Proof of Pick’s Theorem

Pick’s Theorem states that for any polygon formed on a unit-based grid of points

A = ½b + i – 1

where b is the number of points on the border of the polygon, and i is the number of points in the interior of the polygon.

To prove this theorem, we must first prove that Pick’s Theorem is true for all triangles.  Any triangle can be rotated, reflected, and/or translated in such a way that its area will be preserved and one of its vertices A will be its leftmost, lowest, and on (0, 0), as pictured on ΔABC, with B at (x₁, y₁) and C at (x₂, y₂).  (For triangles with horizontal or vertical sides, some of the coordinates for B and C will be 0, but the following argument still applies.)  Then ΔABD, a triangle congruent to ΔABC and rotated 180°, can be constructed to make ▱ABDC, with D at (x₁ + x₂, y₁ + y₂).  The area of ΔABC would be half the area of ▱ABDC.

Using the left picture below, the area of ▱ABDC would then be the area of ▭AFDI minus the areas of ΔABE and ΔCDJ (the two purple triangles), minus the areas of ΔBDG and ΔACH (the two blue triangles), minus the areas of ▭BEFG and ▭CHIJ (the two blue rectangles).  These shapes can be rearranged as in the right picture below, so that the area of ▱ABDC can be calculated as (x₂y₁ – x₁y₂).  Since the area of ΔABC would be half the area of ▱ABDC, this gives us:

Area of ΔABC = ½(x₂y₁ – x₁y₂)

Algebraically, the area of ΔABC

= ½(Area of ▱ABDC)

= ½(Area of ▭AFDI – 2(Area of ΔABE) – 2(Area of ΔACH) – 2(Area of ▭BEFG))

= ½((x₁ + x₂)(y₁ + y₂) – 2(½x₁y₁) – 2(½x₂y₂) – 2(x₁y₂))

= ½((x₁y₁ + x₁y₂ + x₂y₁ + x₂y₂) – x₁y₁ – x₂y₂ – 2x₁y₂)

= ½((x₁y₁ + x₁y₂ + x₂y₁ + x₂y₂) – x₁y₁ – x₂y₂ – 2x₁y₂)

= ½(x₂y₁ – x₁y₂)

(Note: If B and C were arbitrarily switched, the area of ΔABC would be ½(x₁y₂ – x₂y₁) instead, and since area can never be negative, we can adjust the formula so that the area of ΔABC = ½|x₁y₂ – x₂y₁|.)

Now let us examine Pick’s Theorem in terms of x₁, x₂, y₁, and y₂ and show that it yields the same area formula as A = ½(x₂y₁ – x₁y₂).  Let b₁ be the number of border points on AC, b₂ be the number of border points on AC, and b₃ be the number of border points on BC.  Then the number of border points on ΔABC would be b₁ + b₂ + b₃ – (repeated points in A, B, and C), or

b = b₁ + b₂ + b₃ – 3

(Note: b₁, b₂, and b₃ are equal to the greatest common factor of Δx and Δy of its endpoints plus one, but this is not needed in the proof as they will later get crossed out.)

Using the above left picture, the number of interior points of ▱ABDC would be all the points in ▭AFDI minus all the points in ΔABE and ΔCDJ (the two purple triangles), minus all the points in ΔBDG and ΔACH (the two blue triangles), minus all the points in ▭BEFG and ▭CHIJ (the two blue rectangles) minus all the points in b₃ (the center diagonal).  We would also need to add the points that get repeated in this process, including all the repeated points on segments BE and CJ, all the repeated points on segments BG and CH, and all the repeated points on each vertex of ▱ABDC.  Finally, since the number of interior points of ΔABC would be half the number of interior points of ▱ABDC, we would divide everything by 2.  This means that the number of interior points in ΔABC i is:

= ½(points in ▭AFDI – 2(points in ΔABE) – 2(points in ΔACH) – 2(points in ▭BEFG) – b₃ + 2(repeated points on BE) + 2(repeated points on BG) + (repeated points on each vertex of ▱ABDC))

= ½((x₁ + x₂ + 1)(y₁ + y₂ + 1) – 2(½(x₁ + 1)(y₁ + 1) + ½b₂) – 2(½(x₂ + 1)(y₂ + 1) + ½b₁) – 2(x₁ + 1)(y₂ + 1) – b₃ + 2(x₁ + 1) + 2(y₂ + 1) + 4)

= ½((x₁y₁ + x₁y₂ + x₂y₁ + x₂y₂ + x₁ + x₂ + y₁ + y₂ + 1) – (x₁y₁ + x₁ + y₁ + 1 + b₂) – (x₂y₂ + x₂ + y₂ + 1 + b₁) – (2x₁y₂ + 2x₁ + 2y₂ + 2) – b₃ + (2x₁ + 2) + (2y₂ + 2) + 4)

= ½((x₁y₁ + x₁y₂ + x₂y₁ + x₂y₂ + x₁ + x₂ + y₁ + y₂ + 1) – (x₁y₁ + x₁ + y₁ + 1 + b₂) – (x₂y₂ + x₂ + y₂ + 1 + b₁) – (2x₁y₂ + 2x₁ + 2y₂ + 2) – b₃ + (2x₁ + 2) + (2y₂ + 2) + 4)

= ½(x₂y₁ – x₁y₂ – b₁ – b₂ – b₃ + 5)

which gives us:

i = ½(x₂y₁ – x₁y₂ – b₁ – b₂ – b₃ + 5)

We can now substitute b and i and calculate Pick’s Theorem in terms of x₁, x₂, y₁, and y₂ and verify that it is the same as A = ½(x₂y₁ – x₁y₂).  Pick’s Theorem says that area is:

= ½b + i – 1

= ½(b₁ + b₂ + b₃ – 3) + (½(x₂y₁ – x₁y₂ – b₁ – b₂ – b₃ + 5)) – 1

= (½b₁ + ½b₂ + ½b₃ – ³/₂) + (½x₂y₁ – ½x₁y₂ – ½b₁ – ½b₂ – ½b₃ + ⁵/₂) – 1

= (½b₁ + ½b₂ + ½b₃ – ³/₂) + (½x₂y₁ – ½x₁y₂ – ½b₁ – ½b₂ – ½b₃ + ⁵/₂) – 1

= ½(x₂y₁ – x₁y₂)

which shows that Pick’s Theorem is true for any triangle. 

However, we must show that Pick’s Theorem is true for all polygons as well, not just triangles.  But since any polygon with more than three sides can be subdivided into triangles, all that is left to show is that combining a triangle with another triangle (or another polygon in which Pick’s Theorem is already true) preserves Pick’s Theorem.

When two triangles combine to make a quadrilateral, the areas of the two separate triangles in terms of Pick’s Theorem are A = ½b₁ + i₁ – 1 and A = ½b₂ + i₂ – 1 for a combined area of A = ½(b₁ + b₂) + i₁ + i₂ – 2, where b₁ and i₁ are the number of border and interior points on the first triangle, b₂ and i₂ are the number of border and interior points on the second triangle.  If we let s be the number of border points on the shared side (including the endpoints), the resulting quadrilateral has the same number of border and interior points as the original two triangles except that 2(s – 2) border points change to s – 2  interior points, and the two endpoints are repeated.  So the area of the quadrilateral in terms of Pick’s Theorem is:

= ½(b₁ + b₂ – 2(s – 2) – 2) + (i₁ + i₂ + (s – 2)) – 1

= ½b₁ + ½b₂ – (s – 2) – 1 + i₁ + i₂ + (s – 2) – 1

= ½b₁ + ½b₂ – (s – 2) – 1 + i₁ + i₂ + (s – 2) – 1

= ½(b₁ + b₂) + i₁ + i₂ – 2

= (½b₁ + i₁ – 1) + (½b₂ + i₂ – 1)

= (Area of 1^st Triangle) + (Area of 2^nd Triangle)

The area of the quadrilateral is the same as the area of its two subdivided triangles, so Pick’s Theorem is preserved for combining two triangles.  (The same argument can be made for combining a triangle with another polygon in which Pick’s Theorem is already true.)

Since we have shown that Pick’s Theorem is true for all triangles, and that Pick’s Theorem is preserved for combining a triangle with another triangle (or another polygon in which Pick’s Theorem is already true), and since any polygon with more than three sides can be subdivided into triangles, Pick’s Theorem must be true for all polygons as well.