Question

Let ABC 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 $(IO{)}^2=R^2-2Rr$. Further show that the triangle BIO is right angled triangle if and only if b is the arithmetic mean of a and c.

Answer 1

We know the following result from Trigonometry,

$IO=R\sqrt{1-8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}$ ..... (1)

and $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$ ..... (2)

$(IO)^2 \, = \, R^2 \, - \, 8R^2 \, \sin \dfrac{A}{2} \, \sin \dfrac{B}{2} \, \sin

\dfrac{C}{2}$

= $R^2$ - 2R(r) = $R^2$ - 2Rr, by (1) and (2).

2nd part :

For $\Delta$ BIO to be right angled triangled, we have

$B{I}^2+I{O}^2=B{O}^2$

From part (i) above

$BI=4R\sin \frac{A}{2}\sin \frac{C}{2}and(IO{)}^2=R^2-2Rr$

$\therefore 16R^{2}si{n}^2\left(A/2\right){\sin }^2\left(C/2\right)+R^2-2R.\left(4R\sin \right(A/2\left)\sin \right(B/2\left)\sin \right(C/2\left)\right)=R^2$

or $2\sin \left(A/2\right)\sin \left(C/2\right)=\sin \left(B/2\right))=R^2$

Now putting the values of \sin (A/2) etc . in terms of sides . we get

$\frac{2(s-b)}{b}=1$ or $2s=3b$ or $a+c=2b$

i.e.b is A.M. of a and c.