Question

If $I$ is the incentre of the triangle $ABC$; $P_1,P_2$ and $P_3$ are the radii of the circumcircle of the triangles $IBC,ICA$ and $IAB$ respectively, then

• A. $2P_1=asec\left(A/2\right)$
• B. $2P_2=bsec\left(B/2\right)$
• C. $2P_3=csec\left(C/2\right)$
• D. $P_1{P}_2{P}_3=2R^{2}r$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $2P_1=asec\left(A/2\right)$
B $2P_2=bsec\left(B/2\right)$
C $2P_3=csec\left(C/2\right)$
D $P_1{P}_2{P}_3=2R^{2}r$

In $△IBC$
$\angle IBC=\frac{B}{2}$ and $\angle ICB=\frac{C}{2}$
$\Rightarrow \angle BIC=\pi -\left(\frac{B+C}{2}\right)$
From sine rule: $\frac{BC}{\sin \angle BIC}=2P_1$
$\Rightarrow 2P_1=\frac{a}{\sin \left(\frac{B+C}{2}\right)}=a\sec \frac{A}{2}$
Similarly in $\Delta ICA$ and $\Delta IAB$, we get $2P_2=b\sec \frac{B}{2}$ and $2P_3=c\sec \frac{C}{2}$ respectively.
Now, $P_1{P}_2{P}_3=\frac{abc}{8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}=\frac{R^3\sin A\sin B\sin C}{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}$
$\Rightarrow P_1{P}_2{P}_3=2R^2\left(4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\right)=2R^{2}r$
Ans: A,B,C,D