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Difference between revisions of "1976 AHSME Problems/Problem 30"

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== Solution ==
 
== Solution ==
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The first equation suggests the substitution <math>(a,b,c)=(x,2y,4z),</math> from which <math>(x,y,z)=(a,b/2,c/4).</math>
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The first equation suggests the substitution <math>(a,b,c)=(x,2y,4z),</math> from which <math>(x,y,z)=\left(a,\frac b2,\frac c4\right).</math>
  
 
We rewrite the given equations in terms of <math>a,b,</math> and <math>c:</math>
 
We rewrite the given equations in terms of <math>a,b,</math> and <math>c:</math>
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which factors as
 
which factors as
 
<cmath>(x - 2)(x - 4)(x - 6) = 0.</cmath>
 
<cmath>(x - 2)(x - 4)(x - 6) = 0.</cmath>
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It follows that <math>\{a,b,c\}=\{2,4,6\}.</math> Since the substitution <math>(x,y,z)=(a,b/2,c/4)</math> is not symmetric with respect to <math>x,y,</math> and <math>z,</math> we conclude that different ordered triples <math>(a,b,c)</math> generate different ordered triples <math>(x,y,z),</math> as shown below:
+
It follows that <math>\{a,b,c\}=\{2,4,6\}.</math> Since the substitution <math>(x,y,z)=\left(a,\frac b2,\frac c4\right)</math> is not symmetric with respect to <math>x,y,</math> and <math>z,</math> we conclude that different ordered triples <math>(a,b,c)</math> generate different ordered triples <math>(x,y,z),</math> as shown below:
 
<cmath>\begin{array}{c|c|c||c|c|c}
 
<cmath>\begin{array}{c|c|c||c|c|c}
 
& & & & & \\ [-2.5ex]
 
& & & & & \\ [-2.5ex]

Revision as of 09:24, 19 September 2021

Problem 30

How many distinct ordered triples $(x,y,z)$ satisfy the following equations? \begin{align*} x + 2y + 4z &= 12 \\ xy + 4yz + 2xz &= 22 \\ xyz &= 6 \end{align*} $\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$

Solution

The first equation suggests the substitution $(a,b,c)=(x,2y,4z),$ from which $(x,y,z)=\left(a,\frac b2,\frac c4\right).$

We rewrite the given equations in terms of $a,b,$ and $c:$ \begin{align*} a + b + c &= 12, \\ \frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\ \frac{abc}{8} &= 6. \end{align*} We clear fractions in these equations: \begin{align*} a + b + c &= 12, \\ ab + ac + bc &= 44, \\ abc &= 48. \end{align*} By Vieta's Formulas, note that $a,b,$ and $c$ are the roots of the equation \[x^3 - 12x^2 + 44x - 48 = 0,\] which factors as \[(x - 2)(x - 4)(x - 6) = 0.\] It follows that $\{a,b,c\}=\{2,4,6\}.$ Since the substitution $(x,y,z)=\left(a,\frac b2,\frac c4\right)$ is not symmetric with respect to $x,y,$ and $z,$ we conclude that different ordered triples $(a,b,c)$ generate different ordered triples $(x,y,z),$ as shown below: \[\begin{array}{c|c|c||c|c|c} & & & & & \\ [-2.5ex] \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} & \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{z} \\ [0.5ex] \hline & & & & & \\ [-2ex] 2 & 4 & 6 & 2 & 2 & 3/2 \\ 2 & 6 & 4 & 2 & 3 & 1 \\ 4 & 2 & 6 & 4 & 1 & 3/2 \\ 4 & 6 & 2 & 4 & 3 & 1/2 \\ 6 & 2 & 4 & 6 & 1 & 1 \\ 6 & 4 & 2 & 6 & 2 & 1/2 \end{array}\] So, there are $\boxed{\textbf{(E) }6}$ such ordered triples $(x,y,z).$

~MRENTHUSIASM (credit given to AoPS)

See also

1976 AHSME (Problems • Answer Key • Resources)
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