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

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(I highly suspected that someone published copyrighted materials from the "AMC 12 Problem Series" class. So, I alter a little wording while keeping the logic the same. Also, I improved the display of equations and tables.)
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== Solution ==
 
== Solution ==
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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
+
The first equation suggests the substitutions <math>a=x,b=2y,</math> and <math>c=4z.</math> So, we get <math>x=a,y=b/2,</math> and <math>z=c/4,</math> respectively. We rewrite the given equations in terms of <math>a,b,</math> and <math>c:</math>
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<cmath>\begin{align*}
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a + b + c = 12
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a + b + c &= 12, \
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\frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \
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ab + ac + bc = 44
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\frac{abc}{8} &= 6.
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\end{align*}</cmath>
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abc = 48.
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We clear fractions in these equations:
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<cmath>\begin{align*}
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Then by Vieta's formulas, <math>a</math>, <math>b</math>, and <math>c</math> are the roots of the equation
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a + b + c &= 12, \
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<cmath>x^3 - 12x^2 + 44x - 48 = 0,</cmath>
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ab + ac + bc &= 44, \
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abc &= 48.
 +
\end{align*}</cmath>
 +
By Vieta's Formulas, note that <math>a,b,</math> and <math>c</math> are the roots of the equation <cmath>x^3 - 12x^2 + 44x - 48 = 0,</cmath>
 
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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Hence, <math>a</math>, <math>b</math>, and <math>c</math> are equal to 2, 4, and 6 in some order.
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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:
−
 
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<cmath>\begin{array}{c|c|c||c|c|c}
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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:
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& & & & & \ [-2.5ex]
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\boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} & \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{z} \ [0.5ex]
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\hline
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a | b | c | x | y | z
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& & & & & \ [-2ex]
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-----------------------
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2 & 4 & 6 & 2 & 2 & 3/2 \
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2 | 4 | 6 | 2 | 2 | 3/2
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2 & 6 & 4 & 2 & 3 & 1 \
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4 & 2 & 6 & 4 & 1 & 3/2 \
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2 | 6 | 4 | 2 | 3 | 1
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4 & 6 & 2 & 4 & 3 & 1/2 \
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6 & 2 & 4 & 6 & 1 & 1 \
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4 | 2 | 6 | 4 | 1 | 3/2
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6 & 4 & 2 & 6 & 2 & 1/2
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\end{array}</cmath>
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4 | 6 | 2 | 4 | 3 | 1/2
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So, there are <math>\boxed{\textbf{(E) }6}</math> such ordered triples <math>(x,y,z).</math>
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6 | 2 | 4 | 6 | 1 | 1
 
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6 | 4 | 2 | 6 | 2 | 1/2
 
−
 
 
  
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Hence, there are <math>\boxed{6}</math> solutions in <math>(x,y,z)</math>. The answer is (E).
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~MRENTHUSIASM (credit given to AoPS)
  
 
== See also ==
 
== See also ==
 
{{AHSME box|year=1976|n=I|num-b=29|after=Last Problem}}
 
{{AHSME box|year=1976|n=I|num-b=29|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:02, 6 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 substitutions $a=x,b=2y,$ and $c=4z.$ So, we get $x=a,y=b/2,$ and $z=c/4,$ respectively. 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 $(x,y,z)=(a,b/2,c/4)$ 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)
Preceded by
Problem 29 		Followed by
Last Problem
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All AHSME Problems and Solutions

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