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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
parameters:

a
b   f
c   e
d

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
known.

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
area.

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
http://mathforum.org/dr.math/
```

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

Hi,

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.

Regards,
Gaston
```

```
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
http://mathforum.org/dr.math/
```

```
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
question.

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?

Regards,
Gaston
```

```
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
http://mathforum.org/dr.math/
```

```
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
coefficients.

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 +
(-296350*p^8-7819850*p^6-2827467350*p^4-79295259246100*p^2-
79295259246100)*q^2) +
(25313125*p^8+3888271250*p^6+687708700625*p^4-41352642902500*p^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
6.6.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
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:

a----------------b
/                  \
/                    \
d----------------------c

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.

Regards,
Gaston
```

```
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
http://mathforum.org/dr.math/
```

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

Gaston,

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
http://mathforum.org/dr.math/
```
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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