Difference between revisions of "1953 AHSME Problems/Problem 45"

(Created page with "By AM-GM, the answer is <math>\textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}</math>")
 
 
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By AM-GM, the answer is <math>\textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}</math>
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==Problem==
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The lengths of two line segments are <math>a</math> units and <math>b</math> units respectively. Then the correct relation between them is:
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<math>\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad
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\textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad
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\textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad
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\textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}  </math>
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==Solution==
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Since both lengths are positive, the [[AM-GM Inequality]] is satisfied. The correct relationship between <math>a</math> and <math>b</math> is <math>\boxed{\textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}}</math>.
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==See Also==
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{{AHSME 50p box|year=1953|num-b=44|num-a=46}}
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{{MAA Notice}}

Latest revision as of 23:02, 14 February 2020

Problem

The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:

$\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad \textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad \textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad \textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}$

Solution

Since both lengths are positive, the AM-GM Inequality is satisfied. The correct relationship between $a$ and $b$ is $\boxed{\textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}}$.

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 44
Followed by
Problem 46
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All AHSME Problems and Solutions


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