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

$\begin{array}{l}tan\frac{X}{2}+tan\frac{Z}{2}=\frac{2y}{x+y+z}\end{array}$

then which of the following statements is/are TRUE

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

Answer 1

Answer: (B), (C)

$\begin{array}{l}\tan \frac{\mathrm{X}}{2}=\frac{\mathrm{x}}{\mathrm{y}+\mathrm{z}}\end{array}$

Explanation for the correct options:

The length of sides of a triangle opposite to it's angles $x,y\&z$ respectively.

Also, $\begin{array}{l}tan\frac{X}{2}+tan\frac{Z}{2}=\frac{2y}{x+y+z}\end{array}$

Now consider the triangle $∆XYZ$

$$\begin{gathered} \Rightarrow \frac{∆}{s\left(s-x\right)}+\frac{∆}{s\left(s-z\right)}=\frac{2y}{2s}\mathbf{[}\mathbf{\because }\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\left(}\frac{\mathbf{A}}{\mathbf{2}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{△}}{\mathbf{s}\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{a}\mathbf{)}}\mathbf{]} \\ \Rightarrow \frac{∆}{s}\frac{\left(s-z\right)+\left(s-x\right)}{\left(s-z\right)\left(s-x\right)}=\frac{y}{s} \\ \Rightarrow \frac{∆\left(2s-x-z\right)}{\left(s-z\right)\left(s-x\right)}=\left(2s-x-z\right)\mathbf{[}\mathbf{\therefore }\mathbf{}\mathbf{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{s}\mathbf{}\mathbf{-}\mathbf{}\mathbf{x}\mathbf{}\mathbf{-}\mathbf{}\mathbf{z}\mathbf{)}\mathbf{]} \\ \Rightarrow \frac{∆}{\left(s-z\right)\left(s-x\right)}=1 \\ \Rightarrow {∆}^2={\left[\left(s-z\right)\left(s-x\right)\right]}^2 \\ \Rightarrow \left[s\left(s-x\right)\left(s-y\right)\left(s-z\right)\right]={\left[\left(s-x\right)\left(s-z\right)\right]}^2 \\ \Rightarrow s\left(s-y\right)=\left(s-x\right)\left(s-z\right) \\ \Rightarrow s^2-sy=s^2-sx-sz+xz \\ \Rightarrow s\left(x-y+z\right)=xz \\ \Rightarrow \frac{1}{2}\left(x+y+z\right)\left(x+y-z\right)=xz \\ \Rightarrow \frac{1}{2}\left[{\left(x+z\right)}^2-y^2\right]=xz \\ \Rightarrow x^2+z^2+2xz-y^2=2xz \\ \Rightarrow x^2+z^2=y^2....\left(i\right) \\ \Rightarrow \angle y=90° \\ \Rightarrow \angle X+\angle y=90° \\ \Rightarrow Y=X+Z \end{gathered}$$

By using equation (i)

$$\begin{gathered} \Rightarrow \tan \frac{x}{2}=\frac{4\times {\displaystyle \frac{1}{2}}xz}{{\left(y+z\right)}^2-x^2}\mathbf{}\left[\because △=\frac{1}{2}xz,4s(s-x)={\left(y+z\right)}^2-x^2\right] \\ \Rightarrow \tan \frac{x}{2}=\frac{2xz}{y^2+z^2+2yz-x^2} \\ \Rightarrow \tan \frac{x}{2}=\frac{2xz}{x^2+z^2+z^2+2yz-x^2} \\ =\frac{2xz}{2z^2+2yz} \\ =\frac{x}{z+y} \\ \Rightarrow \tan \frac{x}{2}=\frac{x}{z+y} \end{gathered}$$

Therefore, $Y=X+Z\&\tan \frac{x}{2}=\frac{x}{z+y}$ both are true.

Hence, option (B) and (C) are the correct answer.