Q68 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

If $I$ is the incentre of $\triangle A B C$ and $P_1, P_2, P_3$ are respectively the radii of the circumcircles of the $\triangle I B C, \triangle I C A$ and $\triangle I A B$, then $P_1 P_2 P_3=$

  • A
    $2 R r$
  • B
    $2 R r^2$
  • Correct answer
    C
    $2 R^2 r$
  • D
    $\frac{4 R}{r}$

Solution

In $\triangle I B C, \angle B I C=\frac{\pi}{2}-\frac{A}{2}$, let circumcentre of $\triangle I B C$ is $C_1$, then $\angle B C_1 C=\frac{\pi}{2}+\frac{A}{2}$ So, $\quad \frac{a / 2}{P_1}=\sin \left(\frac{\pi}{2}+\frac{A}{2}\right)=\cos \frac{A}{2}$ $$ \Rightarrow \quad P_1=\frac{a}{2 \cos \frac{A}{2}} $$ Similarly, $\quad P_2=\frac{b}{2 \cos \frac{B}{2}}$ and $P_3=\frac{c}{2 \cos \frac{C}{2}}$ So, $\quad P_1 P_2 P_3=\frac{a b c}{8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}$ $$ =\frac{a b c \times \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{\sin A \sin B \sin C}=(2 R)\left(2 R^{\prime}\right)(2 R) \frac{r}{4 R}=2 R^2 r . $$

Hence, the correct answer is option: 'C'.

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