a + \frac{1}{a}$ , $2c = b + \frac{1}{b}$ , $2d = c + \frac{1}{c}$ , and $a = 2d + \frac{1}{d}$ after substituting and cancelling. Notice that $(a, b, c, d)$ must either all be positive or all be negative, and that $(a, b, c, d)$ is a solution if and only if $(-a, -b, -c, -d)$ is a solution.

Suppose $(a, b, c, d)$ are all positive. Then by AM-GM, $2a = d + \frac{1}{d} \geq 2\sqrt{d \cdot \frac{1}{d}} = 2$ , so $a \geq 1$ . By similar logic, $b \geq 1$ , $c \geq 1$ , and $d \geq 1$ . Therefore, $a \geq \frac{1}{a}$ , $b \geq \frac{1}{b}$ , $c \geq \frac{1}{c}$ , and $d \geq \frac{1}{d}$ .

So $2a = d + \frac{1}{d} \leq 2d = c + \frac{1}{c} \leq 2c = b + \frac{1}{b} \leq 2b = a + \frac{1}{a} \leq 2a$ . Thus $a \leq d \leq c \leq b \leq a$ , so $a = b = c = d$ . So $2a = a + \frac{1}{a}$ , that is, $a = 1$ since $a$ is positive. Therefore, the only positive solution is $(a, b, c, d) = (1, 1, 1, 1)$ , and thus the only negative solution is $(a, b, c, d) = (-1, -1, -1, -1)$ .

Since the only solutions for $(a, b, c, d)$ are $(1, 1, 1, 1)$ and $(-1, -1, -1, -1)$ , the only $\boxed{(\mathbf{B})\ 2}$ solutions for $(x, y, z, w)$ are $(\sqrt{17}, \sqrt{17}, \sqrt{17}, \sqrt{17})$ and $(-\sqrt{17}, -\sqrt{17}, -\sqrt{17}, -\sqrt{17})$ .

-j314andrews

1986 AHSME (Problems • Answer Key • Resources)
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1986 AHSME Problems/Problem 30

Contents

Problem

Solution 1

Solution 2

See also