Problem:
Let a,b, and c be real numbers such that
{a+b+c=2, and a2+b2+c2=12
What is the difference between the maximum and minimum possible values of c ?
Answer Choices:
A. 2
B. 310
C. 4
D. 316
E. 320
Solution:
From the equations, a+b=2−c and a2+b2=12−c2. Let x be an arbitrary real number, then (x−a)2+(x−b)2≥0; that is, 2x2−2(a+b)x+(a2+b2)≥0. Thus
2x2−2(2−c)x+(12−c2)≥0
for all real values x. That means the discriminant 4(2−c)2−4⋅2(12−c2)≤0. Simplifying and factoring gives (3c−10)(c+2)≤0. So the range of values of c is −2≤c≤310. Both maximum and minimum are attainable by letting (a,b,c)= (2,2,−2) and (a,b,c)=(−32,−32,310). Therefore the difference between the maximum and minimum possible values of c is 310−(−2)=316.
The problems on this page are the property of the MAA's American Mathematics Competitions