Difference between revisions of "1964 AHSME Problems/Problem 37"

Revision as of 18:46, 26 January 2021

Problem

Given two positive numbers $a$ , $b$ such that $a<b$ . 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:

$\textbf{(A) }\dfrac{(b+a)^2}{ab}\qquad\textbf{(B) }\dfrac{(b+a)^2}{8b}\qquad\textbf{(C) }\dfrac{(b-a)^2}{ab}$

$\textbf{(D) }\dfrac{(b-a)^2}{8a}\qquad \textbf{(E) }\dfrac{(b-a)^2}{8b}$

Solution

$\textbf{\boxed{ (D) }}$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 ==Solution==
+<math>\textbf{\boxed{ (D) }}</math>
 ==See Also==

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Difference between revisions of "1964 AHSME Problems/Problem 37"

Revision as of 18:46, 26 January 2021

Problem

Solution

See Also