Difference between revisions of "1968 AHSME Problems/Problem 22"
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== Solution == | == Solution == | ||
| − | <math>\fbox{E}</math> | + | |
| + | 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}} | + | {{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 17:15, 17 July 2024
Problem
A segment of length 
is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is:
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 
. All other segment lengths are valid, so long as the inequality is satsified. Thus, our answer is 
.
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 21 |
Followed by Problem 23 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
