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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 = 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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Noticing the variables and them being multiplied together, we try to find a good factorization. After trying a few, we stumble upon something in the form of <math>(x+*)(y+*)(z+*(</math> where the blanks should be filled in with numbers. Filling in as <math>(x+4)(y+2)(z+1)</math> gives <math>2x+4y+8z+xy+4yz+2xz+xyz</math>, and all parts happen to be multiples of the given equations.
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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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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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<cmath>x^3 - 12x^2 + 44x - 48 = 0,</cmath>
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which factors as
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<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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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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a | b | c | x | y | z
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-----------------------
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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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4 | 6 | 2 | 4 | 3 | 1/2
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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).

Revision as of 03:48, 15 June 2020

Problem 30

How many distinct ordered triples $(x,y,z)$ satisfy the equations \[x+2y+4z=12\] \[xy+4yz+2xz=22\] \[xyz=6\]


$\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$

Solution

The first equation suggests the substitution $a = x$, $b = 2y$, and $c = 4z$. Then $x = a$, $y = b/2$, and $z = c/4$. Substituting into the given equations, we get


a + b + c = 12

ab + ac + bc = 44

abc = 48.

Then by Vieta's formulas, $a$, $b$, and $c$ are the roots of the equation \[x^3 - 12x^2 + 44x - 48 = 0,\] which factors as \[(x - 2)(x - 4)(x - 6) = 0.\] Hence, $a$, $b$, and $c$ are equal to 2, 4, and 6 in some order.

Since our substitution was not symmetric, each possible solution $(a,b,c)$ leads to a different solution $(x,y,z)$, as follows:


a | b | c | x | y | z

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


Hence, there are $\boxed{6}$ solutions in $(x,y,z)$. The answer is (E).

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