Question

Prove that the least perimeter of an isosceles triangle in which a circle of radius $r$ can be inscribed, is $6\sqrt{3r}$.

Answer 1

REF.Image

Let ABC be isosceles is AB=AC

r is radius of circle

$A{F}^2+B{F}^2=A{B}^2$

$\Rightarrow (3r{)}^2+(y^2)=x^2$ ________ (1)

From $\Delta AOD$

$(2r{)}^2=r^2+(AD{)}^2$

$=3r^2=A{D}^2$

$=AD=\sqrt{3}r$

Now BD=BF & EC=FC (as tangents are equal from an external point)

AD+DB=x

$=\sqrt{3}r+y^2=x$

$\Rightarrow y^2=x=\sqrt{3}r$ -(2)

From (1) & (2)

$\therefore (3r{)}^2+(x-\sqrt{3}r)=x^2$

$9r^2+x^2+3r^2-2\sqrt{3}rx=x^2$

$=9r^2+3r^2-2\sqrt{3}rx=0$

$=12r^2=2\sqrt{3}rx$

$=6r=\sqrt{3}x$

$\therefore x=\frac{6}{\sqrt{3}}r$

From (2)

$y/2=\left(6/\sqrt{3}\right)r-\sqrt{3}r$

$=y/2=\left(3\sqrt{3}/3\right)r\Rightarrow y=2\sqrt{3}r$

perimeter = $2x+y=2\left(6/\sqrt{3}\right)r+2\sqrt{3}r$

$=(12r+6r)/\sqrt{3}$

$=\left(18/\sqrt{3}\right)r$

$=6\sqrt{3}r$

Hence proved.