Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Improved clarity of solution)
 
(4 intermediate revisions by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==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:
+
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
Line 11: Line 11:
 
==Solution==
 
==Solution==
  
−
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>.
+
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>.
 +
 
 +
==Solution 2==
 +
 
 +
<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>
 +
 
 +
-purplepenguin2
 +
 
 +
==See Also==
 +
{{AHSME box|year=1983|num-b=12|num-a=14}}
 +
 
 +
{{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


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_13&oldid=156158"