Expert: Alon Mandes - 11/12/2009

Question
An isosceles triangle is inscribed in a circle of radius R. Find the value of angle theta  that maximizes the area of the triangle.

the only angle of the triangle that is given is the top angle which is 2 theta

Answer
I attached a drawing of the circle & the triangle with all the angles .
Area=S=½X²Sin(Ө). Let's find X as function of Ө :
Sine Rule : X/sin(φ)=R/sin(Ө/2) --> X=Rsin(φ)/sin(Ө/2). We know that 2φ+2Ө=2π , therefore :
φ=π-Ө. Thus, X=Rsin(π-Ө)/sin(Ө/2)=Rsin(Ө)/sin(Ө/2). Because sin(π-Ө)=sin(Ө).
So, we have :
S(Ө)=½[Rsin(Ө)/sin(Ө/2)]²Sin(Ө).
S(Ө)=(½R²)sin²(Ө)/sin²(Ө/2).
We know that : sin(Ө/2)=√[(1-cos(Ө))/2]. Thus,
S(Ө)=(½R²)sin²(Ө)/[(1-cos(Ө))/2]
S(Ө)=R²sin²(Ө)/[1-cos(Ө)] .
to obtain Maximum, we solve S'(Ө)=0 :

         [2sin(Ө)cos(Ө)][1-cos(Ө)]-[sin(Ө)][sin²(Ө)]
S'(Ө)=R² --------------------------------------------
                        [1-cos(Ө)]²

         2sin(Ө)cos(Ө)-2sin(Ө)cos²(Ө)-sin³(Ө)
S'(Ө)=R² ---------------------------------------
                        [1-cos(Ө)]²

S'(Ө)=0 --> 2sin(Ө)cos(Ө)-2sin(Ө)cos²(Ө)-sin³(Ө)=0 --> 2cos(Ө)-2cos²(Ө)-sin²(Ө)=0 -->
2cos(Ө)-cos²(Ө)-cos²(Ө)-sin²(Ө)=0 --> 2cos(Ө)-cos²(Ө)-1=0 . Let set τ=cos(Ө), then :
2-2τ²-1=0 , Solution : τ=½ . --> cos(Ө)=½ --> Ө=60° .

Alon.