Answer :
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
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