Pick's theorem: Difference between revisions

Revision as of 20:55, 12 December 2004

Latest revision as of 15:37, 29 July 2025

(387 intermediate revisions by more than 100 users not shown)

Line 1:

{{good article}}

[[Category:Discrete geometry]] [[Category:Theorems]]

{{short description|Formula for area of a grid polygon}}

{{for|the theorem in complex analysis|Schwarz lemma#Schwarz–Pick theorem}}

[[File:Farey_sunburst_6.svg|thumb|[[Farey sunburst]] of order 6, with 1 interior {{color|red|(red)}} and 96 boundary {{color|green|(green)}} points giving an area of {{nowrap|{{color|red|1}} + {{sfrac|{{color|green|96}}|2}} − 1 {{=}} 48}}{{r|kiradjiev}}]]

In [[geometry]], '''Pick's theorem''' provides a formula for the [[area]] of a [[simple polygon]] with integer [[vertex (geometry)|vertex]] coordinates, in terms of the number of integer points within it and on its boundary. The result was first described by [[Georg Alexander Pick]] in 1899.{{r|pick}} It was popularized in English by [[Hugo Steinhaus]] in the 1950 edition of his book ''Mathematical Snapshots''.{{r|gs|steinhaus}} It has multiple proofs, and can be generalized to formulas for certain kinds of non-simple polygons.

==Formula==

<table align="right"><tr><td>[[image:picks_polygon.png|polygon constructed on a grid of equal-distanced grid points]]</td></tr></table>

[[File:Pick_theorem_simple.svg|thumb|upright=0.6|{{color|red|{{math|''i'' {{=}} 7}}}}, {{color|green|{{math|''b'' {{=}} 8}}}}, {{math|''A'' {{=}} {{color|red|''i''}} + {{sfrac|{{color|green|''b''}}|2}} − 1 {{=}} 10}}]]

Given a [[polygon|simple polygon]] constructed on a grid of equal-distanced points (i.e., points with [[integer]] coordinates) such that all the polygon's vertices are grid points,

'''Pick'sSuppose theorem''' providesthat a simplepolygon has [[formula]]integer forlattice|integer calculating the [[areacoordinates]] ''A''for all of thisits polygonvertices. inLet terms<math>i</math> ofbe the number ''i'' of ''interiorinteger points'' ofinterior to the polygon, and let <math>b</math> be the number ''b'' of ''boundaryinteger points'' on its boundary (including both vertices and points along the sides). Then the [[area]] <math>A</math> of this polygon is:{{r|az|wells|discretely|ball}}

<math display=block>A = i + \frac{b}{2} - 1.</math>

:''A'' = ''i'' + &frac12;''b'' &minus; 1.

The example shown has <math>i=7</math> interior points and <math>b=8</math> boundary points, so its area is <math>A=7+\tfrac{8}{2}-1=10</math> square units.

==Proofs==

In the example shown, we have ''i'' = 9 and ''b'' = 14, so the area is ''A'' = 9 + &frac12;(14) &minus; 1 = 9 + 7 &minus; 1 = 15 (square units).

===Via Euler's formula===

One proof of this theorem involves subdividing the polygon into triangles with three integer vertices and no other integer points. One can then prove that each subdivided triangle has area exactly <math>\tfrac{1}{2}</math>. Therefore, the area of the whole polygon equals half the number of triangles in the subdivision. After relating area to the number of triangles in this way, the proof concludes by using [[Euler's polyhedral formula]] to relate the number of triangles to the number of grid points in the polygon.{{r|az}}

[[File:Pick triangle tessellation.svg|thumb|Tiling of the plane by copies of a triangle with three integer vertices and no other integer points, as used in the proof of Pick's theorem]]

This formula so simple that it has been correctly used by first-grade children, drawing figures on square tiles on floor or wall, or stretching strings from pegs in pegboard. They learn how to add, along with subtraction as "take away". They learn to "halve" by a one-to-one correspondence between counters.

The first part of this proof shows that a triangle with three integer vertices and no other integer points has area exactly <math>\tfrac{1}{2}</math>, as Pick's formula states. The proof uses the fact that all triangles [[tessellation|tile the plane]], with adjacent triangles rotated by 180° from each other around their shared edge.{{r|edward}} For tilings by a triangle with three integer vertices and no other integer points, each point of the integer grid is a vertex of six tiles. Because the number of triangles per grid point (six) is twice the number of grid points per triangle (three), the triangles are twice as dense in the plane as the grid points. Any scaled region of the plane contains twice as many triangles (in the limit as the scale factor goes to infinity) as the number of grid points it contains. Therefore, each triangle has area <math>\tfrac{1}{2}</math>, as needed for the proof.{{r|az}} A different proof that these triangles have area <math>\tfrac{1}{2}</math> is based on the use of [[Minkowski's theorem]] on lattice points in symmetric convex sets.{{r|minkowski}}

[[File:Grid polygon triangulation.svg|thumb|upright|Subdivision of a grid polygon into special triangles]]

Note that the theorem as stated above is only valid for ''simple'' polygons, i.e. ones that consist of a single piece and do not contain "holes". For more general polygons, the "&minus; 1" of the formula has to be replaced with "&minus; &chi;(''P'')", where &chi;(''P'') is the [[Euler characteristic]] of ''P''.

This already proves Pick's formula for a polygon that is one of these special triangles. Any other polygon can be subdivided into special triangles: add non-crossing line segments within the polygon between pairs of grid points until no more line segments can be added. The only polygons that cannot be subdivided in this way are the special triangles considered above; therefore, only special triangles can appear in the resulting subdivision. Because each special triangle has area <math>\tfrac{1}{2}</math>, a polygon of area <math>A</math> will be subdivided into <math>2A</math> special triangles.{{r|az}}

The subdivision of the polygon into triangles forms a [[planar graph]], and Euler's formula <math>V-E+F=2</math> gives an equation that applies to the number of vertices, edges, and faces of any planar graph. The vertices are just the grid points of the polygon; there are <math>V=i+b</math> of them. The faces are the triangles of the subdivision, and the single region of the plane outside of the polygon. The number of triangles is <math>2A</math>, so altogether there are <math>F=2A+1</math> faces. To count the edges, observe that there are <math>6A</math> sides of triangles in the subdivision. Each edge interior to the polygon is the side of two triangles. However, there are <math>b</math> edges of triangles that lie along the polygon's boundary and form part of only one triangle. Therefore, the number of sides of triangles obeys the equation <math>6A=2E-b</math>, from which one can solve for the number of edges, <math>E=\tfrac{6A+b}{2}</math>. Plugging these values for <math>V</math>, <math>E</math>, and <math>F</math> into Euler's formula <math>V-E+F=2</math> gives

The result was first described by [[Georg Pick]] in [[1899]]. It can be generalized to three dimensions and higher by [[Ehrhart polynomial]]s. The formula also generalizes to surfaces of [[polyhedron|polyhedra]].

<math display=block>(i+b) - \frac{6A+b}{2} + (2A+1) = 2.</math>

Pick's formula is obtained by solving this [[linear equation]] for <math>A</math>.{{r|az}} An alternative but similar calculation involves proving that the number of edges of the same subdivision is <math>E=3i+2b-3</math>, leading to the same result.{{r|funkenbusch}}

It is also possible to go the other direction, using Pick's theorem (proved in a different way) as the basis for a proof of Euler's formula.{{r|wells|equivalence}}

===Proof===

===Other proofs===

Consider a polygon ''P'' and a triangle ''T'', with one edge in common with ''P''. Assume Pick's theorem is true for ''P''; we want to show that it is also true to the polygon ''PT'' obtained by adding ''T'' to ''P''. Since ''P'' and ''T'' share an edge, all the boundary points along the edge in common are merged to interior points, except for the two endpoints of the edge, which are merged to boundary points. So, calling the number of boundary points in common ''c'', we have ''i_PT'' = (''i_P'' + ''i_T'') + (''c'' &minus; 2) and ''b_PT'' = (''b_P'' + ''b_T'') &minus; 2(''c'' &minus; 2) &minus; 2.

Alternative proofs of Pick's theorem that do not use Euler's formula include the following.

*One can recursively decompose the given polygon into triangles, allowing some triangles of the subdivision to have area larger than 1/2. Both the area and the counts of points used in Pick's formula add together in the same way as each other, so the truth of Pick's formula for general polygons follows from its truth for triangles. Any triangle subdivides its [[bounding box]] into the triangle itself and additional [[right triangle]]s, and the areas of both the bounding box and the right triangles are easy to compute. Combining these area computations gives Pick's formula for triangles, and combining triangles gives Pick's formula for arbitrary polygons.{{r|discretely|ball|varberg}}

*Alternatively, instead of using grid squares centered on the grid points, it is possible to use grid squares having their vertices at the grid points. These grid squares cut the given polygon into pieces, which can be rearranged (by matching up pairs of squares along each edge of the polygon) into a [[polyomino]] with the same area.{{r|trainin}}

*Pick's theorem may also be proved based on [[complex integration]] of a [[doubly periodic function]] related to [[Weierstrass's elliptic functions|Weierstrass elliptic functions]].{{r|diaz-robins}}

*Applying the [[Poisson summation formula]] to the [[characteristic function]] of the polygon leads to another proof.{{r|bcrt}}

Pick's theorem was included in a 1999 web listing of the "top 100 mathematical theorems", which later became used by Freek Wiedijk as a [[Benchmark (computing)|benchmark]] set to test the power of different [[proof assistant]]s. {{as of|2024}}, Pick's theorem had been formalized and proven in only two of the ten proof assistants recorded by Wiedijk.{{r|wiedijk}}

From the above follows (''i_P'' + ''i_T'') = ''i_PT'' &minus; (''c'' &minus; 2) and (''b_P'' + ''b_T'') = ''b_PT'' + 2(''c'' &minus; 2) + 2.

==Generalizations==

Since we are assuming the theorem for ''P'' and for ''T'' separately,

[[File:Pick_theorem_hole.svg|thumb|upright=0.6|{{color|red|{{math|''i'' {{=}} 2}}}}, {{color|green|{{math|''b'' {{=}} 12}}}}, {{color|grey|{{math|''h'' {{=}} 1}}}}, {{math|''A'' {{=}} {{color|red|''i''}} + {{sfrac|{{color|green|''b''}}|2}} + {{color|grey|''h''}} − 1 {{=}} 8}}]]

Generalizations to Pick's theorem to non-simple polygons are more complicated and require more information than just the number of interior and boundary vertices.{{r|gs|rosenholtz}} For instance, a [[Polygon with holes|polygon with {{mvar|h}} holes]] bounded by simple integer polygons, disjoint from each other and from the boundary, has area{{r|sankri}}

<math display=block>A = i + \frac{b}{2} + h - 1.</math>

It is also possible to generalize Pick's theorem to regions bounded by more complex [[planar straight-line graph]]s with integer vertex coordinates, using additional terms defined using the [[Euler characteristic]] of the region and its boundary,{{r|rosenholtz}} or to polygons with a single boundary polygon that can cross itself, using a formula involving the [[winding number]] of the polygon around each integer point as well as its total winding number.{{r|gs}}

[[File:reeve_tetrahedrons.svg|thumb|upright=1.25|Reeve tetrahedra showing that Pick's theorem does not apply in higher dimensions]]

:''A_PT'' = ''A_P'' + ''A_T''

The [[Reeve tetrahedra]] in three dimensions have four integer points as vertices and contain no other integer points, but do not all have the same volume. Therefore, there does not exist an analogue of Pick's theorem in three dimensions that expresses the volume of a polyhedron as a function only of its numbers of interior and boundary points.{{r|reeve}} However, these volumes can instead be expressed using [[Ehrhart polynomial]]s.{{r|br2|ehrhart}}

:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= ''i_P'' + &frac12;''b_P'' &minus; 1 + ''i_T'' + &frac12;''b_T'' &minus; 1

:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= (''i_P'' + ''i_T'') + &frac12;(''b_P'' + ''b_T'') &minus; 2

:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= ''i_PT'' &minus; (''c'' &minus; 2) + &frac12;(''b_PT'' + 2(''c'' &minus; 2) + 2) &minus; 2

:&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;= ''i_PT'' + &frac12;''b_PT'' &minus; 1.

==Related topics==

Therefore, if the theorem is true for polygons constructed from ''n'' triangles, the theorem is also true for polygons constructed from ''n''+1 triangles.

Several other mathematical topics relate the areas of regions to the numbers of grid points. [[Blichfeldt's theorem]] states that every shape can be translated to contain at least its area in grid points.{{r|olds}} The [[Gauss circle problem]] concerns bounding the error between the areas and numbers of grid points in circles.{{r|guy}} The problem of counting [[integer points in convex polyhedra]] arises in several areas of mathematics and computer science.{{r|barvinok}}

To finish the proof by [[mathematical induction]], it remains to show that the theorem is true for triangles.

In application areas, the [[dot planimeter]] is a transparency-based device for estimating the area of a shape by counting the grid points that it contains.{{r|bellhouse}} The [[Farey sequence]] is an ordered sequence of rational numbers with bounded denominators whose analysis involves Pick's theorem.{{r|bruarc}}

The verification for this case can be done in these short steps:

Another simple method for calculating the area of a polygon is the [[shoelace formula]]. It gives the area of any simple polygon as a sum of terms computed from the coordinates of consecutive pairs of its vertices. Unlike Pick's theorem, the shoelace formula does not require the vertices to have integer coordinates.{{r|braden}}

* directly check that the formula is correct for any [[rectangle]] with sides [[parallel]] to the axes;

* verify from that case that it works for right-angled triangles obtained by cutting such rectangles along a [[diagonal]];

* now any triangle can be turned into a rectangle by attaching (at most three) such right triangles; since the formula is correct for the right triangles and for the rectangle, it also follows for the original triangle.

==References==

The last step uses the fact that if the theorem is true for the polygon ''PT'' and for the triangle ''T'', then it's also true for ''P''; this can be seen by a calculation very much similar to the one shown above.

{{reflist|refs=

<ref name=az>{{cite book | last1 = Aigner | first1 = Martin | author1-link = Martin Aigner | last2 = Ziegler | first2 = Günter M. | author2-link = Günter M. Ziegler | contribution = Three applications of Euler’s formula: Pick's theorem | doi = 10.1007/978-3-662-57265-8 | edition = 6th | isbn = 978-3-662-57265-8 | pages = 93–94 | publisher = Springer | title = Proofs from THE BOOK | title-link = Proofs from THE BOOK | year = 2018}}</ref>

To prove, we shall first show that Pick's theorem has an additive character. Suppose our polygon has more than 3 vertices. Then we can divide the polygon ''P'' into 2 polygons ''P''₁ and ''P''₂ such that their interiors do not meet. Both have fewer vertices than ''P''.

We claim that the validity of Pick's theorem for ''P'' is equivalent to the validity of Pick's theorem for ''P''₁ and ''P''₂.

<ref name=ball>{{cite book | last = Ball | first = Keith | author-link = Keith Martin Ball | contribution = Chapter 2: Counting Dots | isbn = 0-691-11321-1 | mr = 2015451 | pages = 25–40 | publisher = Princeton University Press, Princeton, NJ | title = Strange Curves, Counting Rabbits, and Other Mathematical Explorations | year = 2003}}</ref>

Denote the area, number of interior lattice points and number of boundary lattice points for ''P''_''k'' by ''A''_''k'', ''I''_''k'' and ''O''_''k'', respectively, for ''k'' = 1, 2.

<ref name=barvinok>{{cite book

Clearly <math>A = A_1 + A_2.</math>

| last = Barvinok | first = Alexander | authorlink = Alexander Barvinok

| doi = 10.4171/052

| isbn = 978-3-03719-052-4

| mr = 2455889

| publisher = European Mathematical Society | ___location = Zürich

| series = Zurich Lectures in Advanced Mathematics

| title = Integer Points In Polyhedra

| year = 2008| volume = 11 }}</ref>

<ref name=bcrt>{{cite journal

Also, if we denote the number of lattice points on the edges common to ''P''₁ and ''P''₂ by ''L'', then

| last1 = Brandolini | first1 = L.

| last2 = Colzani | first2 = L.

| last3 = Robins | first3 = S.

| last4 = Travaglini | first4 = G.

| doi = 10.1080/00029890.2021.1839241

| issue = 1

| journal = [[The American Mathematical Monthly]]

| mr = 4200451

| pages = 41–49

| title = Pick's theorem and convergence of multiple Fourier series

| volume = 128

| year = 2021| s2cid = 231624428

}}</ref>

:<math>I = I_1 + I_2 + L - 2</math>

<ref name=bellhouse>{{cite journal

and

| last = Bellhouse | first = D. R.

| doi = 10.2307/2530419

| issue = 2

| journal = Biometrics

| jstor = 2530419

| mr = 673040

| pages = 303–312

| title = Area estimation by point-counting techniques

| volume = 37

| year = 1981}}</ref>

<ref name=braden>{{cite journal|first=Bart|last=Braden|title=The surveyor's area formula|journal=The College Mathematics Journal|volume=17|issue=4|year=1986|pages=326–337|url=http://www.maa.org/sites/default/files/pdf/pubs/Calc_Articles/ma063.pdf|doi=10.2307/2686282|jstor=2686282|access-date=2021-07-04|archive-date=2015-04-06|archive-url=https://web.archive.org/web/20150406152731/http://www.maa.org/sites/default/files/pdf/pubs/Calc_Articles/ma063.pdf|url-status=dead}}</ref>

:<math>O = O_1 + O_2 -2L + 2.</math>

<ref name=bruarc>{{cite journal

Hence

| last1 = Bruckheimer | first1 = Maxim

| last2 = Arcavi | first2 = Abraham

| doi = 10.1007/BF03024792

| issue = 4

| journal = [[The Mathematical Intelligencer]]

| mr = 1365013

| pages = 64–67

| title = Farey series and Pick's area theorem

| volume = 17

| year = 1995| s2cid = 55051527

}}</ref>

<ref name=diaz-robins>{{cite journal

:<math>I + \frac{1}{2}O - 1 = I_1 + I_2 + L - 2 + \frac{1}{2}O_1 + \frac{1}{2}O_2 - L + 1 - 1</math>

| last1 = Diaz | first1 = Ricardo

| last2 = Robins | first2 = Sinai

| doi = 10.2307/2975035

| issue = 5

| journal = [[The American Mathematical Monthly]]

| jstor = 2975035

| mr = 1327788

| pages = 431–437

| title = Pick's formula via the Weierstrass {{nowrap|<math>\wp</math>-function}}

| volume = 102

| year = 1995}}</ref>

<ref name=discretely>{{cite book | last1 = Beck | first1 = Matthias | last2 = Robins | first2 = Sinai | contribution = 2.6 Pick's theorem | doi = 10.1007/978-1-4939-2969-6 | edition = 2nd | isbn = 978-1-4939-2968-9 | mr = 3410115 | pages = 40–43 | publisher = Springer | series = Undergraduate Texts in Mathematics | title = Computing the Continuous Discretely: Integer-Point Enumeration in Polyhedra | year = 2015}}</ref>

:<math>= I_1 + \frac{1}{2}O_1 -1 + I_2 + \frac{1}{2}O_2 -1.</math>

<ref name=br2>{{harvtxt|Beck|Robins|2015}}, 3.6 "From the discrete to the continuous volume of a polytope", pp. 76–77</ref>

<ref name=edward>{{cite book|last=Martin|first=George Edward|doi=10.1007/978-1-4612-5680-9|isbn=0-387-90636-3|mr=718119|at=Theorem 12.1, page 120|publisher=Springer-Verlag|series=Undergraduate Texts in Mathematics|title=Transformation geometry|url=https://books.google.com/books?id=gevlBwAAQBAJ&pg=PA120|year=1982}}</ref>

This proves the claim. Therefore we can triangulate ''P'' and it suffices to prove Pick's theorem

<ref name=ehrhart>{{cite journal | last1 = Diaz | first1 = Ricardo | last2 = Robins | first2 = Sinai | doi = 10.2307/2951842 | issue = 3 | journal = [[Annals of Mathematics]] | mr = 1454701 | pages = 503–518 | series = Second Series | title = The Ehrhart polynomial of a lattice polytope | volume = 145 | year = 1997| jstor = 2951842 }}</ref>

==External links==

<ref name=equivalence>{{cite journal

* [http://www.cut-the-knot.org/ctk/Pick.shtml Pick's Theorem (Java)]

| last1 = DeTemple | first1 = Duane

| last2 = Robertson | first2 = Jack M.

| date = March 1974

| doi = 10.5951/mt.67.3.0222

| issue = 3

| journal = [[The Mathematics Teacher]]

| jstor = 27959631

| mr = 0444503

| pages = 222–226

| title = The equivalence of Euler's and Pick's theorems

| volume = 67}}</ref>

<ref name=funkenbusch>{{cite journal

| last = Funkenbusch | first = W. W.

| department = Classroom Notes

| date = June–July 1974

| doi = 10.2307/2319224

| issue = 6

| journal = [[The American Mathematical Monthly]]

| jstor = 2319224

| mr = 1537447

| pages = 647–648

| title = From Euler's formula to Pick's formula using an edge theorem

| volume = 81}}</ref>

<ref name=gs>{{cite journal | last1 = Grünbaum | first1 = Branko | author1-link = Branko Grünbaum | last2 = Shephard | first2 = G. C. | author2-link = Geoffrey Colin Shephard | date = February 1993 | doi = 10.2307/2323771 | issue = 2 | journal = [[The American Mathematical Monthly]] | jstor = 2323771 | pages = 150–161 | title = Pick's theorem | volume = 100 | mr = 1212401}}</ref>

<ref name=guy>{{cite book

| last = Guy | first = Richard K. | author-link = Richard K. Guy

| contribution = F1: Gauß's lattice point problem

| doi = 10.1007/978-0-387-26677-0

| edition = 3rd

| isbn = 0-387-20860-7

| mr = 2076335

| pages = 365–367

| publisher = Springer-Verlag | ___location = New York

| series = Problem Books in Mathematics

| title = Unsolved Problems in Number Theory

| year = 2004| volume = 1 }}</ref>

<ref name=kiradjiev>{{cite magazine|url=http://www.maths.ox.ac.uk/system/files/attachments/ECMPick.pdf|title=Connecting the dots with Pick's theorem|first=Kristian|last=Kiradjiev|magazine=Mathematics Today|date=October 2018|pages=212–214}}</ref>

<ref name=minkowski>{{cite journal

| last1 = Ram Murty | first1 = M.

| last2 = Thain | first2 = Nithum

| doi = 10.1080/00029890.2007.11920465

| issue = 8

| journal = [[The American Mathematical Monthly]]

| jstor = 27642309

| mr = 2354443

| pages = 732–736

| title = Pick's theorem via Minkowski's theorem

| volume = 114

| year = 2007| s2cid = 38855683

}}</ref>

<ref name=olds>{{cite book

| last1 = Olds | first1 = C. D. | author1-link = Carl D. Olds

| last2 = Lax | first2 = Anneli | author2-link = Anneli Cahn Lax

| last3 = Davidoff | first3 = Giuliana P. | author3-link = Giuliana Davidoff

| contribution = Chapter 9: A new principle in the geometry of numbers

| isbn = 0-88385-643-3

| mr = 1817689

| pages = 119–127

| publisher = Mathematical Association of America, Washington, DC

| series = Anneli Lax New Mathematical Library

| title = The Geometry of Numbers

| title-link = The Geometry of Numbers

| volume = 41

| year = 2000}}</ref>

<ref name=pick>{{cite journal |last=Pick |first=Georg | author-link = Georg Alexander Pick |title=Geometrisches zur Zahlenlehre |journal=Sitzungsberichte des deutschen naturwissenschaftlich-medicinischen Vereines für Böhmen "Lotos" in Prag |series=(Neue Folge) |year=1899 |volume=19 |pages=311–319 |url=https://www.biodiversitylibrary.org/item/50207#page/327 |jfm=33.0216.01 }} [http://citebank.org/node/47270 CiteBank:47270]</ref>

<ref name=reeve>{{cite journal | last = Reeve | first = J. E. | doi = 10.1112/plms/s3-7.1.378 | journal = [[Proceedings of the London Mathematical Society]] | mr = 0095452 | pages = 378–395 | series = Third Series | title = On the volume of lattice polyhedra | volume = 7 | year = 1957}}</ref>

<ref name=rosenholtz>{{cite journal

| last = Rosenholtz | first = Ira

| doi = 10.1080/0025570X.1979.11976797

| issue = 4

| journal = [[Mathematics Magazine]]

| jstor = 2689425

| mr = 1572312

| pages = 252–256

| title = Calculating surface areas from a blueprint

| volume = 52

| year = 1979}}</ref>

<ref name=sankri>{{cite journal

| last1 = Sankar | first1 = P. V.

| last2 = Krishnamurthy | first2 = E. V.

| date = August 1978

| doi = 10.1016/s0146-664x(78)80021-5

| issue = 1

| journal = Computer Graphics and Image Processing

| pages = 136–143

| title = On the compactness of subsets of digital pictures

| volume = 8}}</ref>

<ref name=steinhaus>{{cite book

| last = Steinhaus | first = H. | author-link = Hugo Steinhaus

| mr = 0036005

| page = 76

| publisher = Oxford University Press

| title = Mathematical Snapshots

| year = 1950}}</ref>

<ref name=trainin>{{cite journal |last=Trainin |first=J. |title=An elementary proof of Pick's theorem |journal=[[The Mathematical Gazette]] |volume=91 |issue=522 |date=November 2007 |pages=536–540|doi=10.1017/S0025557200182270 | jstor = 40378436|s2cid=124831432 }}</ref>

<ref name=varberg>{{cite journal

| last = Varberg | first = Dale E.

| doi = 10.2307/2323172

| issue = 8

| journal = [[The American Mathematical Monthly]]

| jstor = 2323172

| mr = 812105

| pages = 584–587

| title = Pick's theorem revisited

| volume = 92

| year = 1985}}</ref>

<ref name=wells>{{cite book | last = Wells | first = David | title = The Penguin Dictionary of Curious and Interesting Geometry | publisher = Penguin Books | year = 1991 | contribution = Pick's theorem | pages = 183–184}}</ref>

<ref name=wiedijk>{{cite web|first=Freek|last=Wiedijk|url=https://www.cs.ru.nl/~freek/100/|title=Formalizing 100 Theorems|publisher=Radboud University Institute for Computing and Information Sciences|access-date=2024-02-20}}</ref>

}}

==External links==

{{commons category}}

* [http://demonstrations.wolfram.com/PicksTheorem/ Pick's Theorem] by [[Ed Pegg, Jr.]], the [[Wolfram Demonstrations Project]].

* [https://www.geogebra.org/m/y2nuDV37 Pi using Pick's Theorem] by Mark Dabbs, [[GeoGebra]]

[[plCategory:WzórDigital Pickageometry]]

[[Category:Lattice points]]

[[Category:Euclidean plane geometry]]

[[Category:Area]]

[[Category:Theorems about polygons]]

[[Category:Articles containing proofs]]

[[Category:Analytic geometry]]