Question

Suppose $a, b, c$ are the lengths of three sides of a $\triangle A B C, a>b>c, 2 b=a+c$ and $b$ is a positive integer. If $a^2+b^2+c^2=84$, then value of $b$ is

• A. 7
• B. 6
• C. 5
• D. 4

Answer 1

Answer: (C)

We have
$a c =\frac{1}{2}\left[(a+c)^2-\left(a^2+c^2\right)\right]=\frac{1}{2}\left[4 b^2-\left(84-b^2\right)\right] $
$ =5 b^2 / 2-42$
Thus, $a$ and $c$ are the roots of the equation
$x^2-2 b x+\left(5 b^2 / 2-42\right)=0$
As $a$ and $c$ are distinct real numbers the discriminant of the above must be positive, that is, $4 b^2-4\left(5 b^2 / 2-42\right)>0$
$\Rightarrow 6 b^2 <168 \text { or } b^2< 28 . $
$\text { Also, } a c >0 \Rightarrow 5 b^2 >84 .$
$\therefore 84 / 5 < b^2< 28 .$
As $b$ is a positive integer, we get $b=5$.