Pick's Theorem, some history.

 

*Wik

 Georg Pick was a Jewish Austrian mathematician (murdered during The Holocaust).  In 1880 he received his PhD from University of Vienna under Leo Konigsberger. In 1899 he published a formula for finding the area of a polygon if all the vertices are on lattice points (points whose x and y coordinates on the plane are both integers, although if they are only rational, you can adapt the theorem.)  He presented the formula he created in 1899, in Geometries zur Zahlenlehre in Prague.  He worked at University of Prague and was on the appointments committee and pushed hard for Einstein to get a position in physics there. He was also instrumental in introducing Einstein to Ricci-Curbastro and Levi-Civita which helped him work out the mathematics in the general theory of relativity.  

Hugo Steinhaus caused a surge in the popularity (and knowledge) of the theorem when he used it in his popular, Mathematical Snapshots, in 1969.  (see Google n-gram below)It probably would have been much more popular in 1899 if graph paper had been more in mathematics education, but thirty to fifty years would pass before the paper became popular in geometry and algebra classes. I also think the appearance of the Geoboard created by Egyptian Educator Caleb Gattegno in the fifties was important in applications of Pick's theorem in elementary and middle schools.  I don't know if these still are used much, or any, so I popped an image of one below. My first geoboard was a crude piece of wood with a 6x6 array nails driven by a terrible carpenter, then and now, the writer. 

Pick would never be aware of this late surge of popularity, he died after two weeks in the Theresiestadt prison camp in 1942.  

*Art of Problem Solving

The formula gives the area in two variables, N and B .  N is the number of lattice points inside the polygon (many teachers, and some books, use I, for inside), and B is the number on the boundary.  The area is given by Area = N + B/2 - 1.  The example at the right shows N=3 for the inside points, B= 14 for the inside points, so our Area = 3 + 13/2 - 1, or 8 1/2 square units.  You can, of course verify this to yourself by counting connecting lines inside and dividing into squares, rectangles, and triangles that are easy to compute.

There is not a higher dimensional analogy of this theorem, counting points inside and on the boundary, but the Ehrhart Polynomial, created by French high school teacher Eugene Ehrhart in the 60's, describes an expression for the volume in terms of the number of interior points in the polytope and dilations.  

You can find more about the theorem and links to some of its extensions at Drexel University where I also found that the original theorem was published in "Geometrisches zur Zahlenlehre" Sitzungber. Lotos, Naturwissen Zeitschrift Prague, Volume 19 (1899) pages 311-319.  

Here are some more sources for information om the method:  

W. W. Funkenbusch
“From Euler’s Formula to Pick’s Formula using an Edge Theorem”
The American Mathematical Monthly
Volume 81 (1974) pages 647-648

In this short paper Funkenbusch shows that Pick's theorem is derived from Euler's Gem using the theorem that Edges = 3I +2B -3, with I and B for inside lattice points and B for boundary points.

Dale E. Varberg
“Pick’s Theorem Revisited”
The American Mathematical Monthly
Volume 92 (1985) pages 584-587

Varberg extends the theorem to cases with polygons with holes in them.  An interesting read.

Branko Grünbaum and G. C. Shephard
“Pick’s Theorem”
The American Mathematical Monthly
Volume 100 (1993) pages 150-161

For young students, the best part of this is the introduction in which the authors describe an applied mathematical approach to timber management, which unknown to the speaker, was Pick's Theorem.

My favorite article to date is by an Eighth grade student, Chris Polis, from Papua, New Guinea in 1991.  He found generalizations of Pick's Thm to several different lattices, Triangular, hexagonal, and a tesselation of the plain by isosceles trapezoids.  He found individual formulas for these, then found a general one for all of them including Pick's Thm, \( A = \frac{e + 2 (i - 1)}{P-2}\) where P is the number of edges in the lattice, e is boundary points and i is interior points.  For a square this would give P=4 and the formula becomes \( A = \frac{e + 2 (i - 1)}{4-2}\) which simplifies to Pick's Thm.  (Wonder what he is doing today?)

You can use this Desmos lattice I plucked from google to try the triangular formula with the 9 boundary points on the edge and 3 in the interior, and confirm that the interior has 13 unit equilateral triangles in area.  

I recently saw another use of the theorem on twitter to answer the following question.  Several people found the right answer with trig, but one saw a different easy solution.  

Draw the lattice and use Pick's Theorem.  

Did you notice the triangle is a right triangle?  

Recently I came across an extension of Pick's theorem that would cover shapes with holes and such that make the original theorem not work (although in many cases a little common sense will still solve them.)  

.The new approach was in the article "Counting Parallel Segments: New Variants of Pick’s Area Theorem" by Alexander Belyaev & Pierre-Alain Fayolle.  I found it in the Mathematical Intelligencer ,Volume 41, pages 1–7, (2019).  

The method counts parallel line segments and parts of segments to find the area of such non-traditional figures.  

The line segments in dark blue are called interior segments,  They contribute one unit each to the area.  In the figure (a) there are eight of them.  The red line segments are called exterior segments, and they add 1/2 a unit.  The light blue segments are partial segments and also count 1/2 unit each. [Notice in the one passing through an interior point (blue) there are two partial segments, one above and one below. ]  So the area would be 8 + 2/2 +4/2 = 11. sq units

Try the (b) image for yourself.  I'll put the answer below after a brief side note; do take time to try one diagram that does work with the regular Pick method, such as the one near the beginning of this article.  It should work for them also.

How many red(edge) lines did you use?  How many drk blue(interior) segments ?  How many partials?

I had two reds on the left edge, two drk blue and one partial on the next vertical row, four drk blue on the third vertical row, and one drk blue on the fourth edge.  That should add up to the same 8 1/2 units as we got the "Pick" way.

OK  so did you get 3 + 2/2 + 4/2 for a total of six square units for the figure (b)?

I think this is an interesting method, and the parallels Don't have to be vertical, horizontal or diagonal at any slope.  try those and see if you get the same area each time.

Good luck!