1976 AMC 12 Problem 30

Question

How many distinct ordered triples

(x,y,z)

satisfy the equations

\begin{aligned} x+2y+4z&=12,\\ xy+4yz+2xz&=22,\\ xyz&=6? \end{aligned}

none

1

2

4

6

Rescale the variables by setting

x=2u,

y=v,

and

z=\frac{w}{2}

The new variables have elementary symmetric sums

6,

11,

and

6

Set

x=2u,

y=v,

and

z=\frac{w}{2}.

Dividing the first two equations by

2

gives

\begin{aligned} u+v+w&=6,\\ uv+vw+uw&=11, \end{aligned}

while

uvw=6.

Thus

u,

v,

and

w

are the three roots of

p(t)=t^3-6t^2+11t-6.

This polynomial factors as

p(t)=(t-1)(t-2)(t-3),

so they are

1,

2,

and

3

in any order. Their

3!=6

permutations produce

6

distinct ordered triples

(x,y,z).

Therefore, the correct answer is E.

Problem 30 in other years