Properties of Triangles 3 Question 18
18. Let $A B C$ be a triangle having $O$ and $I$ as its circumcentre and incentre, respectively. If $R$ and $r$ are the circumradius and the inradius respectively, then prove that $(I O)^{2}=R^{2}-2 R r$. Further show that the $\triangle B I O$ is a right angled triangle if and only if $b$ is the arithmetic mean of $a$ and $c$.
(1999, 10M)
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Solution:
- It is clear from the figure that, $O A=R$
$$ A I=\frac{I F}{\sin (A / 2)} $$
$\because \triangle A I F$ is right angled triangle, so $=\frac{r}{\sin (A / 2)}$
$$ \begin{array}{rlrl} & \text { But } & r & =4 R \sin (A / 2) \sin (B / 2) \sin (C / 2) \\ \therefore & A I & =4 R \sin (B / 2) \sin (C / 2) \\ \text { Again, } & \angle G O A & =B \Rightarrow O A G=90^{\circ}-B \\ \text { Therefore, } \angle I A O & =\angle I A C-\angle O A C \\ & =A / 2-\left(90^{\circ}-B\right)=\frac{1}{2}\left(A+2 B-180^{\circ}\right) \\ & =\frac{1}{2}(A+2 B-A-B-C)=\frac{1}{2}(B-C) \end{array} $$
In $\triangle O A I, O I^{2}=O A^{2}+A I^{2}-2(O A)(A I) \cos (\angle I A O)$
$$ \begin{aligned} & =R^{2}+[4 R \sin (B / 2) \sin (C / 2)]^{2} \\ & \quad-2 R \cdot[4 R \sin (B / 2) \sin (C / 2)] \cos \frac{B-C}{2} \\ & =\left[R^{2}+16 R^{2} \sin ^{2}(B / 2) \sin ^{2}(C / 2)\right. \\ & \quad-8 R^{2} \sin (B / 2) \sin (C / 2) \cos \frac{B-C}{2} \\ & =R^{2}\left[1+16 \sin ^{2}(B / 2) \sin ^{2}(C / 2)\right. \\ & \quad-8 \sin (B / 2) \sin (C / 2) \cos \frac{B-C}{2} \\ & =R^{2}[1+8 \sin (B / 2) \sin (C / 2) \\ & \quad 2 \sin (B / 2) \sin (C / 2)-\cos \frac{B-C}{2} \end{aligned} $$
$$ \begin{aligned} & =R^{2}[1+8 \sin (B / 2) \sin (C / 2) \\ & \quad \cos \frac{B-C}{2}+\cos \frac{B+C}{2}-\cos \frac{B-C}{2} \\ & =R^{2} 1-8 \sin (B / 2) \sin (C / 2) \cos \frac{B+C}{2} \\ & =R^{2} 1-8 \sin (B / 2) \sin (C / 2) \cos \frac{\pi}{2}-\frac{A}{2} \\ & \because \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2} \\ & =R^{2}[1-8 \sin (A / 2) \sin (B / 2) \sin (C / 2)] \\ & =R^{2} 1-8 \frac{r}{4 R}=R^{2}-2 R r \end{aligned} $$
Now, in right angled $\triangle B I O$,
$$ \begin{aligned} & O B^{2}=B I^{2}+I O^{2} \\ & \Rightarrow \quad R^{2}=B I^{2}+R^{2}-2 R r \\ & \Rightarrow \quad 2 R r=B I^{2} \\ & \Rightarrow \quad 2 R r=r^{2} / \sin ^{2}(B / 2) \\ & \Rightarrow \quad 2 R=r / \sin ^{2}(B / 2) \\ & \Rightarrow \quad 2 R \sin ^{2} B / 2=r \\ & \Rightarrow \quad R(1-\cos B)=r \\ & \Rightarrow \quad \frac{a b c}{4 \Delta}(1-\cos B)=\frac{\Delta}{s} \\ & \Rightarrow \quad a b c(1-\cos B)=\frac{4 \Delta^{2}}{s} \\ & \Rightarrow \quad a b c 1-\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{4 \Delta^{2}}{s} \\ & \Rightarrow \quad a b c \frac{2 a c-a^{2}-c^{2}+b^{2}}{2 a c}=\frac{4 \Delta^{2}}{s} \end{aligned} $$
$$ \begin{array}{cc} \Rightarrow & b\left[b^{2}-(a-c)^{2}\right]=\frac{4 \Delta^{2}}{s} \\ \Rightarrow & b\left[b^{2}-(a-c)^{2}\right]=8(s-a)(s-b)(s-c) \\ \Rightarrow & b[{b-(a-c)}{b+(a-c)}] \\ \Rightarrow & =8(s-a)(s-b)(s-c) \\ \Rightarrow & b[(b+c-a)(b+a-c)]=8(s-a)(s-b)(s-c) \\ \Rightarrow & b[(2 s-2 a)(2 s-2 c)]=8(s-a)(s-b)(s-c) \\ \Rightarrow & b[2 \cdot 2(s-a)(s-c)]=8(s-a)(s-b)(s-c) \\ \Rightarrow & b=2 s-2 b \\ \Rightarrow & 2 b=a+c \end{array} $$
which shows that $b$ is arithmetic mean between $a$ and $c$.
