1976 AHSME Problems/Problem 30

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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1976 AHSME Problems/Problem 30

Problem 30

Solution