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.