Three electrical charges +8m(8m-3n)Q, -3n(8m-3n)Q and +13mnQ are respectively situated at the vertices A, B and C of a triangle ABC with AB=15L; AC=13L and BC < 14L. m and n are positive real numbers such that m > (3*n)/8.

Determine the precise length of the side BC such that when the charge +13mnQ (located at C) is shifted to the circum-centre of the triangle ABC; the Net Electric Potential Energy of the new arrangement is equal to zero.

NOTE:
Definition of Net Electric Potential Energy is given here.

In conformity with the above definition, if Qa,Qb and Qc respectively denote the charges of particles a, b and c and the respective separation between the particles a & b, b & c and a & c are R1, R2 and R3, then the Net Electric Potential Energy (U) of the arrangement is given by:

U = k*(Qa*Qb/R1) + k*(Qb*Qc/R2) + k*(Qa*Qc/R3), where k is Coulomb's constant.

Let O be the circumcenter of triangle ABC and A', B', and C'
the charges at A, B, and C respectively. Then
   
    kA'B'     kB'C'     kC'A'
   ------- + ------- + ------- = 0                      (1)
    |AB|      |BO|      |OA|

If R is the circumradius of triangle ABC, then
   
    A'B'     B'C'     C'A'
   ------ + ------ + ------ = 0                         (2)
     15       R        R

             or

        -15C'(A'+B')
   R = --------------                                   (3)
            A'B'

If A' = 8m(8m-3n), B' = -3n(8m-3n), and C' = 13mn, then

        65
   R = ----                                             (4)
        8

If |AB| = c, |BC| = a, and |CA| = b; then

                      abc
   R = ------------------------------------             (5)
        sqrt([a+b+c][a+b-c][b+c-a][c+a-b])

With b = 13 and c = 15, this becomes

                         13*15a
   R = --------------------------------------------     (6)
        sqrt([a+13+15][a+13-15][13+15-a][15+a-13])

Squaring (4) and (6) and setting them equal gives

    65^2         13^2*15^2*a^2
   ------ = ------------------------                    (7)
    8^2      [a+28][a-2][28-a][a+2]

                 or

   a^4 - 212a^2 + 3136 = 0                              (8)

Therefore, a = 4, -4, 14, or -14. Since 0 < a < 14,

   |BC| = a = 4

Posted by Bractals on 2006-05-17 11:40:50