Triangles and series

Question

a, b, c are sides of a triangle, and are in GP loga-log2b, log2b-log3c, and log3c-loga are in AP, then prove that angle A must be obtuse

Answer 1

b/a = c/b  b^2 = a * c 

(log(2b) - log(3c)) - (log(a) - log(2b)) = (log(3c) - log(a)) - (log(2b) - log(3c))  log(2b) + log(2b) - log(3c) - log(a) = log(3c) + log(3c) - log(a) - log(2b)  2 * log(2b) - log(3c) - log(a) = -log(2b) + 2 * log(3c) - log(a)  3 * log(2b) = 3 * log(3c)  log(2b) = log(3c)  2b = 3c  c = (2/3) * b 

c = (2/3) * b  b = (2/3) * a 

a , (2/3) * a , (4/9) * a 

1 , 2/3 , 4/9 

(largest angle will be opposite the side with length of 1)

1^2 = (2/3)^2 + (4/9)^2 - 2 * (2/3) * (4/9) * cos(A)  1 = 4/9 + 16/81 - (16/27) * cos(A)  81/81 = 36/81 + 16/81 - (48/81) * cos(A)  81 = 52 - 48 * cos(A)  29 = -48 * cos(A)  -29/48 = cos(A) 

i.e cos(A) < 0, hence A > 90 degrees 

Answer 2

I have used the cosine formula to prove angle A is obtuse. Hope it helps.