What is the area of the largest isosceles triangle that can be inscribed in a circle of radius 4?

1 Answer
Feb 25, 2016

12sqrt(3)~=20.784

Explanation:

One could start by saying that the isosceles triangle with largest area inscribed in a triangle is also an equilateral triangle.

However if you need a formal demonstration of this statement read the first part of this explanation.

Suppose an isosceles triangle_(ABC) inscribed in a circle with center in D and radius r, like the figure below.

I created this figure using MS ExcelI created this figure using MS Excel

We can obtain the side a in function of r and alpha in this way (Law of Sines applied to triangle_(BCD)):

a/sin(2alpha)=r/sin beta
Since 2alpha+beta+beta=180^@ => beta=90^@-alpha
=> a=rsin (2alpha)/sin(90^@-alpha)=2r(sin alpha *cancel(cos alpha))/cancel(cos alpha)
=> a=2rsin alpha=4rsin (alpha/2)cos (alpha/2)

We can obtain the height h in function of r and alpha in this way:

tan(alpha/2)=(a/2)/h => h=a/(2tan(alpha/2))
Replacing a for its value in function of r and alpha:
h=(4rcancel(sin(alpha/2))cos(alpha/2))/2*cos(alpha/2)/cancel(sin(alpha/2))
=> h=2rcos^2(alpha/2)

Then we can obtain the area of the triangle in function of r and alpha:

S_(triangle_(ABC)) =("base"*"height")/2
S_(triangle_(ABC))=((4rsin(alpha/2)cos(alpha/2))(cancel(2)rcos^2(alpha/2)))/cancel(2)
S_(triangle_(ABC))=4r^2 sin(alpha/2)cos^3(alpha/2)

Since in this problem r is constant, we need to find the derivative relatively to alpha of S_(triangle_(ABC)) and equal it to zero to find the maximum or minimum of the area of the triangle.

So

d/(dalpha)(S_(triangle_(ABC)))=4r^2 cos(alpha/2)*(1/2)cos^3(alpha/2)+4r^2 sin(alpha/2)*3cos^2(alpha/2)(-sin(alpha/2))(1/2)
=2r^2cos^4(alpha/2)-6r^2sin^2(alpha/2)cos^2(alpha/2)
=2r^2cos^2(alpha/2)(cos^2(alpha/2)-3sin^2(alpha/2))
Equating the derivative to zero we get:
cos^2(alpha/2)=0 => cos (alpha/2)=0 => alpha/2=90^@ => alpha=180^@ (this is the point of minimum area)
And
cos^2(alpha/2)-3sin^2(alpha/2)=0 => 3sin^2(alpha/2)=cos^2(alpha/2) => tan(alpha/2)=1/sqrt(3) => alpha/2=30^@ => alpha=60^@ (this is the point of maximum area)
-> But if alpha=60^@ this means that AhatBC=AhatCB=60^@ and the isosceles triangle of maximum area is also an equilateral triangle.

Area of the triangle of maximum area

S_(triangle_(ABC))=4*4^2sin 30^@*cos^3 30^@=4*16*(1/2)(sqrt(3)/2)^3=32*(3sqrt(3))/8=12sqrt(3)~=20.784