prove that the least perimeter of …

Ans.

ABC is an isosceles triangle with sides AB=AC. O is the center of circle of radius inscribed in the triangle. AD is the altitude,

So OD=OF=OE=r

So BD = CD  ............. (1) ( altitude bisect the side)

BD = BE and CD = CF      .............. (2) ( tangents from the same point are equal in length)

From (1) and (2) 

BD=BE=CD=CF ........ (3)

Similariy AE = AF  ........... (4) (tangents from the same point)

Perimeter of triangle (P) = AB + BC + AC 

(P) = AE + EB + BD + CD + CF+AF

(P) = AE + AF + EB + BD + CD + CF

(P) = 2AE + 4BD    (using 3 and 4)

In Right Triangle OEA, 

$AE = \frac{OE}{Tan x}$

=> $AE = \frac{r}{\tan x}$

=> $AE = r cot x .............. \left(5\right)$

In Right Triangle ABD

BD = AD*tanx 

=> BD = (AO + OD)* tan x

=>$BD = \left(\frac{r}{\sin x} + r\right)\tan x$

=> $BD = r secx + r \tan x ............... \left(6\right)$

From 5 and 6 

=> $\left(P\right) = 2r cotx + 4r(sec x + \tan x)$

=> $\left(P\right) = r ( 2 cotx +\hspace{0.17em}4 sec x+ 4 \tan x)$

Differentiate w.r.t. x

we get 

$\frac{dP}{dx} = r( -2\cos e{c}^{2}x + 4 sec x \tan x + 4 se{c}^{2}x) ........ \left(7\right)$

=> $\frac{dP}{dx} = r( -\frac{2}{{\sin }^{2}x} + 4\frac{\sin x}{{\cos }^{2}x} + \frac{4}{{\cos }^{2}x})$

To find critical value, Put $\frac{dP}{dx} = 0$

=> $r( -\frac{2}{{\sin }^{2}x} + 4\frac{\sin x}{{\cos }^{2}x} + \frac{4}{{\cos }^{2}x}) = 0$

=> $( -\frac{2}{{\sin }^{2}x} + 4\frac{\sin x}{{\cos }^{2}x} + \frac{4}{{\cos }^{2}x}) = 0$

=> $\frac{-2{\cos }^{2}x + 4{\sin }^{3}x + 4{\sin }^{2}x }{{\sin }^{2}x \times {\cos }^{2}x } = 0$

=> $-2(1-{\sin }^{2}x)+ 4{\sin }^{3}x + 4{\sin }^{2}x = 0$

=> $-2+2{\sin }^{2}x+ 4{\sin }^{3}x + 4{\sin }^{2}x = 0$

=> $-2+ 4{\sin }^{3}x + 6{\sin }^{2}x = 0$

=>$2{\sin }^{3}x + 3s{\mathrm{in}}^{2}x - 1 = 0$

=>$(2{\sin }^{3}x + 2s{\mathrm{in}}^{2}x) + ({\sin }^{2}x - 1) = 0$

=>$2{\sin }^{2}x(\sin x + 1) + (\sin x +1) (\sin x - 1) = 0$

=> $(\sin x + 1) (2{\sin }^{2}x + \sin x - 1) = 0$

=> $(\sin x + 1) = 0 or (2{\sin }^{2}x + \sin x - 1) = 0$

=> $\sin x = -1 is not possible because x cant be more than 90° so (2{\sin }^{2}x + \sin x - 1) = 0$

=> $2{\sin }^{2}x + 2\sin x - \sin x - 1 = 0$

=>  $2\sin x( \sin x + 1) - 1( \sin x + 1) = 0$

=> $(2\sin x- 1) ( \sin x + 1) = 0$

=> $(2\sin x- 1) = 0 or ( \sin x + 1) = 0$ 

=> $\sin x = -1 is not possible because x cant be more than 90° so (2\sin x - 1) = 0$

=> $2\sin x- 1= 0$

=> $\sin x = \frac{1}{2}$

=> $x = 30°$

at $\sin x = \frac{1}{2}$ sign changes from -ve to +ve. so it is point of local minima, so at this point value ll be least.

Hence least perimeter is 

 $\left(P\right) = r ( 2 cot 30° +\hspace{0.17em}4 sec 30°+ 4 \tan 30°)$

=> $\left(P\right) = r ( 2 \sqrt{3} +\hspace{0.17em}\frac{4\times 2}{\sqrt{3}}+ 4 \frac{1}{\sqrt{3}})$

=> $\left(P\right) = r \frac{\left( 6+ 8+4\right)}{\sqrt{3}}$

=> $\left(P\right) = r \frac{18}{\sqrt{3}}$

=> $\left(P\right) = 6\sqrt{3}r$

Hence Proved