From: Bill Dubuque
Newsgroups: k12.ed.math,sci.math
Subject: Pick's Theorem: Area of Lattice Polygon [was: Geoboards  Pic's Theorem]
Date: Wed, 03 Jun 1998 00:25:19 GMT
John Van't Land wrote to k12.ed.math on 31 May 1998:

 When I first started teaching math over 25 years ago, I remember using
 geoboards and learning about Pic's Theorem or something like that.
 The theorem said that on a geoboard, to find the area of any enclosed
 region (it need not be a quadrilateral) you add the number of nails
 which the rubber band touches, subtract the number of untouched nails in
 the interior, and add one (or something like thisI don't remember the
 exact way it was done).

 Does anyone know what I'm trying to describe. Was it Pic's Theorem,
 Pik's Theorem, Piques' Theorem...?
It's known as Pick's Theorem (G. Pick, 1900; Jbuch 31, 215).
I have appended below some references to expository papers.
A lattice point is a point with integral coordinates, and
a lattice polygon is one whose vertices are lattice points.
Pick's Theorem says that the area of a simple lattice polygon P
is given by I + B/2  1, where I is the number of lattice points
in the Interior of P and B is the number of lattice points on the
Boundary of P.
Pick's Theorem is equivalent to Euler's formula and closely connected
to Farey series; it may be generalized to nonsimple polygons and
to higher dimensions, see the papers cited below.
To help remember the correct formula, you can check it on easy
cases (unit square, small rectangles, etc) or, better, you can
view how it arises from additivity of area. One can view Pick's
formula as weighting each interior point by 1, and each boundary
point by 1/2, except that two boundary points are omitted. Now
suppose we are adjoining two polygons along an edge as in the
diagram below. Let's check that Pick's formula gives the same
result for the union as it does for the sum of the parts (and
thus it gives an additive formula for area, as required).
1/2 1/2 1/2 1/2
...  @ @  ... ...  @ @  ...
/ \ / \ / \ / \
0 @ @ 0 @ 0
 
. 1/2 @ @ 1/2 . . @ 1 .
.  +  . => . .
. 1/2 @ @ 1/2 . . @ 1 .
 
0 @ @ 0 @ 0
\ / \ / \ / \ /
...  @ @  ... ...  @ @  ...
1/2 1/2 1/2 1/2
The edge endpoints we choose as the two omitted boundary points.
The inside points on the edge were each weighted 1/2 + 1/2 on the
left, but are weighted 1 on the right since they become interior.
All other points stay interior or stay boundary points, so their
weight remains the same on both sides. So Pick's formula is additive.
Bill Dubuque
Bruckheimer, Maxim; Arcavi, Abraham.
Farey series and Pick's area theorem.
Math. Intelligencer 17 (1995), no. 4, 6467. MR 96h:01019
Grunbaum, Branko; Shephard, G. C. Pick's theorem.
Amer. Math. Monthly 100 (1993), no. 2, 150161. MR 94j:52012
Morelli, Robert. Pick's theorem and the Todd class of a toric variety.
Adv. Math. 100 (1993), no. 2, 183231. MR 94j:14048
Varberg, Dale E. Pick's theorem revisited.
Amer. Math. Monthly 92 (1985), no. 8, 584587. MR 87a:52015
Liu, Andy C. F. Lattice points and Pick's theorem.
Math. Mag. 52 (1979), no. 4, 232235. MR 82d:10042
Haigh, Gordon. A "natural" approach to Pick's theorem.
Math. Gaz. 64 (1980), no. 429, 173180. MR 82b:51001
DeTemple, Duane; Robertson, Jack M.
The equivalence of Euler's and Pick's theorems.
Math. Teacher 67 (1974), no. 3, 222226. MR 56 #2854
Gaskell, R. W.; Klamkin, M. S.; Watson, P.
Triangulations and Pick's theorem.
Math. Mag. 49 (1976), no. 1, 3537. MR 53 #3881

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==============================================================================
From: Robin Chapman
Newsgroups: k12.ed.math,sci.math
Subject: Re: Pick's Theorem: Area of Lattice Polygon [was: Geoboards  Pic's Theorem]
Date: Fri, 05 Jun 1998 13:36:45 GMT
In article <357b97e5.158500437@news.cyberg8t.com>,
Bill Dubuque wrote:
[large excerpt of previous article deleted  djr]
Another recent reference is
Ricardo Diaz and Sinai Robins
Pick's formula via the Weierstrass Pfunction
American Mathematical Monthly
102 431437 (1995)
This is quite a tour de force, relating Pick to the theory of
elliptic functions.
Robin Chapman + "They did not have proper
Department of Mathematics  palms at home in Exeter."
University of Exeter, EX4 4QE, UK +
rjc@maths.exeter.ac.uk  Peter Carey,
http://www.maths.ex.ac.uk/~rjc/rjc.html + Oscar and Lucinda
== Posted via Deja News, The Leader in Internet Discussion ==
http://www.dejanews.com/ Now offering spamfree webbased newsreading

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From: "William L. Bahn"
Newsgroups: k12.ed.math,sci.math
Subject: Re: Pick's Theorem: Area of Lattice Polygon [was: Geoboards  Pic's Th
Date: Thu, 18 Jun 1998 13:07:18 GMT
Yes and No. There really isn't "A" relationship between perimeter and area.
I can draw lots of polygons that all have the same perimeter and yet have
different areas and, conversely, I can draw lots of polygons that all have
different perimeters and yet have the same area.
What it is based on is that you can take any simple polygon (having lattice
point vertices) and break it up into a set of right triangles (some of which
may be "negative" area) AND you can further break up those right triangles
into smaller right triangles until, eventually, you have a set of right
triangles whose hypotenuses (hypoteni?) passes through NO lattice points
except at the vertices.
So, what is the area of such a triangle:
A = bh/2.
How many lattice points are on the base line?
B = (b+1)
How many lattice points are on the height line?
H = (h+1), but one of them is also on the base line.
How many lattice points are on the hypotentuse?
2, but both are on one of the other two lines.
So now write the area in terms of the number of lattice points on the base
and height lines:
A = (B1)(H1)/2
This is O.K., except we have to distinguish between base and height lattice
points.
How about if we work from the starting point that the more lattice points
are in the interior, the larger the area is going to be? So how many lattice
point are in the interior of our triangle? Well, the number of points in the
interior of a right triangle such as ours is, by symmetry, going to be
exactly half of the number of lattice points in the interior of the
rectangle having sides b and h. How many is this? Well, inspection should
convince you that it is:
I = (B2)(H2)/2
2I = ((B1)1)((H1)1)
2I = (B1)(H1)  (H1)  (B1) + 1
2I = 2A  (H1)  (B1) + 1
2I = 2A  (H + B  3)
A = I + (H + B  3)/2
Now, just like BH was tightly related to the number of interior points, H +
B is something that is related to the perimeter points. So, how many points
are there total on the perimeter?
P = (B) + (H1) + (22) = B + H  1
Notice that I can now write the area as:
A = I + (H + B  1 2)/2
A = I + (P  2)/2
A = I + P/2  1
FWIW: I never heard of Pick's Theorem prior to seeing this thread. So all of
the above is based on trying to see the underlying fundamentals and then
showing that the Theorem follows directly from them.
From this point, it is only necessary to show that things work out when you
starting combining multiple triangles into more complex polygons. However,
notice that if you butt any two of these triangles against each other, the
only thing that happens is that a line segment gets removed. The end points
of this line segment are points that were perimeter points on both triangles
and now they are merely perimeters points on the combined polygon, so the
perimeter point count goes down by two as a result. In addition, any other
lattice points on this line segment become interior points. In particular
notice that for every point that makes this change, you lose two perimeter
points (one from each triangle).
Therefore, if I have two triangular areas that I butt up against each other
I have:
A1 = (I1) + (P1)/2  1
A2 = (I2) + (P2)/2  1
A3 = A1 + A2 = [(I1) + (P1)/2  1] + [(I2) + (P2)/2  1]
If there are N points that become interior points:
I3 = (I1 + I2 + N)
P3 = (P1 + P2  2N  2)
Therefore:
A3 = [(I1) + (I2)] + [(P1) + (P2)]/2  2
A3 = [(I3)  N] + [(P3) + 2N + 2]/2  2
A3 = (I3)  N + (P3)/2 + N + 1  2
A3 = (I3) + (P3)/2  1
So, by induction, the Theorem works for any polygon that can be represented
as the sum of core triangular areas such as we started with. But not all
simple polygons can be represented this way. However, we can represent them
if we are willing to use additive and subtractive core triangles  the proof
that the Theorem remains valid under that extension is left as an exercise
for the reader.
Jason R. Weiss <346LUEK@cmuvm.csv.cmich.edu> wrote in message
<358e8fe1.458779@news.wenet.net>...
>Is Pick's formula based on the relationship between perimeter and area?
>

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