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Largest quadrilateral in a circle?

  1. Sep 1, 2006 #1
    i believe that a square is the largest quadrilateral that can be fit inside a circle, but how would you prove it?
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  3. Sep 1, 2006 #2


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    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 lenght L for which A has a maximum (dA/dL=0).
    Last edited: Sep 1, 2006
  4. Sep 1, 2006 #3


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    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.
  5. Sep 1, 2006 #4


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    Here's another way. First, what is the area of the polygon with vertices at (x1, y1), (x2, y2), ...(xn, yn)? 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 (xk, yk) to (xk+1, yk+1) is just:

    [tex]\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)[/tex]

    So the total area is:

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

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

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

    Now if we look at four points on the unit circle, labelled by their angle in polar coordinates, [itex]\theta_k[/tex], the area of the corresponding inscribed quadrilateral is:

    [tex]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)[/tex]

    We clearly can't do any better than if the angles between adjacent vertices is pi/2 (90 degrees), so a square maximizes the area.
    Last edited: Sep 1, 2006
  6. Sep 1, 2006 #5
    Why not simply use differential? I think its simpler. For an quadrilateral that touches the edges of the circle (if it dosen'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.
    Last edited: Sep 1, 2006
  7. Oct 21, 2006 #6
    check these out:

    Last edited: Oct 21, 2006
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