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  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

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