Question

Given an isosceles triangle with equal sides of length b, base angle $\alpha <\pi /4,$ R, r the radii and O, I the centres of the circumcirlce and incircle, respectively. Then

• A. $R=\frac{1}{2}bsec\alpha$
• B. $\Delta =2b^{2}sin2\alpha$
• C. $r=\frac{bsin2\alpha }{2(1+cos\alpha )}$
• D. $OI=\left|\frac{bcos\left(3\alpha /2\right)}{2sin\alpha cos\left(\alpha /2\right)}\right|$

Answer 1

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

$R=\frac{1}{2}bsec\alpha$
$r=\frac{bsin2\alpha }{2(1+cos\alpha )}$
$OI=\left|\frac{bcos\left(3\alpha /2\right)}{2sin\alpha cos\left(\alpha /2\right)}\right|$

Let ABC be the isosceles triangle with $AB=AC=b$ and $\angle B=\angle C=\alpha$. Let AD be the perpendicular bisector of the side BC. Since $\alpha$ABC is isosceles, AD is also the bisector of angle A, so that O and I both lies on AD. We have $OB=R$ and $KID=r$. Also, since O is the circumcentre, we get $OA=OB=R$. Therefore, from isosceles triangle OAB
$\frac{OB}{sin({90}^o-\alpha )}=\frac{AB}{sin2\alpha }$
$\Rightarrow R=\frac{bcos\alpha }{2sin\alpha cos\alpha }=\frac{1}{2}bcosec\alpha$
so that (a) is correct.
Again $\Delta =BD.AD=bcos\alpha .bsin\alpha =\frac{1}{2}{b}^{2}sin2\alpha$ so that (b) is not correct.
Also$,r=\frac{\Delta }{s}=\frac{\frac{1}{2}{b}^{2}sin2\alpha }{\frac{1}{2}(b+b+2bcos\alpha )}=\frac{bsin2\alpha }{2(1+cos\alpha )}$
so that (c) is correct.
Further $OI=OD+DI|=|OD+r|$
because $\alpha <\pi /4,A>\pi /2$ and O lies on AD produced.
Now, from right-angled triangle ODB, we get
$O{D}^2=O{B}^2-B{D}^2=R^2-(bcos\alpha {)}^2$
$=\frac{1}{4}\frac{b^2}{si{n}^2\alpha }-b^{2}co{s}^2\alpha$
$=\frac{b^2(1-4si{n}^2\alpha co{s}^2\alpha )}{4si{n}^2\alpha }$ [from (a)]
$=\frac{b^2(co{s}^2\alpha -si{n}^2\alpha {)}^2}{4si{n}^2\alpha }$
$=\frac{b^{2}co{s}^{2}2\alpha }{(2sin\alpha {)}^2}$
$\Rightarrow OD=\frac{bcos2\alpha }{2sin\alpha }$
$\therefore OI=|\frac{bsin2\alpha }{2(1+cos\alpha )}+\frac{bcos2\alpha }{2sin\alpha }|$
$=|\frac{bsin2\alpha }{4co{s}^2\left(\alpha /2\right)}+\frac{bcos2\alpha }{4sin\left(\alpha /2\right)cos\left(\alpha /2\right)}|$
$=|\frac{b}{4cos\left(\alpha /2\right)}.\frac{sin2\alpha sin\left(\alpha /2\right)+cos2\alpha cos\left(\alpha /2\right)}{sin\left(\alpha /2\right)cos\left(\alpha /2\right)}|$
$=\left|\frac{bcos\left(3\alpha /2\right)}{2sin\alpha cos\left(\alpha /2\right)}\right|$
Thus, (D) is also correct.