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 $△$ 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
$AI=\frac{IF}{sin(A/2)}$
$\because △$ 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 - $(90^{\circ }-B)=\frac{1}{2}(A+2B-180^{\circ })$
= $\frac{1}{2}(A+2B-A-B-C)=\frac{1}{2}(B-C)$
In $△OAI,O{I}^2=O{A}^2+A{I}^2-2(OA)(AI)cos(\angle IAO)$
= $R^2+[4Rsin(B/2)sin(C/2){]}^2$
$-2R.[4Rsin(B/2)sin(C/2)cos(\frac{B-C}{2})$
= [ $R^2+16R^{2}si{n}^2(B/2)si{n}^2(C/2)$
- 8 $R^{2}sin(B/2)sin(C/2)cos\left(\frac{B-C}{2}\right)]$
= $R^2[1+16si{n}^2(B/2)si{n}^2(C/2)$
- 8 sin (B/2) sin (C/2) cos $\left(\frac{B-C}{2}\right)]$
= $R^2[1+8sin(B/2)sin(C/2)$
$\{2sin(B/2)sin(C/2)-cos(\frac{B-C}{2}\left)\right\}]$
= $R^2[1+8sin(B/2)sin(C/2)$
$\left\{cos\right(\frac{B-C}{2})+cos(\frac{B+C}{2})-cos(\frac{B-C}{2}\left)\right\}]$
= $R^2[1-8sin(B/2)sin(C/2cos\left(\frac{B+C}{2}\right)]$
= $R^2[1-8sin(B/2)sin(C/2)cos(\frac{\pi }{2}-\frac{A}{2}\left)\right]$
$[\because \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi }{2}]$
= $R^2[1-8sin(A/2)sin(B/2)sin(C/2)]$
= $R^2[1-8(\frac{r}{4R}\left)\right]=R^2-2Rr$
Now, in right angled $△BIO$,
$O{B}^2=B{I}^2+I{O}^2$
$\Rightarrow R^2=B{I}^2+R^{@}2-2Rr$
$\Rightarrow 2Rr=B{I}^2$
$\Rightarrow 2Rr=r^2/si{n}^2(B/2)$
$\Rightarrow 2Rr=r/si{n}^2(B/2)$
$\Rightarrow 2Rsi{n}^2\frac{B}{2}=r$
$\Rightarrow R(1-cosB)=r$
$\Rightarrow \frac{abc}{4△}(1-cosB)=\frac{△}{8}$
$\Rightarrow abc(1-cosB)=\frac{4{△}^2}{8}$
$\Rightarrow abc[1-\frac{a^2+c^2-b^2}{2ac}]=\frac{4{△}^2}{s}$
$\Rightarrow abc\left[\frac{2ac-a^2-c^2+b^2}{2ac}\right]=\frac{4△}{s}$
$\Rightarrow b[b^2-(a-c{)}^2]=\frac{4{△}^2}{s}$
$\Rightarrow b[b^2-(a-c{)}^2]=8(s-a)(s-b)(s-c)$
$\Rightarrow b[\{b-(a-c)\}\{b+(a-c)\}]$
= 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.