Difference between revisions of "1986 AHSME Problems/Problem 30"
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\textbf{(E)}\ 16 </math> | \textbf{(E)}\ 16 </math> | ||
− | ==Solution== | + | ==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>. | ||
+ | Now suppose <math>x > \sqrt{17}</math>. Then <math>y - \sqrt{17} = \frac{x^{2}+17}{2x} - \sqrt{17} = (\frac{x-\sqrt{17}}{2x})(x-\sqrt{17}) < \frac{1}{2}(x-\sqrt{17})</math>, so that <math>x > y</math>. Similarly, we can get <math>y > z</math>, <math>z > w</math>, and <math>w > x</math>, and combining these gives <math>x > x</math>, an obvious contradiction. | ||
+ | |||
+ | Thus we must have <math>x \geq \sqrt{17}</math>, but <math>x \ngtr \sqrt{17}</math>, so if <math>x > 0</math>, the only possibility is <math>x = \sqrt{17}</math>, and analogously from the other equations we get <math>x = y = z = w = \sqrt{17}</math>; indeed, by substituting, we verify that this works. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | -j314andrews | ||
== See also == | == See also == |
Latest revision as of 02:35, 5 July 2025
Contents
Problem
The number of real solutions of the simultaneous equations
is
Solution 1
Consider the cases and
, and also note that by AM-GM, for any positive number
, we have
, with equality only if
. Thus, if
, considering each equation in turn, we get that
, and finally
.
Now suppose . Then
, so that
. Similarly, we can get
,
, and
, and combining these gives
, an obvious contradiction.
Thus we must have , but
, so if
, the only possibility is
, and analogously from the other equations we get
; indeed, by substituting, we verify that this works.
As for the other case, , notice that
is a solution if and only if
is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is
, so that we have
solutions in total, and therefore the answer is
.
Solution 2
Let . Then these equations become
,
,
, and
after substituting and cancelling. Notice that
must either all be positive or all be negative, and that
is a solution if and only if
is a solution.
Suppose are all positive. Then by AM-GM,
, so
. By similar logic,
,
, and
. Therefore,
,
,
, and
.
So . Thus
, so
. So
, that is,
since
is positive. Therefore, the only positive solution is
, and thus the only negative solution is
.
Since the only solutions for are
and
, the only
solutions for
are
and
.
-j314andrews
See also
1986 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.