Question

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

Answer 1

`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.

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`