Difference between revisions of "1976 AHSME Problems/Problem 30"
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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{center} | |
| − | \ | ||
\begin{tabular}{c|c|c|c|c|c} | \begin{tabular}{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{tabular} | \end{tabular} | ||
| − | \ | + | \end{center} |
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 02:39, 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
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{center} \begin{tabular}{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{tabular} \end{center}
Hence, there are 
solutions in . The answer is (E).
