Difference between revisions of "1986 AHSME Problems/Problem 30"
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==Solution== | ==Solution== | ||
| + | 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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| + | 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. | ||
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| + | 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. | ||
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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>. | ||
== See also == | == See also == | ||
Latest revision as of 14:06, 21 June 2019
Problem
The number of real solutions 
of the simultaneous equations
is
Solution
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 
.
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 | ||
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