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

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== Problem 30 ==
 
== Problem 30 ==
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How many distinct ordered triples <math>(x,y,z)</math> satisfy the following equations?
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How many distinct ordered triples <math>(x,y,z)</math> satisfy the equations  
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<cmath>\begin{align*}
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<cmath>x+2y+4z=12</cmath>
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x + 2y + 4z &= 12 \\
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<cmath>xy+4yz+2xz=22</cmath>
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xy + 4yz + 2xz &= 22 \\
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<cmath>xyz=6</cmath>
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xyz &= 6
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\end{align*}</cmath>
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<math>\textbf{(A) }\text{none}\qquad
 
<math>\textbf{(A) }\text{none}\qquad
 
\textbf{(B) }1\qquad
 
\textbf{(B) }1\qquad
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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
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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>
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a + b + c = 12
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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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ab + ac + bc = 44
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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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abc = 48.
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\frac{abc}{8} &= 6.
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\end{align*}</cmath>
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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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We clear fractions in these equations:
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<cmath>x^3 - 12x^2 + 44x - 48 = 0,</cmath>
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<cmath>\begin{align*}
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a + b + c &= 12, \\
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ab + ac + bc &= 44, \\
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abc &= 48.
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\end{align*}</cmath>
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By Vieta's Formulas, note that <math>a,b,</math> and <math>c</math> are the roots of the equation <cmath>r^3 - 12r^2 + 44r - 48 = 0,</cmath>
 
which factors as
 
which factors as
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<cmath>(x - 2)(x - 4)(x - 6) = 0.</cmath>
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<cmath>(r - 2)(r - 4)(r - 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)=\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:
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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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\[
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& & & & & \\ [-2ex]
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\begin{tabular}{c|c|c|c|c|c}
 
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a & b & c & x & y & z \\ \hline
 
 
2 & 4 & 6 & 2 & 2 & 3/2 \\
 
2 & 4 & 6 & 2 & 2 & 3/2 \\
 
2 & 6 & 4 & 2 & 3 & 1 \\
 
2 & 6 & 4 & 2 & 3 & 1 \\
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6 & 2 & 4 & 6 & 1 & 1 \\
 
6 & 2 & 4 & 6 & 1 & 1 \\
 
6 & 4 & 2 & 6 & 2 & 1/2
 
6 & 4 & 2 & 6 & 2 & 1/2
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\end{tabular}
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\end{array}</cmath>
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\]
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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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~MRENTHUSIASM (credit given to AoPS)
  
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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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== See also ==
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{{AHSME box|year=1976|n=I|num-b=29|after=Last Problem}}
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{{MAA Notice}}

Latest revision as of 09:36, 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 \[r^3 - 12r^2 + 44r - 48 = 0,\] which factors as \[(r - 2)(r - 4)(r - 6) = 0.\] 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)
Preceded by
Problem 29 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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