Difference between revisions of "2000 AMC 12 Problems/Problem 20"
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If <math>x,y,</math> and <math>z</math> are positive numbers satisfying | If <math>x,y,</math> and <math>z</math> are positive numbers satisfying | ||
− | <cmath>x + 1 | + | <cmath>x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}</cmath> |
− | <math> | + | Then what is the value of <math>xyz</math> ? |
− | + | <math>\text {(A)}\ \frac{2}{3} \qquad \text {(B)}\ 1 \qquad \text {(C)}\ \frac{4}{3} \qquad \text {(D)}\ 2 \qquad \text {(E)}\ \frac{7}{3}</math> | |
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− | <cmath>\ | + | == Solution 1 == |
− | + | We multiply all given expressions to get: | |
− | + | <cmath>(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}</cmath> | |
− | + | Adding all the given expressions gives that | |
− | + | <cmath>(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}</cmath> | |
− | \ | + | We subtract <math>(2)</math> from <math>(1)</math> to get that <math>xyz + \frac{1}{xyz} = 2</math>. Hence, by inspection, <math>\boxed{xyz = 1 \rightarrow B}</math>. |
+ | <cmath></cmath> | ||
+ | ~AopsUser101 | ||
− | + | == Solution 2 == | |
− | |||
− | |||
We have a system of three equations and three variables, so we can apply repeated substitution. | We have a system of three equations and three variables, so we can apply repeated substitution. | ||
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Multiplying out the denominator and simplification yields <math>4(4x-3) = x(4x-3) + 7x - 3 \Longrightarrow (2x-3)^2 = 0</math>, so <math>x = \frac{3}{2}</math>. Substituting leads to <math>y = \frac{2}{5}, z = \frac{5}{3}</math>, and the product of these three variables is <math>1</math>. | Multiplying out the denominator and simplification yields <math>4(4x-3) = x(4x-3) + 7x - 3 \Longrightarrow (2x-3)^2 = 0</math>, so <math>x = \frac{3}{2}</math>. Substituting leads to <math>y = \frac{2}{5}, z = \frac{5}{3}</math>, and the product of these three variables is <math>1</math>. | ||
− | == See | + | == Video Solution by OmegaLearn == |
+ | https://www.youtube.com/watch?v=SpSuqWY01SE&t=374s | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/ph8o017pw_o | ||
+ | |||
+ | == See Also == | ||
{{AMC12 box|year=2000|num-b=19|num-a=21}} | {{AMC12 box|year=2000|num-b=19|num-a=21}} | ||
− | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 07:34, 4 November 2022
Contents
Problem
If and
are positive numbers satisfying
Then what is the value of ?
Solution 1
We multiply all given expressions to get:
Adding all the given expressions gives that
We subtract
from
to get that
. Hence, by inspection,
.
~AopsUser101
Solution 2
We have a system of three equations and three variables, so we can apply repeated substitution.
Multiplying out the denominator and simplification yields , so
. Substituting leads to
, and the product of these three variables is
.
Video Solution by OmegaLearn
https://www.youtube.com/watch?v=SpSuqWY01SE&t=374s
~ pi_is_3.14
Video Solution
See Also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.