Difference between revisions of "1976 AHSME Problems/Problem 30"
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<cmath>xy+4yz+2xz=22</cmath> | <cmath>xy+4yz+2xz=22</cmath> | ||
<cmath>xyz=6</cmath> | <cmath>xyz=6</cmath> | ||
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<math>\textbf{(A) }\text{none}\qquad | <math>\textbf{(A) }\text{none}\qquad | ||
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== Solution == | == Solution == | ||
The first equation suggests the substitution <math>a = x</math>, <math>b = 2y</math>, and <math>c = 4z</math>. Then <math>x = a</math>, <math>y = b/2</math>, and <math>z = c/4</math>. Substituting into the given equations, we get | The first equation suggests the substitution <math>a = x</math>, <math>b = 2y</math>, and <math>c = 4z</math>. Then <math>x = a</math>, <math>y = b/2</math>, and <math>z = c/4</math>. Substituting into the given equations, we get | ||
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a + b + c = 12 | a + b + c = 12 | ||
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Hence, there are <math>\boxed{6}</math> solutions in <math>(x,y,z)</math>. The answer is (E). | Hence, there are <math>\boxed{6}</math> solutions in <math>(x,y,z)</math>. The answer is (E). | ||
| + | |||
| + | == See also == | ||
| + | {{AHSME box|year=1976|n=I|num-b=29|after=Last Problem}} | ||
| + | {{MAA Notice}} | ||
Revision as of 21:51, 5 September 2021
Problem 30
How many distinct ordered triples 
satisfy the equations
![\[x+2y+4z=12\]](http://latex.artofproblemsolving.com/e/5/3/e53f6d4f56a9fd1f01ffe12c1a29b46a0f06c1dd.png)
![\[xy+4yz+2xz=22\]](http://latex.artofproblemsolving.com/6/a/3/6a38bcbbdb99991b176c33335bd1a64cf6e13970.png)
![\[xyz=6\]](http://latex.artofproblemsolving.com/b/6/2/b62793631f6681941ffdba570d7401ab0f39f1e9.png)
Solution
The first equation suggests the substitution 
, 
, and 
. Then 
, 
, and 
. Substituting into the given equations, we get
a + b + c = 12
ab + ac + bc = 44
abc = 48.
Then by Vieta's formulas, 
, 
, and 
are the roots of the equation
![\[x^3 - 12x^2 + 44x - 48 = 0,\]](http://latex.artofproblemsolving.com/3/b/d/3bd4922c814ebf6abbc97b645d61322a1e4470a6.png)
which factors as
![\[(x - 2)(x - 4)(x - 6) = 0.\]](http://latex.artofproblemsolving.com/f/5/8/f58ffd9f6197951df620229151fa219a117edbd3.png)
Hence, , , and are equal to 2, 4, and 6 in some order.
Since our substitution was not symmetric, each possible solution 
leads to a different solution , as follows:
a | b | c | x | y | z
2 | 4 | 6 | 2 | 2 | 3/2
2 | 6 | 4 | 2 | 3 | 1
4 | 2 | 6 | 4 | 1 | 3/2
4 | 6 | 2 | 4 | 3 | 1/2
6 | 2 | 4 | 6 | 1 | 1
6 | 4 | 2 | 6 | 2 | 1/2
Hence, there are 
solutions in . The answer is (E).
See also
| 1976 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 29 |
Followed by Last Problem | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
