1976 AHSME Problems/Problem 30
Problem 30
How many distinct ordered triples satisfy the equations
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
which factors as
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).