Prove that the le

Let ABC be the isosceles triangle with AB = AC and the circle be inscribed in it of radius r and centre O

The centre of incircle is intersections of altitudes and in isosceles triangle the altitude from vertex A is also the bisector hence AF bisects BC at F hence BF = FC

Now we require some parameter based on which the perimeter of triangle will change

Let that parameter be the angle θ as shown

We have to find a θ such that the perimeter of ΔABC is minimum and also the minimum perimeter

let P be the perimeter

P = AB + BC + AC

⇒ P = 2AB + BC

From figure BC = BF + FC but BF = FC hence BC = 2BF

⇒ P = 2AB + 2BF

Consider ΔABF

Hence

⇒ P = 2AB + 2(ABsinθ)

⇒ P = 2AB(1 + sinθ) …(a)

Again consider ΔABF

But from figure AF = AO + OF and OF is radius r hence AF = AO + r

Now consider ΔADO

But OD is the radius r hence

Put in (b)

Put this value of AB in (a)

We know that sin2θ = 2sinθcosθ

Now we have found P in terms of theta, to find minimum value of P we have to first find the critical points where

Differentiate P with respect to θ and equate to 0

Differentiating using the rule which says that

Here u = (1 + sinθ)² and v = sin2θ

As θ is the angle of triangle shown it has to be greater than 0 and less than 90° hence sin2θ is not 0

Hence

⇒ 0 = 2(1 + sinθ)cosθsin2θ – 2(1 + sinθ)²cos2θ

⇒ cosθsin2θ = (1 + sinθ)cos2θ

Since sin2θ = 2sinθcosθ and cos2θ = 1 – 2sin²θ

⇒ 2sinθcos²θ = (1 + sinθ)(1 – 2sin²θ)

Since sin²θ + cos²θ = 1

⇒ 2sinθ(1 – sin²θ) = (1 + sinθ)(1 – 2sin²θ)

⇒ 2sinθ(1 + sinθ)(1 – sinθ) = (1 + sinθ)(1 – 2sin²θ)

⇒ 2sinθ(1 – sinθ) = 1 – 2sin²θ

⇒ 2sinθ – 2sin²θ = 1 – 2sin²θ

⇒ 2sinθ = 1

⇒ sinθ = 1/2

⇒ θ = 30°

Consider θ = 30° on number line we start with negative from left and for each point on which we got we change the sign. Here we have only one point 30° on which the sign for changes from minus to plus hence 30° is point of minima

Put θ = 30° in P to get the minimum perimeter

⇒ P = 6√3r

Hence proved the minimum perimeter is 6√3r