1968 AHSME Problems/Problem 22
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 | |
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