Question

The sides of a triangle are given by $x^2+x+1,2x+1,1-x^2$, where $0<x<1$. If the maximum possible angle of the triangle is given by $\theta$ and $\underset{x\to 1}{lim}\cos \frac{\theta }{2}=\frac{a}{b}$, then the least possible value of $a^2+b^2$ is

• A. $5$
• B. $7$
• C. $3$
• D. $0$

Answer 1

The correct option is A $5$

If $x^2+x+1>2x+1$
$\Rightarrow x^2>x$ which is not possible as $0<x<1$
$\therefore 2x+1>x^2+x+1$
And $2x+1>1-x^2$
So, the largest angle will be opposite to side $2x+1$ and we can write,

$\cos \theta =\frac{(1-x^2{)}^2+(x^2+x+1{)}^2-(2x+1{)}^2}{2(1-x^2)(x^2+x+1)}\Rightarrow \cos \theta =\frac{(1-x^2{)}^2+(x^2+x+1-2x-1\left)\right(x^2+x+1+2x+1)}{2(1-x^2)(x^2+x+1)}\Rightarrow \cos \theta =\frac{(1-x^2{)}^2+(x^2-x\left)\right(x^2+3x+2)}{2(1-x^2)(x^2+x+1)}\Rightarrow \cos \theta =\frac{(1-x^2)-\left(x\right)(x+2)}{2(x^2+x+1)}\Rightarrow \cos \theta =\frac{1-2x^2-2x}{2(x^2+x+1)}\Rightarrow \cos \theta =\frac{1-2x^2-2x-2+2}{2(x^2+x+1)}\Rightarrow \cos \theta =\frac{3}{2(x^2+x+1)}-1\Rightarrow 1+\cos \theta =\frac{3}{2(x^2+x+1)}\Rightarrow 2{\cos }^2\frac{\theta }{2}=\frac{3}{2(x^2+x+1)}\Rightarrow \cos \frac{\theta }{2}=\frac{\sqrt{3}}{2(x^2+x+1{)}^{1/2}}$
$\underset{x\to 1}{lim}\cos \frac{\theta }{2}=\frac{1}{2}=\frac{a}{b}\therefore a^2+b^2=1+4=5$