Question #6a798

Making #b=k_1a# and #c=k_2 a# we have

#a/b=1/k_1#, #b/c=k_1/k_2# and #c/a=k_2#

then

#a+k_1a+k_2a=4sqrt3# and
#a^2+k_1^2a^2+k_2^2a^2=16#

solving

#{(1+k_1+k_2=4sqrt3/a),(1+k_1^2+k_2^2=16/a^2):}#

and considering only positive values for #k_1,k_2# we obtain

#k_1 = (4 sqrt[3] a - a^2 + sqrt[a^2 (8 sqrt[3] a - 3 a^2-16)])/( 2 a^2)#

#k_2 = (4 sqrt[3] a - a^2 - sqrt[a^2 (8 sqrt[3] a - 3 a^2-16)])/( 2 a^2)#

Analyzing the discriminant

#sqrt[a^2 (8 sqrt[3] a - 3 a^2-16)]# and choosing #a# such that

#8 sqrt[3] a - 3 a^2-16 ge 0# to avoid complex solutions, we conclude that the only solution is for #a=4/sqrt(3)# (a double root) so making #a = 4/sqrt(3)# we obtain

#k_1=1# and #k_2=1# so

#a/b=1, b/c=1,c/a=1#

and #a=b=c=4/sqrt(3)#

#a+b+c=4sqrt3 rArr (a+b+c)^2=48#

#rArr a^2+b^2+c^2+2ab+2bc+2ca=48#

#rArr 16+2(ab+bc+ca)=48#

#rArr (ab+bc+ca)=(48-16)/2=16#

Now, #(a-b)^2+(b-c)^2+(c-a)^2#

#=2{(a^2+b^2+c^2)-(ab+bc+ca)}#

#=2{16-16}=0#

#:. (a-b)=(b-c)=(c-a)=0," i.e., to say, "a=b=c#.

Hence, #a:b:c=1:1:1#.

In fact, #a+b+c=4sqrt3, &, a=b=c" (proved)"rArr a=b=c=4/sqrt3#.

Enjoy Maths.!