Difference between revisions of "1964 AHSME Problems/Problem 37"
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==Problem== | ==Problem== | ||
| − | Given two positive number <math>a</math>, <math>b</math> such that <math>a<b</math>. Let A.M. be their arithmetic mean and let | + | Given two positive number <math>a</math>, <math>b</math> such that <math>a<b</math>. Let A.M. be their arithmetic mean and let G.M. be their positive geometric mean. Then A.M. minus G.M. is always less than: |
| − | < | + | <math>\textbf{(A) }\dfrac{(b+a)^2}{ab}\qquad\textbf{(B) }\dfrac{(b+a)^2}{8b}\qquad\textbf{(C) }\dfrac{(b-a)^2}{ab}</math> |
| − | < | + | <math>\textbf{(D) }\dfrac{(b-a)^2}{8a}\qquad \textbf{(E) }\dfrac{(b-a)^2}{8b}</math> |
| + | |||
| + | ==Solution== | ||
| + | |||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1964|num-b=36|num-a=38}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 23:19, 24 July 2019
Problem
Given two positive number 
, 
such that 
. Let A.M. be their arithmetic mean and let G.M. be their positive geometric mean. Then A.M. minus G.M. is always less than:
Solution
See Also
| 1964 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 36 |
Followed by Problem 38 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
