Question

In a triangle $XYZ$, let $x,y,z,$ be the lengths of sides opposite to the angles $X,Y,Z,$ respectively, and $2s=x+y+z$. If $\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}$ and the area of the incircle of the triangle $XYZ$ is $\frac{8\pi }{3},$ then:

• A. Area of the triangle $XYZ$ is $6\sqrt{6}$
• B. The radius of the circumcircle of the triangle $XYZ$ is $\frac{35}{6}\sqrt{6}$
• C. $\sin \frac{X}{2}\sin \frac{Y}{2}\sin \frac{Z}{2}=\frac{4}{35}$
• D. ${\sin }^2\left(\frac{X+Y}{2}\right)=\frac{3}{5}$

Answer 1

Answer: (A), (C), (D)

${\sin }^2\left(\frac{X+Y}{2}\right)=\frac{3}{5}$

Let $\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=k$
$\Rightarrow s-x=4k,s-y=3k,s-z=2k$
$\Rightarrow s=9k\left(\text{add the above equations}\right)$
$x=5k,y=6k,z=7k$

Area of the incircle $=\frac{8\pi }{3}$
$\Rightarrow r^2=\frac{8}{3}$
$\frac{{\Delta }^2}{s^2}=\frac{8}{3}(\because r=\frac{\Delta }{s})$
$\Rightarrow k=1(\text{put}\Delta =\sqrt{s(s-x)(s-y)(s-z)})$

$s=9,x=5,y=6,z=7$
$\Delta =\sqrt{s(s-x)(s-y)(s-z)}$
$\Rightarrow \Delta =6\sqrt{6}$

$R=\frac{xyz}{4\Delta }$
$\Rightarrow R=\frac{35}{4\sqrt{6}}$

$r=4R\sin \frac{X}{2}\sin \frac{Y}{2}\sin \frac{Z}{2}$
$\Rightarrow \sin \frac{X}{2}\sin \frac{Y}{2}\sin \frac{Z}{2}=\frac{4}{35}$

${\sin }^2\left(\frac{X+Y}{2}\right)={\cos }^2\frac{Z}{2}=\frac{s(s-z)}{xy}=\frac{3}{5}$