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
$\tan \frac{X}{2}+\tan \frac{Z}{2}=\frac{2y}{x+y+z}$,
then which of the following statements is/are TRUE?

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

Answer 1

Answer: (B), (C)

$\tan \frac{X}{2}=\frac{x}{y+z}$


$\tan \frac{X}{2}+\tan \frac{Z}{2}=\frac{2y}{x+y+z}$
$\Rightarrow \frac{\Delta }{S(S-x)}+\frac{\Delta }{S(S-z)}=\frac{2y}{2S}$
$\Rightarrow \frac{\Delta }{S}\left(\frac{2S-(x+z)}{(S-x)(S-z)}\right)=\frac{y}{S}$
$\Rightarrow \frac{\Delta \cdot y}{S(S-x)(S-z)}=\frac{y}{S}$
$\Rightarrow {\Delta }^2=(S-x{)}^2(S-z{)}^2$
$\Rightarrow S(S-y)=(S-x)(S-z)$
$\Rightarrow (x+y+z)(x+z-y)=(y+z-x)(x+y-z)$
$\Rightarrow (x+z{)}^2-y^2=y^2-(z-x{)}^2$
$\Rightarrow (x+z{)}^2+(x-z{)}^2=2y^2$
$\Rightarrow x^2+z^2=y^2\Rightarrow \angle Y=\frac{\pi }{2}$
$\Rightarrow \angle Y=\angle X+\angle Z$
$\tan \frac{X}{2}=\frac{\Delta }{S(S-x)}$
$\tan \frac{X}{2}=\frac{\frac{1}{2}xz}{\frac{(y+z{)}^2-x^2}{4}}$
$\tan \frac{X}{2}=\frac{2xz}{y^2+z^2+2yz-x^2}$
$\tan \frac{X}{2}=\frac{2xz}{2z^2+2yz}$
(using $y^2=x^2+z^2)$
$\tan \frac{X}{2}=\frac{x}{y+z}$