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