Dear Ed,
I just remembered that I have a copy of Cajori's _History of Mathematics_
(Macmillan, 1894). Cajori says of Brahmagupta:
At that time flourished Brahmagupta (born 598). In 628 he wrote his
_Brahma-sphuta-siddhanta_ ("The Revised System of Brahma"), of which
the twelfth and eighteenth chapters belong to mathematics.
Later:
Of interest is the formula given by Brahmagupta for the area of a
triangle in terms of its sides. In the great work attributed to
Heron the Elder this formula is first found. Whether the Indians
themselves invented it, or whether they borrowed it from Heron, is
a disputed question. Several theorems are given by Brahmagupta on
quadrilaterals which are true only of those which can be inscribed
on a circle--a limitation which he omits to state.
I would guess that his formula for the maximum area of a quadrilateral
in terms of its sides is one of those!
--George Sicherman
"Michael Rios sent me an interesting problem. What is the largest possible
area of a quadrilateral with sides 1, 2, 3, and 4? Send Answer. If you can
figure that out, try solving (2,3,4,5) and (3,4,5,6). See a pattern? A
general solution eludes me."
The answer to each is sqrt(24), sqrt(120), & sqrt(360).
The general solution is as follows:
Maximum area = sqrt[4*(A*B+C*D)^2-(A^2+B^2-C^2-D^2)^2]/4,
where A, B, C, & D are the sides of the quadrilateral in order. However,
the expression inside the brackets can be expanded to another expression in
which any two variables can be exchanged to create a new equation that is
identical to the first equation. Therefore, the order in which A, B, C, & D
are represented in the quadrilateral is irrelevant, and the maximum area
will remain the same.
In the course of my calculations, I have also found that the area is at
maximum when any two opposing angles add up to 180 degrees.
Alan Lemm
I was playing with the equation I got for the area (btw, I think I may have
made a simple arithmetic error last time, and the maximum area for the
(1,2,3,4) quadrilateral is Sqrt[24])
The equation simplifies down to A = Sqrt[n(n+1)(n+2)(n+3)] or the square
root of the product of the sides
as n increases to infinity, the quadrilateral approximates a square, and the
maximum area goes to n^2
-Travis Taylor
one/half = 1/2 --> 534/1068
Clint Weaver
1-2-3-4 = 4.89898
2-3-4-5 = 10.9545
3-4-5-6 = 18.9737
4-5-6-7 = 28.9828
5-6-7-8 = 40.9878
6-7-8-9 = 54.9909
Attached are C++ program listing and executable (no, I wouldn't run it
either without doing at least an email check first). The method
combines Heron's formula for finding the area of a triangle and brute
force.
One of my former APCS students came by and pointed out your site and
the puzzle and we played arround with it - we get it right?
Bud Crain
Now, I have a formula for the area for a quadrilateral with sides
(n,n+1,n+2,n+3)
A = Sqrt[n^4 + 6n^3 + 11n^2 + 6n]
For (1,2,3,4) the area is Sqrt[24]
(2,3,4,5); A = Sqrt[120]
(3,4,5,6); A = Sqrt[360]
-travis_r_taylor
If I've done my research right - the largest area is when the quadrilateral
is inscribed in a circle - in which case the area can be found by
Brahmagupta's formula.
This works out as root(24) for 1,2,3,4
And generalised for x,x+1,x+2,x+3 to
Root(x^4 + 6x^3 + 11x^2 + 6x)
Regards
Matthew Newell
For the first case (1,2,3,4) the maximum area is about y = 4.8989795
where the shorter diagonal has length of about x = 3.5765
I solved it as the maximum of the ugly function
y = (-1/16x^4 + 5/4x^2 - 4)^.5 + (-1/16x^4 + 5/2x^2 - 9)^.5
Interestingly, the order of the sides doesn't seem to matter. The area of
(1,2,4,3) came out the same.
The general solution to (a,a+1,a+2,a+3) evades me too, but it appears
using geometers sketchpad that the maximun area occurs when the
quadrilateral is inscribed in a circle.
Possibly is can be (or has been) proved that the maximum area of any
quadrilateral (w,x,y,z) occurs when it is inscribed.
If this is true it would probably yield a geneal solution pretty easily.
I'll work on this just in case.
-Jeremy Galvagn
Hi Ed,
Michael Rios's problem is an interesting one, although it has a simple, if
not elegant, solution.
Draw a diagonal across the quadrilateral and call this length a.
We can now define a function f(a) for the area of each triangle, and hence
the whole quadrilateral.
Start with the Cosine rule:
a^2 = b^2 + c^2 - 2bc.cos(A),
cos(A) = (b^2+c^2-a^2)/2bc
[ cos^2(A)+sin^2(A)=1, or, sinA = sqrt(1 - cos^2(A)) ]
sin(A) = sqrt( 1 - ( (b^2+c^2-a^2)/2bc )^2 )
[ Area of triangle = bc.sin(A)/2 ]
Area of each triangle = bc.sqrt( 1 - ( (b^2+c^2-a^2)/2bc )^2 )/2 [note: b &
c are known, so we have f(a)]
Now, say we have lengths {1,4,a} make up one triangle, and {2,3,a} make up
the other, we get:
Area{1,2,3,4} = 2.sqrt(1-((17-a^2)/8)^2) + 3.sqrt(1-((13-a^2)/12)^2)
I don't know of a simple reduction of this formula, but it is sufficient to
adjust a until we have maximal Area.
Perhaps this can be differentiated (beyond me), to drop out our required a
automatically, and hence calculate Area.
The maximum of each triangle is at a=sqrt(17) and a=sqrt(13), so a(max) will
be between these 2 values.
Using recursive techniques, I find:
a = 3.9242833 for an area of: 4.8989795.
For quadrilateral {2,3,4,5} we get (using 2,5,a and 3,4,a):
Area(2,3,4,5} = 5.sqrt(1-((29-a^2)/20)^2) + 6.sqrt(1-((25-a^2)/24)^2).
a = 5.2136183 for an area of: 10.95445115.
The same procedure can be used for any quadrilateral.
I wondered at first if it made any difference which way the lines segments
were ordered.
ie, 1,2,3,4; 1,3,2,4; 1,4,2,3 etc, but no, it doesn't matter! Since we are
looking at the sum of the area of 2 triangles, and the diagonal we choose is
arbitrary, we can always choose a diagonal, and reflect one of the triangles
formed to bring us back to the {1,2,3,4} case.
eg. starting with {1,3,4,2}. Choose a diagonal to leave {1,2,a} and {3,4,a}.
reflecting the latter brings us back to having a {1,2,3,4} quadrilateral
again.
Regards,
Alan O'Donnell
You'll probably get plenty of responses about the largest quadrilateral
with given sides! Here's one more:
My father was a civil lawyer, and one day he brought home a problem
for me. He was handling the purchase of an irregular quadrilateral of
land, and he suspected that the advertised acreage was exaggerated.
I studied the problem, found a general formula, and found that he was
right. A few weeks later I stumbled on the problem in an old book of
mathematical recreations!
The largest quadrilateral with given sides is an inscribed quadrilateral.
The formula for the area of an inscribed quadrilateral is due to Brahma-
gupta (598-?) and is a straightforward generalization of Heron's Formula
for the area of a triangle.
Heron's Formula:
KK = s(s - a)(s - b)(s - c)
Brahmagupta's Formula:
KK = (s - a)(s - b)(s - c)(s - d)
The symbol s denotes the semiperimeter. There are similar, more
complex, formulas for the size of an n-dimensional simplex in terms
of the lengths of the edges.
--George
I suppose the maximum area to be about 4,9.
I solved this by using Heron's formula and simple numeric methods.
First of all it doesn't matter in which sequence the four sides are linked together. So take any and divide the quadrilateral into two triangles.
Is there a way to solve the problem with elementary methods? I'd like to know this. Well, there IS an elementary solution for some special cases as for example the numbers 1, 4, 7 and 8 ...
Of course my computer solution works with any four numbers which can be used as sides of a quadrilateral, but unfortunately I wasn't able to see any pattern.
Happy puzzling,
Franjo Schulte
I'm assuming that the problem is for a quadrilateral ABCD where the length
of AB = n, BC = n+1, CD = n+2, and DA = n+3 (as opposed to being able to
permute the order the lengths are in) and I haven't checked for other cases,
but I think that the maximum area occurs when the length of AC is
(2n+1)(2n+5)*Sqrt(2n^2+6n+6)/(4n^2+12n+12)
-Travis Taylor
Concerning the problem of the largest area for a quadrilateral with sides
(1,2,3,4), I would signal you a noticeable result (Theorem of Huigens &
Cramer)
<< If we fix the lengths of the sides of a polygone, the largest
possible area corrisponds to an inscribed poligone. >>
This holds true for any number of sides; in particular, for a quadrilateral,
many geometrical constructions for the solution are known.
From the algebric point of view, let (a,b,c,d) be the lengths of the sides;
then in the inscribed quadrilateral the diagonal separating the couple (a,b)
from the couple (c,d) has a length given by the square root of the ratio
[ a b (c^2+d^2) + c d (a^2+b^2) ] / [ a b + c d ]
while, concerning the value A of the largest area, we have an interesting
generalization of the Hero formula for the area of a triangle: for the
inscribed quadrilateral, one has:
16 A^2 = (a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)
Regards, Claudio Baiocchi
Ed:
For sides of 1,2,3, and 4, my answer is an area of ~ 4.8989794+
For certain* sides of a,b,c, and d, my answer is:
SQRT [(s-a)(s-b)(s-c)(s-d)] where s=(a+b+c+d)/2
* NOTE: The sides must be of length such that they form a cyclic
quadrilateral.
Robert Wainwright
To maximize the area of a quadrilateral it must be
cyclic (I remember this rule from a math quiz i did
in high school) and there is a formula - called
Brahmagupta's Formula (this I had to find online!) -
which gives the area of a cyclic quadrilateral as
where s is = (1/2) * (a+b+c+d)
For the set of quadrilaterals that you chose, the maximum
area always works out as (s-a)=d, (s-b) = c, (s-c) = b and (s-d) = a!
A(max) = sqrt(a*b*c*d)
because (s-a)=d, (s-b) = c, (s-c) = b and (s-d) = a
Nice!
Alasdair Trotter
Good afternoon
Let ABCD the quadrilateral of sides 1,2,3,4.
A priori,there are two possible configurations either the two largest sides 3 and 4 are adjacent or they are opposite, the other ones come down to these two cases.
1) AB=4, BC=3, CD=1 and DA=2
Let angle(AB,BC) = a and angle(CD,AD) = b
Area of the triangle ABC = 1/2 * 4 * 3*sin(a) = 6 *sin(a)
Area of the triangle ACD = 1/2 * 1 * 2*sin(b) = sin(b)
Then the area S of the quadrilateral ABCD is equal to 6*sin(a) + sin(b)
On the other hand AC^2 = AB^2+BC^2-2*AB*BC*cos(a)=25 - 24*cos(a)
and AC^2 = CD^2+AD^2-2*CD*AD*cos(b) = 5 - 4*cos(b)
Therefore, a and b are linked by the relation cos(b) = 6*cos(a) - 5
S is a maximum if dS/da=0 ==> 6*cos(a) + cos(b)*db/da = 0
As db/da = 6*sin(a)/sin(b) ==> dS/da = 0 if a+b=pi. The 4 points A,B,C,D are on the same circle. This is an expected result.
Then cos(a)=5/7 and S = 2*sqrt(6)
2) AB=4, BC=2, CD=3 and DA=1
With the same notations, we get S = 4*sin(a) +3/2*sin(b) with 3*cos'b)=8*cos(a)-5
The maximum of S is again obtained with the four points on the same circle such that S=2*sqrt(6)
Generalization
Let the quadrilateral ABCD of sides n, n+1, n+2, and n+3
As above, it is easy to check that the area of the quadrilateral is the largest possible when the quadrilateral is cyclic.
In this case, the area of the quadrilateral is given by the Brahmagupta formula, S=sqrt[s*(s-n)*(s-n-1)*(s-n-2)*(s-n-3)]
with s=semi-perimeter of the quadrilateral equal to 1/2*[n+n+1+n+2+n+3]=2*n+3
Then S=sqrt[n*(n+1)*(n+2)*(n+3)].
For n=2 and 3 we find respectively S= sqrt(120)=2*sqrt(30) and sqrt(360)=6*sqrt(10)
Philippe Fondanaiche