Problem:
Let a,b,c,d,e,f,g and h be distinct elements in the set
{−7,−5,−3,−2,2,4,6,13}
What is the minimum possible value of
(a+b+c+d)2+(e+f+g+h)2?
Answer Choices:
A. 30
B. 32
C. 34
D. 40
E. 50
Solution:
Note that the sum of the elements in the set is 8 . Let x=a+b+c+d, so e+f+g+h=8−x. Then
(a+b+c+d)2+(e+f+g+h)2=x2+(8−x)2=2x2−16x+64=2(x−4)2+32≥32
The value of 32 can be attained if and only if x=4. However, it may be assumed without loss of generality that a=13, and no choice of b,c, and d gives a total of 4 for x. Thus (x−4)2≥1, and
(a+b+c+d)2+(e+f+g+h)2=2(x−4)2+32≥34
A total of 34 can be attained by letting a,b,c, and d be distinct elements in the set {−7,−5,2,13}.
The problems on this page are the property of the MAA's American Mathematics Competitions