JEE Advanced 2019 · Paper 1, Mathematics question 8

Angles. By the sine rule with circumradius 1:
$$\sin P=\frac{\sqrt3}2,\qquad\sin Q=\frac12.$$
$P=60^\circ$ would make $R=90^\circ$, which is excluded. So $P=120^\circ$ and $Q=R=30^\circ$, giving $r=q=1$.

(A) Median.
$$RS^2=\frac{2p^2+2q^2-r^2}4=\frac74\implies RS=\frac{\sqrt7}2.$$
True.

Coordinates. Put $Q=(0,0)$, $R=(\sqrt3,0)$ and $P=\left(\frac{\sqrt3}2,\frac12\right)$. Then $E=\left(\frac{\sqrt3}2,0\right)$ and $S=\left(\frac{\sqrt3}4,\frac14\right)$. The line RS meets $x=\frac{\sqrt3}2$ at $O=\left(\frac{\sqrt3}2,\frac16\right)$.

(C) $OE=\frac16$. True.

(B) Area $SOE=\frac12\cdot\frac16\cdot\frac{\sqrt3}4=\frac{\sqrt3}{48}$, not $\frac{\sqrt3}{12}$. False.

(D) Inradius. $\Delta=\frac12\sin120^\circ=\frac{\sqrt3}4$ and $s=\frac{2+\sqrt3}2$, so
$$r=\frac\Delta s=\frac{\sqrt3}{2(2+\sqrt3)}=\frac{\sqrt3}2(2-\sqrt3).$$
True.

(A), (C), (D)