Question

Question #6a798

Answer 1

`a:b:c=1:1:1`

Explanation:

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)`

Answer 2

`a:b:c=1:1:1`.

`"In fact, "a=b=c=4/sqrt3`.

Explanation:

`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.!