Largest quadrilateral in a circle?

[jnorman]

i believe that a square is the largest quadrilateral that can be fit inside a circle, but how would you prove it?

 

[quasar987]

Calculus. The area of a rectangle is A=hL. Moreover, if the rectangle is inscribed in the circle, its sides are related to the radius of the circle by r²=h²+L². write A as a function of L only. Find the length L for which A has a maximum (dA/dL=0).

 

[0rthodontist]

I don't think that works.

There is a geometric argument:
Let such a quadrilateral be ABCD. First consider side AC. If the quadrilateral is maximum-area, B and D must be placed so that the two triangles ABC and ADC are each maximum-area. This occurs when the height of these triangles is maximum--which is when BD is a diameter of the circle and perpendicular to AC. Similarly AC must be a diameter of the circle and perpendicular to BD, so ABCD is a square.

 

[StatusX]

Here's another way. First, what is the area of the polygon with vertices at (x₁, y₁), (x₂, y₂), ...(x_n, y_n)? That's not as hard as it sounds. Just use green's theorem. You need to integrate y dx (or x dy) around the edges of the polygon, returning to your starting point (keeping the interior on your left side to have a positive orientation, or just taking the absolute value at the end). The integral from (x_k, y_k) to (x_k+1, y_k+1) is just:

\int_{x_k}^{x_{k+1}} \left( \frac{y_{k+1}-y_k}{x_{k+1}-x_k} (x-x_k) + y_k \right) dx= \frac{1}{2} (y_{k+1}+y_k)(x_{k+1}-x_k)

So the total area is:

A=\frac{1}{2} \sum_{k=1}^{n} (y_{k+1}+y_k)(x_{k+1}-x_k)

where we look at k (mod n), and after expanding and cancelling this simplifies to:

A=\frac{1}{2} \sum_{k=1}^{n} (x_k y_{k+1} - x_{k+1} y_k )

Now if we look at four points on the unit circle, labelled by their angle in polar coordinates, \theta_k[/tex], the area of the corresponding inscribed quadrilateral is:<br /> <br /> A=\frac{1}{2} \sum_{k=1}^{4} \cos(\theta_k) \sin(\theta_{k+1}) - \cos(\theta_{k+1}) \sin(\theta_k) =\frac{1}{2} \sum_{k=1}^{4} \sin(\theta_{k+1}-\theta_k)<br /> <br /> We clearly can&#039;t do any better than if the angles between adjacent vertices is pi/2 (90 degrees), so a square maximizes the area.

 

[Werg22]

Why not simply use differential? I think its simpler. For an quadrilateral that touches the edges of the circle (if it doesn't it is fairly simple to prove there exist another quadrilateral with a bigger area, so this case is not consider) we consider the point a on the interval x = 0 to x = r (then again it is fairly easy to prove that for any quadrilateral which edges are not in all the "quarters" of the circle, there exist a quadrilateral with bigger area). Now by symetry we have, L = 2a and W = 2(r^2-a^2)^1/2. Thus the area is A(x) = 4x(r^2-x^2)^1/2. Now simply use the derivative to find a zero on the interval (0,r), and prove that for the solution found a = (r^2-a^2)^1/2 and the task is done.

PS. Sorry this method cannot be use since it only takes account of rectangular quadirlaterals. My bad.

 

[murshid_islam]

check these out:

http://en.wikipedia.org/wiki/Bretschneider's_formula
http://en.wikipedia.org/wiki/Brahmagupta's_formula