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Difference between revisions of "1983 AHSME Problems/Problem 13"

(Created page with "==Problem== If <math>xy = a, xz =b,</math> and <math>yz = c</math>, and none of these quantities is zero, then <math>x^2+y^2+z^2</math> equals: <math>\textbf{(A)}\ \frac{ab+...")
 
 
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==Problem==
 
==Problem==
  
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If <math>xy = a, xz =b,</math> and <math>yz = c</math>, and none of these quantities is zero, then <math>x^2+y^2+z^2</math> equals:
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If <math>xy = a, xz =b,</math> and <math>yz = c</math>, and none of these quantities is <math>0</math>, then <math>x^2+y^2+z^2</math> equals
  
 
<math>\textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad
 
<math>\textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad
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==Solution==
 
==Solution==
  
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Using the equations, you can get <math>x = \frac{a}{y}, z = \frac{b}{x},</math> and <math>y = \frac{c}{z}</math>. Substitute into the expression given. You get <math>\frac{a^2}{y^2} + \frac{b^2}{x^2} + \frac{c^2}{z^2}</math>. Combine into one large fraction to get <math>\frac{a^2x^2z^2+b^2y^2z^2+c^2y^2x^2}{x^2y^2z^2}</math>. Make everything in terms of a,b, c. <math>\frac{a^2b^2+b^2c^2+c^2a^2}{x^2y^2z^2}</math>. So the answer is <math>\boxed{\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}}</math>.
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From the equations, we deduce <math>x = \frac{a}{y}, z = \frac{b}{x},</math> and <math>y = \frac{c}{z}</math>. Substituting these into the expression <math>x^2+y^2+z^2</math> thus gives <math>\frac{a^2}{y^2} + \frac{b^2}{x^2} + \frac{c^2}{z^2} = \frac{a^2x^2z^2+b^2y^2z^2+c^2y^2x^2}{x^2y^2z^2} = \frac{a^2b^2+b^2c^2+c^2a^2}{x^2y^2z^2}</math>, so the answer is <math>\boxed{\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}}</math>.
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==Solution 2==
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<math>x^2</math> is <math>\frac{abc}{c^2}</math>, <math>y^2</math> is <math>\frac{abc}{b^2}</math>, and <math>z^2</math> is <math>\frac{abc}{a^2}</math>, so <math>x^2+y^2+z^2</math> is <math>\frac{abc}{c^2}+\frac{abc}{b^2}+\frac{abc}{a^2}= \boxed{\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}}</math>
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-purplepenguin2
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==See Also==
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{{AHSME box|year=1983|num-b=12|num-a=14}}
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{{MAA Notice}}

Latest revision as of 16:55, 16 June 2021

Problem

If $xy = a, xz =b,$ and $yz = c$, and none of these quantities is $0$, then $x^2+y^2+z^2$ equals

$\textbf{(A)}\ \frac{ab+ac+bc}{abc}\qquad \textbf{(B)}\ \frac{a^2+b^2+c^2}{abc}\qquad \textbf{(C)}\ \frac{(a+b+c)^2}{abc}\qquad \textbf{(D)}\ \frac{(ab+ac+bc)^2}{abc}\qquad \textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}$

Solution

From the equations, we deduce $x = \frac{a}{y}, z = \frac{b}{x},$ and $y = \frac{c}{z}$. Substituting these into the expression $x^2+y^2+z^2$ thus gives $\frac{a^2}{y^2} + \frac{b^2}{x^2} + \frac{c^2}{z^2} = \frac{a^2x^2z^2+b^2y^2z^2+c^2y^2x^2}{x^2y^2z^2} = \frac{a^2b^2+b^2c^2+c^2a^2}{x^2y^2z^2}$, so the answer is $\boxed{\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}}$.

Solution 2

$x^2$ is $\frac{abc}{c^2}$, $y^2$ is $\frac{abc}{b^2}$, and $z^2$ is $\frac{abc}{a^2}$, so $x^2+y^2+z^2$ is $\frac{abc}{c^2}+\frac{abc}{b^2}+\frac{abc}{a^2}= \boxed{\textbf{(E)}\ \frac{(ab)^2+(ac)^2+(bc)^2}{abc}}$

-purplepenguin2

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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