Difference between revisions of "1976 AHSME Problems/Problem 30"
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== Problem 30 == | == Problem 30 == | ||
− | + | How many distinct ordered triples <math>(x,y,z)</math> satisfy the following equations? | |
− | How many distinct ordered triples <math>(x,y,z)</math> satisfy the equations | + | <cmath>\begin{align*} |
− | <cmath>x+2y+4z=12 | + | x + 2y + 4z &= 12 \\ |
− | + | xy + 4yz + 2xz &= 22 \\ | |
− | + | xyz &= 6 | |
− | + | \end{align*}</cmath> | |
− | |||
<math>\textbf{(A) }\text{none}\qquad | <math>\textbf{(A) }\text{none}\qquad | ||
\textbf{(B) }1\qquad | \textbf{(B) }1\qquad | ||
Line 14: | Line 13: | ||
== Solution == | == Solution == | ||
− | The first equation suggests the substitution <math>a = x | + | 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> | ||
+ | <cmath>\begin{align*} | ||
a + b + c &= 12, \\ | a + b + c &= 12, \\ | ||
\frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\ | \frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\ | ||
− | \frac{abc}{8} = 6 | + | \frac{abc}{8} &= 6. |
− | \end{align*} | + | \end{align*}</cmath> |
− | + | We clear fractions in these equations: | |
− | \begin{align*} | + | <cmath>\begin{align*} |
a + b + c &= 12, \\ | a + b + c &= 12, \\ | ||
ab + ac + bc &= 44, \\ | ab + ac + bc &= 44, \\ | ||
abc &= 48. | 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>r^3 - 12r^2 + 44r - 48 = 0,</cmath> | |
− | <cmath> | ||
which factors as | which factors as | ||
− | <cmath>( | + | <cmath>(r - 2)(r - 4)(r - 6) = 0.</cmath> |
− | + | 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} | |
− | + | & & & & & \\ [-2.5ex] | |
− | + | \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} & \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{z} \\ [0.5ex] | |
− | + | \hline | |
− | \begin{array}{c|c|c|c|c|c} | + | & & & & & \\ [-2ex] |
− | 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 \\ | ||
Line 42: | Line 42: | ||
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 | ||
− | \end{array} | + | \end{array}</cmath> |
+ | So, there are <math>\boxed{\textbf{(E) }6}</math> such ordered triples <math>(x,y,z).</math> | ||
+ | |||
+ | ~MRENTHUSIASM (credit given to AoPS) | ||
− | + | == See also == | |
+ | {{AHSME box|year=1976|n=I|num-b=29|after=Last Problem}} | ||
+ | {{MAA Notice}} |
Latest revision as of 10:36, 19 September 2021
Problem 30
How many distinct ordered triples satisfy the following equations?
Solution
The first equation suggests the substitution from which
We rewrite the given equations in terms of and
We clear fractions in these equations:
By Vieta's Formulas, note that
and
are the roots of the equation
which factors as
It follows that
Since the substitution
is not symmetric with respect to
and
we conclude that different ordered triples
generate different ordered triples
as shown below:
So, there are
such ordered triples
~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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