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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>
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>Rumus Pick
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 ===
Bukti-bukti teorema Pick lain tanpa menggunakan rumus Euler, diantaranya sebagai berikut:
* 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}}
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