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Difference between revisions of "1986 AHSME Problems/Problem 30"

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\textbf{(E)}\ 16  </math>
 
\textbf{(E)}\ 16  </math>
  
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==Solution==
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==Solution 1==
 
Consider the cases <math>x>0</math> and <math>x<0</math>, and also note that by AM-GM, for any positive number <math>a</math>, we have <math>a+\frac{17}{a} \geq 2\sqrt{17}</math>, with equality only if <math>a = \sqrt{17}</math>. Thus, if <math>x>0</math>, considering each equation in turn, we get that <math>y \geq \sqrt{17}, z \geq \sqrt{17}, w \geq \sqrt{17}</math>, and finally <math>x \geq \sqrt{17}</math>.  
 
Consider the cases <math>x>0</math> and <math>x<0</math>, and also note that by AM-GM, for any positive number <math>a</math>, we have <math>a+\frac{17}{a} \geq 2\sqrt{17}</math>, with equality only if <math>a = \sqrt{17}</math>. Thus, if <math>x>0</math>, considering each equation in turn, we get that <math>y \geq \sqrt{17}, z \geq \sqrt{17}, w \geq \sqrt{17}</math>, and finally <math>x \geq \sqrt{17}</math>.  
  
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As for the other case, <math>x < 0</math>, notice that <math>(x,y,z,w)</math> is a solution if and only if <math>(-x,-y,-z,-w)</math> is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is <math>x = y = z = w = -\sqrt{17}</math>, so that we have <math>2</math> solutions in total, and therefore the answer is <math>\boxed{B}</math>.
 
As for the other case, <math>x < 0</math>, notice that <math>(x,y,z,w)</math> is a solution if and only if <math>(-x,-y,-z,-w)</math> is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is <math>x = y = z = w = -\sqrt{17}</math>, so that we have <math>2</math> solutions in total, and therefore the answer is <math>\boxed{B}</math>.
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==Solution 2==
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Let <math>(a, b, c, d) = \left(\frac{x}{\sqrt{17}}, \frac{y}{\sqrt{17}}, \frac{z}{\sqrt{17}}, \frac{w}{\sqrt{17}}\right)</math>.  Then these equations become <math>2b = a + \frac{1}{a}</math>, <math>2c = b + \frac{1}{b}</math>, <math>2d = c + \frac{1}{c}</math>, and <math>a = 2d + \frac{1}{d}</math> after substituting and cancelling.  Notice that <math>(a, b, c, d)</math> must either all be positive or all be negative, and that <math>(a, b, c, d)</math> is a solution if and only if <math>(-a, -b, -c, -d)</math> is a solution.
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Suppose <math>(a, b, c, d)</math> are all positive.  Then by AM-GM, <math>2a = d + \frac{1}{d} \geq 2\sqrt{d \cdot \frac{1}{d}} = 2</math>, so <math>a \geq 1</math>.  By similar logic, <math>b \geq 1</math>, <math>c \geq 1</math>, and <math>d \geq 1</math>.  Therefore, <math>a \geq \frac{1}{a}</math>, <math>b \geq \frac{1}{b}</math>, <math>c \geq \frac{1}{c}</math>, and <math>d \geq \frac{1}{d}</math>.
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So <math>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</math>.  Thus <math>a \leq d \leq c \leq b \leq a</math>, so <math>a = b = c = d</math>.  So <math>2a = a + \frac{1}{a}</math>, that is, <math>a = 1</math> since <math>a</math> is positive.  Therefore, the only positive solution is <math>(a, b, c, d) = (1, 1, 1, 1)</math>, and thus the only negative solution is <math>(a, b, c, d) = (-1, -1, -1, -1)</math>.
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Since the only solutions for <math>(a, b, c, d)</math> are <math>(1, 1, 1, 1)</math> and <math>(-1, -1, -1, -1)</math>, the only <math>\boxed{(\mathbf{B})\ 2}</math> solutions for <math>(x, y, z, w)</math> are <math>(\sqrt{17}, \sqrt{17}, \sqrt{17}, \sqrt{17})</math> and <math>(-\sqrt{17}, -\sqrt{17}, -\sqrt{17}, -\sqrt{17})</math>.
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-j314andrews
  
 
== See also ==
 
== See also ==

Latest revision as of 02:35, 5 July 2025

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

See also

1986 AHSME (Problems • Answer Key • Resources)
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