Let a, b, c be the lengths of three sides of a triangle satisfying the condition (a2+b2 )x2-2b(a+c)x+ (b2+c2 )=0. If the set of all possible values of x is the interval ( , ), then 12 ( 2+ 2 ) is equal to .
Answer
36
Solution
(a2x2-2abx+b2 )+ (b2x2-2bcx+c2 )=0 (ax-b)2+(bx-c)2=0
b=ax, c=bx=ax2: sides a, ax, ax2 (x>0)
Triangle inequalities: 1+x>x2 and x+x2>1 (1+x2>x always holds)
x2-x-1<0 and x2+x-1>0
5-12<x< 5+12
12 ( 2+ 2 )=12 (6-2 5)+(6+2 5)4=36
Answer: 36