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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
 +
\begin{align*}
 +
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.
 +
\end{align*}
 +
 +
Then by Vieta's formulas, <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the equation
 +
<cmath>x^3 - 12x^2 + 44x - 48 = 0,</cmath>
 +
which factors as
 +
<cmath>(x - 2)(x - 4)(x - 6) = 0.</cmath>
 +
Hence, <math>a</math>, <math>b</math>, and <math>c</math> are equal to 2, 4, and 6 in some order.
 +
 +
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:
 +
 +
\[
 +
\begin{array}{c|c|c|c|c|c}
 +
a & b & c & x & y & z \\ \hline
 +
2 & 4 & 6 & 2 & 2 & 3/2 \\
 +
2 & 6 & 4 & 2 & 3 & 1 \\
 +
4 & 2 & 6 & 4 & 1 & 3/2 \\
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4 & 6 & 2 & 4 & 3 & 1/2 \\
 +
6 & 2 & 4 & 6 & 1 & 1 \\
 +
6 & 4 & 2 & 6 & 2 & 1/2
 +
\end{array}
 +
\]
 +
 +
Hence, there are <math>\boxed{6}</math> solutions in <math>(x,y,z)</math>. The answer is (E).

Revision as of 02:31, 15 June 2020

Problem 30

How many distinct ordered triples $(x,y,z)$ satisfy the equations \[x+2y+4z=12\] \[xy+4yz+2xz=22\] \[xyz=6\]


$\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$

Solution

The first equation suggests the substitution $a = x$, $b = 2y$, and $c = 4z$. Then $x = a$, $y = b/2$, and $z = c/4$. Substituting into the given equations, we get \begin{align*} 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. \end{align*}

Then by Vieta's formulas, $a$, $b$, and $c$ are the roots of the equation \[x^3 - 12x^2 + 44x - 48 = 0,\] which factors as \[(x - 2)(x - 4)(x - 6) = 0.\] Hence, $a$, $b$, and $c$ are equal to 2, 4, and 6 in some order.

Since our substitution was not symmetric, each possible solution $(a,b,c)$ leads to a different solution $(x,y,z)$, 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 $\boxed{6}$ solutions in $(x,y,z)$. The answer is (E).

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