Difference between revisions of "1976 AHSME Problems/Problem 30"
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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 |
| − | |||
a + b + c &= 12, \\ | a + b + c &= 12, \\ | ||
\frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\ | \frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\ | ||
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ab + ac + bc &= 44, \\ | ab + ac + bc &= 44, \\ | ||
abc &= 48. | abc &= 48. | ||
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Then by Vieta's formulas, <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the equation | Then by Vieta's formulas, <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the equation | ||
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Since our substitution was not symmetric, each possible solution <math>(a,b,c)</math> leads to a different solution <math>(x,y,z)</math>, as follows: | Since our substitution was not symmetric, each possible solution <math>(a,b,c)</math> leads to a different solution <math>(x,y,z)</math>, as follows: | ||
| − | + | ||
\begin{array}{c|c|c|c|c|c} | \begin{array}{c|c|c|c|c|c} | ||
a & b & c & x & y & z \\ \hline | a & b & c & x & y & z \\ \hline | ||
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6 & 4 & 2 & 6 & 2 & 1/2 | 6 & 4 & 2 & 6 & 2 & 1/2 | ||
\end{array} | \end{array} | ||
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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). | ||
Revision as of 02:33, 15 June 2020
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, \\
\frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\
\frac{abc}{8} = 6,
\end{align*}
which simplifies to
\begin{align*}
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:
\begin{array}{c|c|c|c|c|c}
a & b & c & x & y & z \\ \hline
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
\end{array}
Hence, there are 
solutions in . The answer is (E).
