# Question 6a798

Jan 3, 2017

$a : b : c = 1 : 1 : 1$

#### Explanation:

Making $b = {k}_{1} a$ and $c = {k}_{2} a$ we have

$\frac{a}{b} = \frac{1}{k} _ 1$, $\frac{b}{c} = {k}_{1} / {k}_{2}$ and $\frac{c}{a} = {k}_{2}$

then

$a + {k}_{1} a + {k}_{2} a = 4 \sqrt{3}$ and
${a}^{2} + {k}_{1}^{2} {a}^{2} + {k}_{2}^{2} {a}^{2} = 16$

solving

$\left\{\begin{matrix}1 + {k}_{1} + {k}_{2} = 4 \frac{\sqrt{3}}{a} \\ 1 + {k}_{1}^{2} + {k}_{2}^{2} = \frac{16}{a} ^ 2\end{matrix}\right.$

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

$\sqrt{{a}^{2} \left(8 \sqrt{3} a - 3 {a}^{2} - 16\right)}$ 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 $a = \frac{4}{\sqrt{3}}$ (a double root) so making $a = \frac{4}{\sqrt{3}}$ we obtain

${k}_{1} = 1$ and ${k}_{2} = 1$ so

$\frac{a}{b} = 1 , \frac{b}{c} = 1 , \frac{c}{a} = 1$

and $a = b = c = \frac{4}{\sqrt{3}}$

Jan 30, 2017

$a : b : c = 1 : 1 : 1$.

$\text{In fact, } a = b = c = \frac{4}{\sqrt{3}}$.

#### Explanation:

$a + b + c = 4 \sqrt{3} \Rightarrow {\left(a + b + c\right)}^{2} = 48$

$\Rightarrow {a}^{2} + {b}^{2} + {c}^{2} + 2 a b + 2 b c + 2 c a = 48$

$\Rightarrow 16 + 2 \left(a b + b c + c a\right) = 48$

$\Rightarrow \left(a b + b c + c a\right) = \frac{48 - 16}{2} = 16$

Now, ${\left(a - b\right)}^{2} + {\left(b - c\right)}^{2} + {\left(c - a\right)}^{2}$

$= 2 \left\{\left({a}^{2} + {b}^{2} + {c}^{2}\right) - \left(a b + b c + c a\right)\right\}$

$= 2 \left\{16 - 16\right\} = 0$

$\therefore \left(a - b\right) = \left(b - c\right) = \left(c - a\right) = 0 , \text{ i.e., to say, } a = b = c$.

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.!