Question

In a triangle $ABC,$ the minimum value if the sum of the squares of sides is ($\Delta$ is the area of the triangle $ABC$)

• A. $3\sqrt{3}\Delta$
• B. $4\sqrt{3}\Delta$
• C. $2\sqrt{3}\Delta$
• D. $5\sqrt{3}\Delta$

Answer 1

The correct option is B $4\sqrt{3}\Delta$

Let the sides $a,b,c$
$3(a^2{b}^2{c}^2{)}^{1/3}\le a^2+b^2+c^2\le 4R^2({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C\left)3\right(a^2{b}^2{c}^2{)}^{1/3}\le a^2+b^2+c^2\le 9R^{2}3(a^2{b}^2{c}^2{)}^{1/3}\le a^2+b^2+c^2\le 9\left(\frac{a^2{b}^2{c}^2}{16{\Delta }^2}\right)$
The minimum value will occur when,
$3(a^2{b}^2{c}^2{)}^{1/3}=9\left(\frac{a^2{b}^2{c}^2}{16{\Delta }^2}\right)\Rightarrow 3(abc{)}^{2/3}=\frac{9(abc{)}^2}{16{\Delta }^2}\Rightarrow abc={\left(\frac{16{\Delta }^2}{3}\right)}^{3/4}$

So the minimum value is,
$a^2+b^2+c^2=3(abc{)}^{2/3}\Rightarrow a^2+b^2+c^2=3{\left(\frac{16{\Delta }^2}{3}\right)}^{1/2}\Rightarrow a^2+b^2+c^2=4\sqrt{3}\Delta$