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Uniquely Determining a Polygon

Date: 02/05/2001 at 04:29:22
From: Gaston
Subject: Polygon area and shape

I am trying to prove whether a polygon is uniquely determined if the 
following parameters are known:

   - area of the polygon
   - number of sides of the polygon
   - length of each side of the polygon
   - which angles of the polygon are obtuse or acute

I included the last parameter because it is easy to prove that the 
statement is not true without it. Using only the three first 

     b   f
     c   e
Take a convex n-sided polygon with equal side lengths, n > 4. (The 
above is a regular hexagon.) Choose any two non-adjacent vertices and 
slide them both "inside" the polygon in such a way that the distance 
between them and their neighbors remains the same. (For example, move 
a and d.) The above statement meant that the distances from a to the 
line bf and from d to the line cd remain the same after the slide. As 
the side lengths, being known, must remain constant, there is only one 
possible concave resulting polygon.

Now return one vertex to its original position and choose a vertex 
non-adjacent to the one that hasn't been replaced. (For example, 
replace d and choose e.) If we do the same "sliding" operation on the 
new vertex, we have another polygon with the same area as after the 
first operation (as all sides are of equal length) but of a different 
shape (the "inverted" points are at different locations relative to 
each other).

So this is why I have added that the angle's "obtuseness" must be 

In fact, the problem reduces to proving that for any given polygon, if 
the side lengths and the obtuseness or acuteness of the angles are 
known, changing one or more angles in the polygon does change the 

Am I simply wrong, or how can I prove this?

Thanks a lot in advance,
Gaston Gloesener

Date: 02/05/2001 at 12:39:59
From: Doctor Rob
Subject: Re: Polygon area and shape

Thanks for writing to Ask Dr. Math, Gaston.

You will have trouble proving the statement at the beginning of the 
question, because it is already false for quadrilaterals. This can be 
seen by taking a nonrectangular parallelogram, cutting along the long 
diagonal, flipping one half over, and rejoining along that diagonal. 
You will get a figure called a kite, which is definitely different 
from the starting parallelogram, yet has the same area and side 
lengths, and has the same number of obtuse and acute angles: two of 
each, forming opposite pairs. The side lengths are in a different 
order, of course.

The flexibility of a general quadrilateral, which is demonstrated by 
drawing a diagonal and varying its length while keeping the lengths of 
the sides the same, can also produce examples where the areas are 
equal although the angles are not.

- Doctor Rob, The Math Forum   

Date: 02/05/2001 at 16:49:37
From: Gloesener, Gaston
Subject: Re: Polygon area and shape


Maybe I was not clear in my statement. By "knowing the side lengths" I 
meant that I know also the order in which they come. This means that, 
for example, I can give the side lengths moving clockwise around the 
polygon. Also, the positions of the obtuse and acute angles are known 
relative to these side lengths. This means that the rough shape of the 
polygon is known.


Date: 02/05/2001 at 12:50:06
From: Doctor Peterson
Subject: Re: Polygon area and shape

Hi, Gaston.

I don't think the situation is nearly as simple as you imagine. Taking 
a simple case, draw an isosceles trapezoid like:

        /                   \
       /                     \
      /                       \

Now imagine it is made of fixed-length sticks, and jiggle the top a 
little while keeping the bottom fixed. I believe the trapezoid will 
have the greatest area for any quadrilateral with these particular 
side lengths, so on either side of this position (that is, with the 
top tipped a little to the right and to the left), you will find a 
slightly smaller pair of alternate shapes with equal areas. 

Because the "jiggle" is not large, the top angles will still be 
obtuse, and the bottom angles acute, so your conditions are met, but 
there are at least two different polygons with the same area. This is 
not just a matter of having the same shape flipped over, because if 
the sides were not exactly equal, the same thing would still happen. 
With more vertices, it will get far more complex, and undoubtedly 
there will be many such equal-area configurations.

- Doctor Peterson, The Math Forum   

Date: 02/05/2001 at 17:12:14
From: Gloesener, Gaston
Subject: Re: Polygon area and shape

You are right! Didn't think about that. Then let's extend the 

The only thing I could think to add is knowing some of the angles. 
Because angle measurement is not precise and errors are multiplied by 
the length, it's best to require the minimum possible number of angles 
to be known. So for a polygon of n sides, knowing all the parameters I 
mentioned before: how many angles have to be known, and what do their 
relative positions have to be? By relative position, I mean do they 
have to be adjacent, or interleaved with unknown angles?


Date: 02/05/2001 at 22:41:44
From: Doctor Peterson
Subject: Re: Polygon area and shape

Hi, Gaston.

If you picture constructing a polygon, you can do so if you know, say, 
all but one of the side lengths and all but two of the angles. (The 
last pair will be determined for you.) That says that it takes 2n-3 
numbers to determine an n-gon. (For example, with n = 3 this says that 
a triangle needs 3 numbers, in agreement with SSS, SAS, etc.)

If you know the area, I would expect that to account for one of the 
numbers you need; so if you know that plus n lengths, you will need 
n-4 angles. (Similarly, if you know the area and two sides of a 
triangle, it will almost determine the triangle.) This is vague 
reasoning, of course, and I'm not guaranteeing its accuracy, but I'm 
pretty sure something like this will be true. Which numbers you know 
will make some difference, and there may be some combinations that 
would not work as well as others (as in the case of AAA and SSA in 
triangles), but in general you will find that it takes at least this 
much information.

- Doctor Peterson, The Math Forum   

Date: 02/06/2001 at 12:45:07
From: Doctor Rob
Subject: Re: Polygon area and shape

Thanks for writing back, Gaston.

If you know the side lengths (in order) and the area of a 
quadrilateral, it is uniquely determined.

I proved that by drawing a diagonal of length p forming two triangles 
with sides a, b, and p, and c, d, and p. Then, using Hero's Formula, I 
got the following equation for p:

     sqrt[(a+b+p)*(a+b-p)*(a-b+p)*(-a+b+p)] +
       sqrt[(c+d+p)*(c+d-p)*(c-d+p)*(-c+d+p)] = 4*K

where K is the area of the quadrilateral. I knew everything in this 
equation except p. To solve for p, I subtracted one of the radicals 
from both sides of the equation, squared both sides, and simplified. 
Then I isolated the remaining radical on one side of the equation, 
squared again, and simplified. By bringing all terms to one side of 
the equation, I obtained a polynomial equation in p with known 

It turns out that this equation is quadratic in p^2. This has two 
roots, one of which is extraneous and introduced by the repeated 
squaring process described above. When a, b, c, and d do form an 
actual quadrilateral, the remaining root is a positive real number. 
Its negative square root is not meaningful geometrically, so I 
rejected that. That left just one solution, p. That proved that the 
value of p was uniquely determined by a, b, c, d, and K. That means 
that the quadrilateral was also uniquely determined.

The situation is a bit different for pentagons, however. If we have 
sides a, b, c, d, and e (in order), and area K, we can set up a 
similar equation using the lengths of two diagonals, p and q, which 
meet at a vertex. Using Hero's Formula on the three triangles, adding 
up their areas, and setting the sum equal to K, gives one equation 
relating p and q.  Actually it involves only p^2 and q^2, and it is a 
quartic equation in these two variables. The coefficients are 
polynomials in a, b, c, d, e, and K. Now the solutions to this 
equation form not a finite set of points, as in the quadrilateral 
case, but a continuous curve, with infinitely many points (p,q) lying 
on it. Any two of these pairs that are close enough together will give 
you a counterexample to your statement.

For example, if you take a = 4, b = 5, c = 6, d = 7, e = 8 and K = 50, 
you get the polynomial equation:

 (2581*p^4-308158*p^2+33046861)*q^8 +
 (-5486*p^6+106666*p^4-10294634*p^2-8464088546)*q^6 +
 (3229*p^8+258874*p^6+45028254*p^4+1846432954*p^2+1160490651689)*q^4 +
   79295259246100)*q^2) +
   291966196122500) = 0

When p = 7.5, one of the solutions is q = 6.571206739720809... This 
leads to a pentagon whose angles are

     <ab = 112.411 degrees
     <bc =  86.502 degrees
     <cd = 145.256 degrees
     <de =  51.436 degrees
     <ea = 144.395 degrees
When p = 8, one of the solutions is q = 6.687691392266601... This 
leads to a pentagon whose angles are

     <ab = 125.100 degrees
     <bc =  78.993 degrees
     <cd = 149.460 degrees
     <de =  52.440 degrees
     <ea = 134.008 degrees

This pair of pentagons provides a counterexample to your statement, 
and there are infinitely many other such pairs gotten by taking any 
value of p between 7.5 and 8, and the corresponding value of q near 

- Doctor Rob, The Math Forum   

Date: 02/06/2001 at 17:41:19
From: Gloesener, Gaston
Subject: Re: Polygon area and shape

Hi Dr. Rob,

Well, I got an answer that proved the opposite in a much easier way.

Consider a trapezoid:

       /                  \
      /                    \

Now, imagine you rotate point a slightly clockwise (it will be a 
rotation since the side length remains constant). You will have a 
second trapezoid with area a1. Now take the same starting trapezoid 
(above) but this time rotate point b for the same amount that you 
rotated a before. The resulting trapezoid will not be the same (but 
mirrored) but have the same area.

This is why I would like to know how to extend my parameters in order 
to uniquely determine the polygon. One idea was to see how many angles 
must be known in addition to all the previous parameters.

Maybe your demonstration can be used to prove that for any irregular 
polygon (no two sides equal) my parameters would be enough.


Date: 02/07/2001 at 12:10:59
From: Doctor Rob
Subject: Re: Polygon area and shape

Thanks for writing again, Gaston.

The two quadrilaterals you describe are congruent, being reflections 
of each other, I think. Thus they are not really "different." The side 
lengths of one in clockwise order are the side lengths of the other in 
counterclockwise order.

- Doctor Rob, The Math Forum   

Date: 02/08/2001 at 10:13:23
From: Doctor Rob
Subject: Re: Polygon area and shape


Doctor Peterson has pointed out to me that with a quadrilateral, when 
there are two positive real roots of the biquadratic equation in p I 
gave, it may be that one is extraneous and so there is a unique 
solution, or it may be that none is extraneous and so there are two 
solutions. Never are there more than two. Of course you have to check 
that the two solutions both satisfy your angle condition of preserving 
acuteness or obtuseness, which may or may not be so. In the case when 
it is so, you do get two quadrilaterals with the same side lengths (in 
order) and the same area, and with the angle condition satisfied.

I leave it to you to find a pair of quadrilaterals satisfying all 
these conditions using the technique I described earlier.

- Doctor Rob, The Math Forum   
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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