# Thread: 4-sided polygon largest area inscribed in a circle

1. ## 4-sided polygon largest area inscribed in a circle

Hello all, can someone please provide me some help with an optimal design question. Any sort of help would be awesome.

Q1) Show that among all the 4-sided polygons inscribed in a circle the squares have the largest area.
Q2) And for fixed n which is the n-gon of largest area inscribed in a circle?

For Q1 I know I should focus on showing that all sides of a quadrilateral of maximal area have to be of equal length. So I would use Brahmagupta's formula I think, but how would I show that the squares have the largest area?

Q2) I don't know how to approach this.

Thank you all for any help.

2. You need to find a relationship between the circle and the 4-sided polygon to help establish a function for area in terms of either a side length or an angle. after this find where A' = 0.

3. Originally Posted by Nguyen
Q2) And for fixed n which is the n-gon of largest area inscribed in a circle?
For Q2, the regular n-gon "feels" like it should be the right answer, but as for proof I'm also at a loss. Perhaps we can prove that the center of the circle must lie inside the polygon, then gain something by considering the triangles defined by the center and pairs of adjacent vertices of the polygon? (The sum of areas of these triangles equals the area of the polygon.)

Edit: Maybe something like: for a triple of adjacent vertices, the sum of areas of the two triangles is maximized when the middle vertex bisects the arc connecting the outer vertices. Thus any candidate polygon which has a triple of adjacent vertices such that the middle vertex does not bisect the associated arc can't be optimal. Thus the only possibility is the regular n-gon.

4. Originally Posted by pickslides
You need to find a relationship between the circle and the 4-sided polygon to help establish a function for area in terms of either a side length or an angle. after this find where A' = 0.

Thanks, but how do I find the relationship using the 4-sided polygon? I could find it for say a square, but not a general 4-sided polygon.

Originally Posted by undefined
For Q2, the regular n-gon "feels" like it should be the right answer, but as for proof I'm also at a loss. Perhaps we can prove that the center of the circle must lie inside the polygon, then gain something by considering the triangles defined by the center and pairs of adjacent vertices of the polygon? (The sum of areas of these triangles equals the area of the polygon.)

Edit: Maybe something like: for a triple of adjacent vertices, the sum of areas of the two triangles is maximized when the middle vertex bisects the arc connecting the outer vertices. Thus any candidate polygon which has a triple of adjacent vertices such that the middle vertex does not bisect the associated arc can't be optimal. Thus the only possibility is the regular n-gon.
I am sorry for being such a noob but I don't understand this.

5. Consider a 4-sided polygon in a circle. Ler r be the radius. Divide the polygon in to four triangles with one vertex as the center. Let θ1, θ2, θ3 and θ4 be the angles at that vertex.

Area of the polygon = sum of the areas of the triangles.

Area of the triangle is gives by 1/2*r*rsinθ = 1/2*r^2*sinθ

Area of the polygon = 1/2*r^2(sinθ1 +sinθ2 + sinθ3 + sinθ4)

= 1/2*r^2{2sin[(θ1+θ2)/2]cos[(θ1-θ2)/2] + 2sin[(θ3+θ4)/2]cos[(θ3-θ4)/2]}

The area will be maximum, when cos values will be maximum i.e. 1. So θ1 = θ2 and θ3 = θ4.

Now the area of the polygon = 1/2*r^2*2(sinθ1 + sinθ3) = r^2{2sin[(θ1+θ3)/2]cos[(θ1-θ3)/2]}

Again this area is maximum, when θ1 = θ3.

Hence for the area of 4-sides polygon is maximum when θ1 = θ2 = θ3 = θ3 = π/2.

That polygon is a square.