Problem: A segment of length 1 is divided into four segments. Then there exists a simple quadrilateral with the four segments as sides if and only if each segment is:
Answer Choices:
A. equal to 41​
B. equal to or greater than 81​ and less than 21​
C. greater than 81​ and less than 21​
D. greater than 81​ and less than 41​
E. less than 21​
Solution:
Let the sides be s1​,s2​,s3​,s4​. Since it is given that s1​+s2​+s3​+s4​=1,s1​+s2​+s3​ =1−s4​. But s1​+s2​+s3​>s4​. Therefore, 0<1−2s4​,s4​<21​. Similar reasoning applied to s1​,s2​,s3​, in turn, shows that each side is less than 21​. Conversely, a quadrilateral exists if s1​+s2​+s3​>s4​,s1​+s2​+s4​>s3​,s1​+s3​+s4​>s2​, and s2​+s3​ +s4​>s1​. It is given that s1​+s2​+s3​+s4​=1 and s1​<21​,s2​<21​,s3​<21​,s4​<21​. Therefore, s2​+s3​+s4​>21​ and so s2​+s3​+s4​>s1​. In a similar manner we prove the other three cases.