Question

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

$\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=k $ (Let)
$\Rightarrow x=s-4 k, y=s-3 k, z=s-2 k$
Now $2 s=x+y+z=3 s-9 k $
$\therefore k=\frac{s}{9}$
Area of triangle
$\Delta=\sqrt{s(s-x)(s-y)(s-z)}$
$=\sqrt{s \cdot 4 k \cdot 3 k \cdot 2 k}=\sqrt{24 \frac{s^{4}}{729}}$
$=\frac{2}{9} \sqrt{\frac{2}{3}} s^{2}$
Area of incircle $\pi r^{2}=\frac{8 \pi}{3}$
$\therefore r=2 \sqrt{\frac{2}{3}}=\frac{\Delta}{s}=\frac{2 \sqrt{2}}{9 \sqrt{3}} s$
$\therefore s=9 $ and $ k =1$
$\therefore $ Area of triangle $\Delta=\frac{2 \sqrt{2}}{9 \sqrt{3}} \times 81=6 \sqrt{6}$
$R =\frac{ xyz }{4 \Delta}=\frac{5 \times 6 \times 7}{4 \times 6 \sqrt{6}}=\frac{35}{24} \sqrt{6}$
$\sin \frac{ X }{2} \sin \frac{ Y }{2} \sin \frac{ Z }{2}$
$=\sqrt{\frac{( s - y )( s - z )( s - x )( s - z )( s - x )( s - y )}{ yz }}$
$=\frac{(s-x)(s-y)(s-z)}{x y z}$
$=\frac{4 \times 3 \times 2}{5 \times 6 \times 7}=\frac{4}{35}$
$\sin ^{2}\left(\frac{ X + Y }{2}\right)=\frac{1-\cos ( X + Y )}{2}$
$=\frac{1-\cos (\pi-Z)}{2}=\frac{1+\cos Z}{2}$
$=\frac{1}{2}+\frac{1}{2}\left(\frac{x^{2}+y^{2}-z^{2}}{2 x y}\right)$
$=\frac{1}{2}+\frac{1}{2}\left(\frac{25+36-49}{2 \times 5 \times 6}\right) $
$=\frac{6}{10}=\frac{3}{5}$