An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. Show that the area of the triangle is maximum when .

Consider the circle with centre O and radius a and the isosceles triangle as ABC having AB = AC and ∠BAC = 2θ

BC is the chord for circle and angle subtended by chord at centre is double the angle subtended by same chord at a point on circle

Hence ∠BOC = 4θ

But as the altitude from vertex A for isosceles triangle is also the angle bisector and O lies on the line AD hence OD is also the angle bisector for ∠BOC

Hence ∠BOD = 2θ

Now area of triangle is given by 1/2 × base × height

⇒ A = 1/2 × BC × AD

As AD is also the perpendicular bisector hence BD = DC hence BC = BD + DC = 2BD

And from figure AD = AO + OD

Hence the area becomes

⇒ A = 1/2 × 2BD × (AO + OD)

⇒ A = BD × (a + OD)

Consider ΔOBD

Put these values of BD and OD in area A

⇒ A = asin2θ × (a + acos2θ)

⇒ A = a²(sin2θ + sin2θcos2θ)

We know that sin2x = 2sinxcosx

⇒ A = a²(sin2θ + 1/2(sin4θ))

We have to find θ such that A is maximum hence we have to first find the critical points where

Differentiate A with respect to θ and equate to 0

⇒ 0 = a²(2cos2θ + 2cos4θ)

⇒ cos2θ + cos4θ = 0

We know that cos2x = 2cos²x – 1

⇒ cos2θ + 2cos²2θ – 1 = 0

⇒ 2cos²2θ + cos2θ – 1 = 0

⇒ 2cos²2θ + 2cos2θ - cos2θ – 1 = 0

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

⇒ (2cos2θ – 1)(cos2θ + 1) = 0

⇒ cos2θ = 1/2 or cos2θ = -1

We have to prove now that is a point of maxima

For to be point of maxima

Differentiate (i)

Put

Now a² is positive because of square hence

Hence is a point of maxima

Hence area of triangle is maximum when