Difference between revisions of "1983 AHSME Problems/Problem 13"
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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>. | 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 | ||
==See Also== | ==See Also== |
Latest revision as of 17:55, 16 June 2021
Contents
Problem
If and , and none of these quantities is , then equals
Solution
From the equations, we deduce and . Substituting these into the expression thus gives , so the answer is .
Solution 2
is , is , and is , so is
-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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