Difference between revisions of "1968 AHSME Problems/Problem 22"

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== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
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As a consequence of the [[Triangle Inequality]], we can form a quadrilateral from the four segments iff the sum of any three sides is greater than the sum of the fourth side. Thus, no segment can have a length <math>\geq \frac{1}{2}</math>. All other segment lengths are valid, so long as the inequality is satsified. Thus, our answer is <math>\fbox{(E) less than 1/2}</math>.
  
 
== See also ==
 
== See also ==
{{AHSME box|year=1968|num-b=21|num-a=23}}   
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{{AHSME 35p box|year=1968|num-b=21|num-a=23}}   
  
 
[[Category: Intermediate Geometry Problems]]
 
[[Category: Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:15, 17 July 2024

Problem

A segment of length $1$ is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is:

$\text{(A) equal to } \frac{1}{4}\quad\\ \text{(B) equal to or greater than } \frac{1}{8} \text{ and less than }\frac{1}{2}\quad\\ \text{(C) greater than } \frac{1}{8} \text{ and less than }\frac{1}{2}\quad\\ \text{(D) equal to or greater than } \frac{1}{8} \text{ and less than }\frac{1}{4}\quad\\ \text{(E) less than }\frac{1}{2}$


Solution

As a consequence of the Triangle Inequality, we can form a quadrilateral from the four segments iff the sum of any three sides is greater than the sum of the fourth side. Thus, no segment can have a length $\geq \frac{1}{2}$. All other segment lengths are valid, so long as the inequality is satsified. Thus, our answer is $\fbox{(E) less than 1/2}$.

See also

1968 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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