Question

Let $x, y$ and $z$ be positive real numbers. Suppose $x, y$ and $z$ are the lengths of the sides of a triangle opposite to its angles $X, Y$ and $Z$, respectively. If
$\tan \frac{x}{2}+\tan \frac{z}{2}=\frac{2 y}{x+y+z}$
then which of the following statements is/are TRUE?

• A. $2 Y=X+Z$
• B. $Y=X+Z$
• C. $\tan \frac{X}{2}=\frac{x}{y+z}$
• D. $x^{2}+z^{2}-y^{2}=x z$

Answer 1

Answer: (B), (C)

$\because \tan \frac{x}{2}+\tan \frac{z}{2}=\frac{2 y}{x+y+z} $
let $x+y+z=2 s$
$\sqrt{\frac{(s-y)(s-z)}{s(s-x)}+\sqrt{\frac{(s-x)(s-y)}{s(s-z)}}}=\frac{y}{s}$
$\Rightarrow \sqrt{\frac{s-y}{s}\left(\frac{y}{\sqrt{(s-x)(s-z))}}\right)}=\frac{y}{s}$
$\Rightarrow \sqrt{\frac{s(s-y)}{(s-x)(s-z)}}=1$
$\Rightarrow \tan \frac{Y}{2}=1$
$\Rightarrow Y=\frac{\pi}{2}$

Clearly $Y=X+Z$
and $x^{2}+z^{2}=y^{2}$
Also, $\tan \frac{x}{2}=\sqrt{\frac{1-\cos x}{1+\cos x}}$
$=\sqrt{\frac{1-\frac{z}{y}}{1+\frac{z}{y}}}$
$=\sqrt{\frac{y-z}{y+z}}=\sqrt{\frac{y^{2}-z^{2}}{(y+z)^{2}}}$
$=\frac{x}{y+z}$