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 - 2 Rr$. Further show that
the $\triangle$ BIO is a right angled triangle if and only if b is
the arithmetic mean of a and c.

Answer 1

It is clear from the figure that, OA = R
$ AI = \frac{ IF}{ sin \, (A/2)} $
$\because \triangle $ AIF is right angled triangle, so = $ \frac{ r}{ sin \, (A / 2)} $
But r = 4R sin (A /2) sin ( B /2) sin (C/2)
$\because $ AI = 4 R sin (B / 2) sin (C / 2)
Again, $ \angle GOA = B \Rightarrow OAG = 90^\circ - B $
Therefore, $ \angle IAO = \angle IAC - \angle OAC $
= A/2 - $ (90^\circ - B) = \frac{1}{2} (A + 2 B - 180^\circ) $
= $ \frac{1}{2} (A + 2 B - A - B - C ) = \frac{1}{2} (B - C)$
In $ \triangle OAI, \, OI^2 = OA^2 + AI^2 - 2 (OA) \, (AI) \, cos \, (\angle IAO)$
= $ R^2 + [ 4 R \, sin \, (B/2) \, sin \, (C/2)]^2 $
$ - 2 R. [ 4 R \, sin \, (B / 2) \, sin (C/2) \, cos \, \bigg( \frac{ B - C }{ 2} \bigg)$
= [ $ R^2 + 16 R^2 \, sin^2 \, (B/2) \, sin^2 (C/2) $
- 8 $ R^2 \, sin \, (B/2) \, sin \, (C/2) \, cos \bigg( \frac{ B - C }{ 2} \bigg) \bigg ] $
= $ R^2 [ 1 + 16 \, sin^2 \, (B/2) \, sin^2 (C/2) $
- 8 sin (B/2) sin (C/2) cos $ \bigg( \frac{ B - C }{ 2} \bigg) \bigg ] $
= $ R^2 [ 1 + 8 \, sin \, (B/2) \, sin (C/2)$
$\Bigg \{ 2 \, sin \, (B/2) \, sin \, (C/2) - cos \bigg( \frac{ B - C }{ 2} \bigg) \Bigg \} \Bigg ] $
= $ R^2 [ 1 + 8 \, sin \, (B/2) \, sin \, (C/2) $
$ \bigg \{ cos \bigg( \frac{ B - C }{ 2} \bigg) + cos \, \bigg( \frac{ B + C }{ 2} \bigg) - \, cos \bigg( \frac{ B - C }{ 2} \bigg) \bigg \} \bigg ] $
= $ R^2 \bigg [ 1 - 8 \, sin \, (B/2) \, sin \, (C/2 cos \bigg( \frac{ B + C }{ 2} \bigg) \bigg ] $
= $ R^2 \bigg [ 1 - 8 \, sin \, (B/2) sin \, (C/2) \, cos \, \bigg( \frac{ \pi}{ 2 } - \frac{ A }{ 2} \bigg) \bigg ] $
$$ $ \bigg [ \because \frac{A}{2} + \frac{B}{ 2} + \frac{C}{2} = \frac{\pi}{2} \bigg ] $
= $ R^2 [ 1 - 8 sin \, (A/ 2) \, sin \, (B/2) \, sin \, (C/ 2) ] $
= $ R^2 \bigg [ 1 - 8 \bigg( \frac{ r}{ 4 R} \bigg) \bigg ] = R^2 - 2Rr $
Now, in right angled $ \triangle BIO$,
$ OB^2 = BI^2 + IO^2 $
$\Rightarrow R^2 = BI^2 + R^@2 - 2 Rr$
$\Rightarrow 2 Rr = BI^2 $
$\Rightarrow 2 Rr = r^2 / sin^2 \, (B/2)$
$\Rightarrow 2 Rr = r / sin^2 (B/2)$
$\Rightarrow 2Rsin^2\frac{B}{2}= r $
$\Rightarrow R(1-cosB) = r $
$\Rightarrow \frac{abc}{4 \triangle}(1-cos \, B)= \frac{\triangle}{8} $
$\Rightarrow abc(1 - \, cos B) = \frac{4 \, \triangle^2}{8} $
$\Rightarrow abc \Bigg[1- \frac{a^2 + c^2 - b^2}{ 2ac } \bigg ] = \frac{ 4 \triangle^2 }{ s} $
$\Rightarrow abc \bigg [ \frac{ 2ac - a^2 - c^2 + b^2 }{ 2ac} \bigg ] = \frac{ 4 \triangle}{s} $
$\Rightarrow b [ b^2 - (a - c)^2 ] = \frac{ 4 \, \triangle^2 }{ s} $
$\Rightarrow b [ b^2 - (a - c)^2 ] = 8 ( s - a) \, (s - b) \, ( s - c) $
$\Rightarrow b [ \{ b - (a - c) \} \{ b + (a - c) \} ] $
$ $ = 8 (s - a) (s - b) (s - c) $\Rightarrow $ b [ (b + c - a ) ( b + a - c) ] = 8 (s - a) (s - b) (s - c)
$\Rightarrow $ b [(2s - 2a)(2s - 2c)] = 8(s - a)(s - b)(s - c)
$\Rightarrow $ b [2 . 2 (s - a)(s - c)] = 8(s - a)(s - b)(s - c)
$\Rightarrow b = 2s - 2b \Rightarrow \, 2b = a + c $
which shows that b is arithmetic mean between a and c.