Conference rusure::math

    If the figure is a parallelogram, then the answer is 90 degrees.

    Off hand, it would appear that the area is maximized when
    
    		BAD = ABC + ADC - BCD
    
    -Gary

I assume the sides are of FIXED length such that no side is > the sum of the
other 3. 

I believe the maximum area is acheived when the quadrilateral can be inscribed
in a circle, although I don't have a proof yet. The angles are thus a function
of the lengths of all 4 sides, and the expression I suspect to be complex.
Where is Stan Rabinowitz when we need him?

    In general there will be two minima (as can be seen by taking a
    triangle off on side and flipping it over, which will not alter
    the area.)

    .-1 may be right about the circle.

    - Jim

    Chemical Rubber Company says the area of a quadrilateral is (tan AOD)
    (AB^2+CD^2-DA^2-BC^2)/2, where O is the intersection of AC and BD.
    Since everything is fixed except AOD, we want to make AOD as close to
    ninety degrees as possible.
    
    
    				-- edp 

    AC^2 = AB^2 - 2*AC*BC*COS(ABC) + BC^2 = AD^2 - 2*AD*CD*COS(ADC) + CD^2

    should read

    AC^2 = AB^2 - 2*AB*BC*COS(ABC) + BC^2 = AD^2 - 2*AD*CD*COS(ADC) + CD^2

    - Jim

    Re .6:
    
    If AB = AD and BD = CD, then AB^2+CD^2-DA^2-BC^2 is zero, tan AOD is
    "infinite", and the formula degenerates.  But for quadrilaterals for
    which AB^2+CD^2-DA^2-BC^2 is not zero, it is impossible to arrange them
    so that AOD is a right angle.  So you want to get as close to ninety
    degrees as you can. 
    
    Quadrilaterals for which AOD is always a right angle will have to be
    handled separately. 
                       
    
    				-- edp

Test case: 2,2,3,5.

With the two 2's aligned as a 4, we get a right triangle with area = 6.

With the 2&3 aligned as a 5, we get a symmetric triangle with area = 4.8990.

With the 5 and 2 parallel, we get a trapezoid with area 6.5996+

With the corners inscribed in a circle, the area is 4*sqrt(3) = 6.9282+.

    It's not hard to show that the quadrilateral which maximizes area
    should be inscribed in a circle as Lynn has claimed.

    Let the corner angles be A, B, C, D and for convenience let the
    sides be a = AB, b = BC, c = CD, d = DA, none zero or otherwise
    'impossible'.

    Recall that when a quadrilateral is inscribed in a circle the opposite
    angles sum to 180 degrees.

    Now the area is

	Area(B,D) = 1/2 * (a*b*sin(B) + c*d*sin(D))

    and the length of the diagonal subtending angles B and D is

	Length(B,D)^2 = a^2 - 2*a*b*cos(B) + b^2 = c^2 - 2*c*d*cos(D) + d^2

    Taking differentials dB AND dD of B and D and equating to zero we have

	a*b*cos(B)*dB + c*d*cos(D)*dD = 0
	a*b*sin(B)*dB - c*d*sin(D)*dD = 0

    So for this homogenous system to be consistent under perturbations of
    B and D the determinant must vanish and

	cos(B)*sin(D) + cos(D)*sin(B) = 0

	tan(B) = -tan(D)

    The last being satisfied when B+D = 180 degrees.

    - Jim

    So .2 (area is max when A = B + D - C) is correct right?
    Thanks.
    
    -Gary

    Can we generalize this problem?

    What n-gon in the plane with sides of lengths s1, s2, s3, ... sn
    inscribes the largest area?

    Dwayne

    BTW, I'm extremely suspicious of the formulation in 761.12. Given
    
    AB=3, BC=2, CD=6, and DA=10, then
    
    cos(BAD)= -69/84. However, I believe BAD must be acute - a
    contradiction.
    
    Dwayne
    

>    What n-gon in the plane with sides of lengths s1, s2, s3, ... sn
>    inscribes the largest area?

If the order is as given, then again it's a circle with chords as given. If 
the order can be rearranged, then it may be a different circle; max over 
permutations of 1..n is a harder problem.