Very nice, heureka !!!!
No...we have 25 pairs (by my count) ....for example CKJ and IKD is the same pair as IKD and CKJ
To have one real solution the discriminant must = 0 ....so...
(2b)^2 - 4 (1) (a - b) = 0
4b^2 - 4a + 4b = 0
b^2 - a + b = 0
b^2 + b = a
An easier - and more straightforward - way to do this
log x 32 = 5/2
This says that
x^(5/2) = 32 take each side to the (2/5) power
x = 32^(2/5) = [ 32^(1/5) } ^2 = [ 2]^2 = 4
Using the change of base rule we have that
log 32 5
_____ = ___ rearrange as
log x 2
(2/5) log 32 = log x
This says that 10^[ (2/5)log 32 ] = x = 4
Angle CKJ is paired wth angle IKD
Sorry....I don't remember.....!!!
My first answer was incorrect, tom......maybe someone else has another approach/answer
EP is correct!!!!
BTW..... we can verify that s = 14 produces a minimum surface area by taking the derivative of
2s - 5488s^(-2) =
2 + 10976s^(-3)
When s = 14 this is > 0 which means that s = 14 produces a minimum surface area
These angles will have a corresponding pair
AKE , IKA, GKI, DKG , FKD
EKI, AKG, IKD, GKF, DKB
EKG, AKD IKF, GKB , DKJ
EKD , AKF, IKB, GKJ, DKH
EKF, AKB, IKJ, GKH,DKC
I get n^2 = 25, as well, tom......