Question

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

• A. Area of the triangle $XYZ$ is $6\sqrt{6}$
• B. The radius of 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)

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

$2S=x+y+z\Rightarrow \frac{S-x}{4}=\frac{s-y}{3}=\frac{S-z}{2}=\lambda$
$S-x=4\lambda$
$S-y=3\lambda$
$S-z=2\lambda$
____________
$S=9\lambda$

Adding all, we get
$S=9\lambda ,x=5\lambda ,y=6\lambda ,=z=7\lambda$

$\pi r^2=\frac{8\pi }{3}\Rightarrow r^2=\frac{8}{3}$

$\Delta =\sqrt{S(S-x)(S-y)(S-z)}$

$\Rightarrow \Delta =\sqrt{9\lambda .4\lambda .3\lambda .2\lambda }=6\sqrt{6}{\lambda }^2$

$R=\frac{xyz}{4\Delta }=\frac{5\lambda .6\lambda .7\lambda }{4.6\sqrt{6}{\lambda }^2}=\frac{35}{4\sqrt{6}}\lambda$

we get $\lambda =1$

(A) $\Delta =6\sqrt{6}$

(B) $R=\frac{35}{4\sqrt{6}}\lambda =\frac{35}{4\sqrt{6}}$

(C) $r=4R\sin \frac{X}{2}\sin \frac{Y}{2}\sin \frac{Z}{2}\Rightarrow \frac{2\sqrt{2}}{\sqrt{3}}=4.\frac{35}{4\sqrt{6}}.\sin \frac{x}{2}\sin \frac{Y}{2}\sin \frac{z}{2}$

$\frac{4}{35}=\sin \frac{x}{2}\sin \frac{y}{2}\sin \frac{z}{2}$

(D) ${\sin }^2\left(\frac{X+Y}{2}\right)={\cos }^2\frac{Z}{2}=\frac{S(S-z)}{xy}=\frac{9.2}{5.6}=\frac{3}{5}$