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{{Periksa terjemahan|en|Pick's theorem}}
{{for|the theorem in complex analysis|Schwarz lemma#Schwarz–Pick theorem}}
[[Berkas:Pick-theorem.svg|jmpl|{{color|red|{{math|''i'' {{=}} 7}}}}, {{color|green|{{math|''b'' {{=}} 8}}}}, {{math|''A'' {{=}} {{color|red|''i''}} + {{sfrac|{{color|green|''b''}}|2}} − 1 {{=}} 10}}]]
Dalam [[geometri]], '''teorema Pick''' merupakan sebuah rumus luas [[poligon sederhana]] dengan koordinat simpul berupa bilangan bulat dengan menjumlahkan titik-titik bilangan bulat dalam poligon dan batasnya. Hasil teorema ini dijelaskan pertama kali oleh [[Georg Alexander Pick]] pada tahun 1899.
== Rumus ==
Tinjau bahwa sebuah poligon memiliki koordinat bilangan bulat untuk semua simpul pada poligon. Misalkan <math>i</math> adalah jumlah titik bilangan bulat yang ada di dalam poligon, dan misalkan <math>b</math> adalah jumlah titik bilangan bulat pada batas poligon (termasuk verteks dan juga titik di sisi-sisi poligon). Maka, luas poligon <math>A</math> adalah:
:
Contoh pada gambar di atas menunjukkan bahwa titik dalam dan titik luar pada poligon adalah <math>i = 7</math> dan <math>b = 8</math>, sehingga luas poligonnya adalah <math display="inline">A = 7 + \frac{8}{2} - 1 = 10</math> satuan persegi.
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=== Melalui rumus Vieta ===
Salah satu bukti teorema ini melibatkan subpembagian poligon menjadi menjadi segitiga dengan tiga verteks bilangan bulat dan tidak ada titik bilangan bulat lain. Lalu, rumus ini dapat membuktikan bahwa setiap subpembagian segitiga memiliki luas setidaknya <math>\tfrac{1}{2}</math>. Oleh karena itu, luas seluruh poligon sama dengan setengah jumlah segitiga yang dibagi. Setelah mengaitkan luas dengan jumlah segitiga, bukti teorema ini dapat diselesaikan dengan mengaitkan jumlah segitiga dengan jumlah titik kisi dalam poligon melalui [[rumus polihedron Euler]].{{r|az}}
[[Berkas:Pick_triangle_tessellation.svg|jmpl|Pengubinan bidang melalui salinan segitiga dengan tiga simpul bilangan bulat dan tidak ada titik bilangan bulat lain. Ini dipakai dalam membuktikan teorema Pick.]]
Bagian pertama mengenai bukti ini memperlihatkan bahwa segitiga dengan tiga verteks bilangan bulat dan tidak ada titik bilangan bulat lain memiliki setidaknya <math>\tfrac{1}{2}</math>, seperti yang dijelaskan melalui rumus Pick. Faktanya, bukti ini menggunakan semua segitiga yang [[Teselasi|mengubin di bidang]]<u>,</u> dengan segitiga yang berdampingan berputar 180° <u>from each other around their shared edge.</u>{{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}}
[[Berkas:Grid_polygon_triangulation.svg|jmpl|Subdivision of a grid polygon into special triangles]]
Bukti ini sudah membuktikan rumus Pick untuk poligon yang merupakan salah satu dari segitiga-segitiga khusus tersebut. Suatu poligon lain dapat dibagi lagi menjadi segitiga khusus. <u>To do so, add non-crossing line segments within the polygon between pairs of grid points until no more line segments can be added.</u> The only polygons that cannot be subdivided into smaller shapes 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}}
Poligon yang dapat dibagi menjadi segitiga membentuk [[graf planar]], dan rumus Euler <math>V - E + F = 2</math> memberikan persamaan yang berlaku untuk jumlah simpul, tepi dan wajah suatu poligon. Simpul poligon tersebut hanya berupa jumlah kisi dari poligon, yang berjumlahkan <math>V = i + b</math>. 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 boundary of the polygon, and form part of only one triangle. Therefore, the number of sides of triangles obeys an 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<math display="block">(i+b) - \frac{6A+b}{2} + (2A+1) = 2.</math>Pick's formula can be obtained by simplifying this [[linear equation]] and solving for <math>A</math>.{{r|az}} An alternative calculation along the same lines 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}}
=== Bukti lainnya ===
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|right triangles]], 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]].{{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 web listing of the "top 100 mathematical theorems", dating from 1999, which later became used by Freek Wiedijk as a [[Benchmark (computing)|benchmark]] set to test the power of different [[Proof assistant|proof assistants]]. {{as of|2021}}, a proof of Pick's theorem had been formalized in only one of the ten proof assistants recorded by Wiedijk.{{r|wiedijk}}
== Perumuman ==
Generalizations to Pick's theorem to non-simple polygons are possible, but are more complicated and require more information than just the number of interior and boundary vertices.{{r|gs|rosenholtz}} For instance, a polygon with <math>h</math> 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|planar straight-line graphs]] 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}}
The [[Reeve tetrahedron|Reeve tetrahedra]] in three dimensions have four integer points as vertices and contain no other integer points. However, they do not all have the same volume as each other. Therefore, there can be no analogue of Pick's theorem in three dimensions that expresses the volume of a polytope as a function only of its numbers of interior and boundary points.{{r|reeve}} However, these volumes can instead be expressed using [[Ehrhart polynomial|Ehrhart polynomials]].{{r|br2|ehrhart}}
== Topik yang berkaitan ==
Ada beberapa topik dalam matematika yang mengaitkan luas daerah dengan jumlah titik kisi, di antaranya: [[teorema Blichfeldt]], yang mengatakan bahwa setiap bentuk yang dapat ditranslasikan memiliki setidaknya luas bentuk tersebut dalam titik kisi;
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 vertices of the polygon. Unlike Pick's theorem, it does not require the vertices to have integer coordinates.{{r|braden}}
== Rujukan ==
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== Pranala luar ==
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