JEE Advanced 2020 · Paper 1, Mathematics question 10

Half-angle tangents. With semi-perimeter $s$ and inradius $r$, $\tan\frac X2=\frac r{s-x}$ and $\tan\frac Z2=\frac r{s-z}$. So
$$\tan\frac X2+\tan\frac Z2=\frac{r\,y}{(s-x)(s-z)}.$$

The condition. It says this equals $\frac{2y}{2s}=\frac ys$, so $rs=(s-x)(s-z)$.

Finding $Y$. Since $\Delta=rs$ and $\Delta^2=s(s-x)(s-y)(s-z)$:
$$(s-x)(s-z)=s(s-y)\implies\tan^2\frac Y2=\frac{(s-x)(s-z)}{s(s-y)}=1\implies Y=90^\circ.$$

The options.

• (B) $Y=X+Z$. True.
• (A) would need $Y=60^\circ$. False.
• (C) $\tan\frac X2=\frac{\sin X}{1+\cos X}=\frac{x/y}{1+z/y}=\frac x{y+z}$. True.
• (D) $x^2+z^2-y^2=0$, not $xz$. False.

(B), (C)