Pick’s Theorem

Pick’s Theorem, first described by the Jewish-Austrian mathematician Georg Alexander Pick in 1899, finds the area of any polygon formed on a unit-based grid of points.  It states that

                                                                                                                                  

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.

Here are a few examples of Pick’s Theorem applied to some polygons:

Pick’s Theorem Examples
b = 12 i = 4 A = ½b + i – 1 A = ½·12 + 4 – 1 = 9	b = 9 i = 1 A = ½b + i – 1 A = ½·9 + 1 – 1 = 4½	b = 10 i = 2 A = ½b + i – 1 A = ½·10 + 2 – 1 = 6

These three examples can all be verified by common area formulas.  The first example, the square, has an area of A = s², so A = 3² = 9.  The second example, the triangle, has an area of A = ½bh, so A = ½·3·3 = 4½.  The last example, the house-shaped polygon, is made up of 5 unit squares for the bottom and 2 half unit squares for the roof, so A = 5(1) + 2(½) = 6.  Pick’s Theorem was able to correctly calculate the area of each of these shapes.

Restrictions

Amazingly, Pick’s Theorem works for any polygon as long as the following restrictions are met:

1)  All vertices must be on one of the unit grid points.

2)   All border points must connect to exactly two segments.

3)  There are no holes.

4)  There are no curved edges.

All of these restrictions, except for the first, are derived by the very definition of a polygon itself.

Pick’s Theorem Restrictions
All vertices must be on  one of unit grid points.	All border points must connect to two segments.
No holes.	No curves.

Proof

The proof for Pick’s Theorem is quite involved.  The first step is to show that Pick’s Theorem is true for any triangle using variable coordinates.  The second step is to show that combining a triangle with another triangle (or another polygon in which Pick’s Theorem is already true) preserves Pick’s Theorem.  Since any polygon with more than three sides can be subdivided into triangles, this shows that Pick’s Theorem is true for any polygon as well.  The complete proof can be found here.

                                                                                           

Variation of Pick’s Theorem – Triangular Grid

                 

There are several variations to Pick’s Theorem that are also worth mentioning, and the first variation is to use a triangular grid of points instead of a square grid of points.  Surprisingly, the formula for Pick’s Theorem remains nearly the same for a triangular grid of points except that it is multiplied by a factor of ^√3/₂, the height of an equilateral triangle for a unit triangle.  Therefore, Pick’s Theorem for the area of any polygon formed on a unit-based grid of triangular points is

                                                                                                                                  

A = ^√3/₂(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.

This variation of Pick’s Theorem for a triangular grid can be demonstrated through a series of transformations that changes a square unit grid to a triangular unit grid.  The first step is to start off with a square unit grid and skew each row of points half a unit to the right (or left).  Since skewing a shape preserves area (for example, a rectangle and a parallelogram both have an area of A = bh), the area formula remains as A = ½b + i – 1.  The second step is to shrink the height between rows by a factor of ^√3/₂, the height of an equilateral triangle for a unit triangle.  This new transformation changes the area by its shrinking factor, namely ^√3/₂, so the new formula becomes A = ^√3/₂(½b + i – 1).

Transforming a Square Unit Grid to a Triangular Unit Grid
Start with a square unit grid	Skew each row of points half a unit to the right	Shrink the ht. between rows by a factor of √3/2

If both the height and the width are shrunk so that each triangle has a unit area (instead of each side having a unit area), then the formula for Pick’s Theorem is multiplied by a factor of 2 (the area of each parallelogram formed by two adjacent unit triangles).  In this variation, Pick’s Theorem would be A = 2(½b + i – 1), or

A = b + 2i – 2

where once again 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.

Here are some examples of this variation of Pick’s Theorem for a triangular grid in which each triangle has a unit area:

Variation of Pick’s Theorem – Triangular Grid
b = 12 i = 3 A = b + 2i – 2 A = 12 + 2·3 – 2 = 16	b = 12 i = 1 A = b + 2i – 2 A = 12 + 2·1 – 2 = 12

These two examples can be verified by counting the number of unit triangles inside each shape.  The first example, the large triangle, is made up of 16 unit triangles, so it has an area of 16.  The second example, the star, is made up of 12 unit triangles, so it has an area of 12.  This variation of Pick’s Theorem for a triangular grid was able to correctly calculate the area of each of these shapes.

More information on this variation of Pick’s Theorem for a triangular grid can be found at http://www.drking.org.uk/hexagons/pick/index.html.

Variation of Pick’s Theorem – Shapes with Holes

Although one of the restrictions for using the regular version of Pick’s Theorem is that there cannot be any holes in the polygon, this restriction can be lifted by using the adjusted equation

A = ½b + i + h – 1

where b is the number of border points, i is the number of interior points, and h is the number of holes.

Recall that the proof for the regular version of Pick’s Theorem makes use of the fact that any polygon can be built by adding one triangle at a time on a shared side and still preserve Pick’s Theorem.  However, when building a shape with a hole in it, there must be a quadrilateral added in the process that acts as a bridge with two shared sides instead of one, which will result in a contradiction of Pick’s Theorem.  The discrepancy can be accounted for by adding the number of holes in the shape.

For example, in the picture above, the dark gray square is bridging the gap on the light gray polygon to form one hole.  According to Pick’s Theorem, the two areas are 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 dark gray square, b₂ and i₂ are the number of border and interior points on the light gray polygon.  Letting s₁ and s₂ be the number of border points on the two shared sides (including the endpoints), the resulting shape has the same number of border and interior points as the original two polygons except that 2(s₁ – 2) border points change to s₁ – 2 interior points, the two endpoints of s₁ are repeated, 2(s₂ – 2) border points change to s₂ – 2 interior points, and the two endpoints of s₂ are repeated.  So the area of the resulting shape in terms of Pick’s Theorem is:

= ½(b₁ + b₂ – 2(s₁ – 2) – 2 – 2(s₂ – 2) – 2) + (i₁ + i₂ + (s₁ – 2) + (s₂ – 2)) – 1

= ½b₁ + ½b₂ – (s₁ – 2) – 1 – (s₂ – 2) – 1 + i₁ + i₂ + (s₁ – 2) + (s₂ – 2) – 1

= ½b₁ + ½b₂ – (s₁ – 2) – 1 – (s₂ – 2) – 1 + i₁ + i₂ + (s₁ – 2) + (s₂ – 2) – 1

= ½(b₁ + b₂) + i₁ + i₂ – 3

= (½b₁ + i₁ – 1) + (½b₂ + i₂ – 1) – 1

= (area of the dark gray square) + (area of the light gray polygon) – 1

For every hole that is created, and extra 1 is subtracted, so the equation can be balanced by adding the number of holes in the shape.

Here are some examples of this variation of Pick’s Theorem for shapes with holes:

Variation of Pick’s Theorem – Shapes with Holes
b = 32 i = 16 h = 1 A = ½b + i + h – 1 A = ½·32 + 16 + 1 – 1 = 32	b = 40 i = 8 h = 5 A = ½b + i + h – 1 A = ½·40 + 8 + 5 – 1 = 32

These two examples can be verified by common area formulas.  The first example, the square with a smaller square taken out, has an area of A = 6² – 2² = 36 – 4 = 32.  The second example, the square with four triangles and a diamond taken out, has an area of A = 6² – 4·½·1·1 – ½·2·2 = 36 – 2 – 2 = 32.  This variation of Pick’s Theorem for shapes with holes was also able to correctly calculate the area of both of these shapes.

More information on this variation of Pick’s Theorem for shapes with holes can be found at http://jwilson.coe.uga.edu/emat6680fa05/schultz/6690/pick/pick_main.htm.

Variation of Pick’s Theorem – Three Dimensions

There are many mathematicians who believe that all attempts to extrapolate Pick’s Theorem to the third dimension result in failure.  Indeed, the volume of a polyhedron formed by points on a three-dimensional unit grid cannot be determined by the number of its interior points and border points.  One quick counterexample is to compare the smallest triangular prism that can be formed on a unit grid and the smallest octahedron that can be formed on a unit grid. 

Both polyhedrons have the same number of interior and border points (0 interior points and 6 border points), but their volumes are different (the triangular prism has a volume of ½ and the octahedron has a volume of ²/₃).

However, if all the restrictions to Pick’s Theorem are bumped up a dimension as well, it is possible to derive a variation of Pick’s Theorem for finding volumes in the third dimension.  The new restrictions are as follows:

1)  All sides must be on one of the unit grid segments.

2)  All border sides must connect to exactly two faces.

3)  There are no topological holes or hollow parts.

4)  There are no curved faces.

These new restrictions force all eligible polyhedrons to be blocky, which admittedly is not as elegant as the two dimensional version of Pick’s Theorem, but if these new restrictions are met, the volume of a polyhedron can be determined by

V = ⅛(b_e + 4i_e – 4i_p – 4)

where b_e is the number of edges on the border of the polyhedron, i_e is the number of edges on the interior of the polyhedron, and i_p is the number of points on the interior of the polyhedron.

Here are some examples of this variation of Pick’s Theorem for three dimensions:

Variation of Pick’s Theorem – Three Dimensions
be = 48, ie = 6, ip = 1 V = ⅛(be + 4ie – 4ip – 4) V = ⅛(48 + 4·6 – 4·1 – 4) = 8	be = 108, ie = 36, ip = 8 V = ⅛(be + 4ie – 4ip – 4) V = ⅛(108 + 4·36 – 4·8 – 4) = 27
be = 44, ie = 2, ip = 0 V = ⅛(be + 4ie – 4ip – 4) V = ⅛(44 + 4·2 – 4·0 – 4) = 6	be = 60, ie = 0, ip = 0 V = ⅛(be + 4ie – 4ip – 4) V = ⅛(60 + 4·0 – 4·0 – 4) = 7

These examples can be verified by counting the number of unit blocks inside each shape.  The first example, the 2 x 2 x 2 cube, is made up of 8 unit blocks, so it has a volume of 8.  The second example, the 3 x 3 x 3 cube, is made up of 27 unit blocks, so it has a volume of 27.  The third example, the staircase, is made up of 6 unit blocks, so it has a volume of 6.  Finally, the last example, the letter H, is made up of 7 unit blocks, so it has a volume of 7.  This variation of Pick’s Theorem for three dimensions was also able to correctly calculate the volume of each of these shapes.

The proof for this variation of Pick’s Theorem for three dimensions is also quite long and involved, and can be found here.

Conclusion

Pick’s Theorem is not a very well-known theorem other than to a few math trivia buffs.  It does not appear in many Geometry textbooks, nor is it taught in many Geometry classes, and yet it is easy to understand (since it is based on a unit grid), flexible (there are many variations), and useful (it can be used to find areas and volumes).  For these reasons, Pick’s Theorem is one of the most beautiful and elegant theorems in mathematics.

Proof of Pick’s Theorem for Three Dimensions

If a polyhedron is formed on a unit grid such that all of its sides are on one of the unit grid segments, all of its border sides connect to exactly two faces, there are no topological holes or hollow parts, and there are no curved faces, then the volume of that polyhedron can be determined by

V = ⅛(b_e + 4i_e – 4i_p – 4)

where b_e is the number of edges on the border of the polyhedron, i_e is the number of edges on the interior of the polyhedron, and i_p is the number of points on the interior of the polyhedron.

Since the first restriction is that all sides are on one of the unit grid segments, this means that all eligible polyhedrons must be made out of unit blocks.  So to prove this variation of Pick’s Theorem for three dimensions, the first step is to show that it is true for a single block, and the second step is to show that combining a unit block to another eligible polyhedron made up of unit blocks preserves this theorem.

                                            

The first step is to show that this variation of Pick’s Theorem for three dimensions is true for a single block.  This can be verified easily enough.  A single block has 12 border edges, 0 interior edges, 0 interior points, and 0 holes.  Therefore, using the formula we can calculate its volume to be V = ⅛(b_e + 4i_e – 4i_p + 4h – 4) = ⅛(12 + 4·0 – 4·0 + 4·0 – 4) = 1, which is exactly the volume we would expect.

The second step is to show that combining a unit block to another eligible polyhedron made up of unit blocks preserves this variation of Pick’s Theorem.  Depending on the already existing polyhedron, the unit block can be combined so that it shares 1 face, 2 faces (adjacently), 3 faces (adjacently in a corner or adjacently in a row), 4 faces (adjacently in a corner), or 5 faces.  As the table below shows, all of these combinations keep the equation to the theorem balanced:

1 shared face	8 new border edges: beN = beO + 8 0 new interior edges: ieN = ieO 0 new interior points: ipN = ipO volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) (VO+1) = ⅛((beO+8)+4ieO–4ipO–4) VN = ⅛(beN+4ieN–4ipN–4)
2 shared  adjacent faces	5 – 1 = 4 new border edges: beN = beO + 4 1 new interior edge: ieN = ieO + 1 0 new interior points: ipN = ipO volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(beO+4ieO–4ipO–4+4+4) VO+1 = ⅛(beO+4+4ieO+4–4ipO–4) VO+1 = ⅛((beO+4)+4(ieO+1)–4ipO–4) VN = ⅛(beN+4ieN–4ipN–4)
3 shared adjacent faces in a corner	3 – 3 = 0 new border edges: beN = beO 3 new interior edges: ieN = ieO + 3 1 new interior point: ipN = ipO + 1 volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(beO+4ieO–4ipO–4+12–4) VO+1 = ⅛(beO+4ieO+12–4ipO–4–4) VO+1 = ⅛(beO+4(ieO+3)–4(ipO+1)–4) VN = ⅛(beN+4ieN–4ipN–4)
3 shared  adjacent faces in a row	2 – 2 = 0 new border edges: beN = beO 2 new interior edges: ieN = ieO + 2 0 new interior points: ipN = ipO volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(be+4ieO+8–4ipO–4) VO+1 = ⅛(beO+4+4(ieO+2)–4ipO+4(hO+1)–4) VN = ⅛(beN+4ieN–4ipN–4)
4 shared  adjacent faces in a corner	1 – 5 = 4 less border edges: beN = beO – 4 5 new interior edges: ieN = ieO + 5 2 new interior points: ipN = ipO + 2 volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(beO+4ieO–4ipO–4–4+20–8) VO+1 = ⅛(beO–4+4ieO+20–4ipO–8–4) VO+1 = ⅛(beO–4+4(ieO+5)–4(ipO+2)–4) VN = ⅛(beN+4ieN–4ipN–4)
5 shared faces	8 less border edges: beN = beO – 8 8 new interior edges: ieN = ieO + 8 4 new interior points: ipN = ipO + 4 volume increases by 1: VN = VO  + 1	VO = ⅛(beO+4ieO–4ipO–4) VO+1 = ⅛(beO+4ieO–4ipO–4)+1 VO+1 = ⅛(beO+4ieO–4ipO–4+8) VO+1 = ⅛(beO+4ieO–4ipO–4–8+32–16) VO+1 = ⅛(beO–8+4ieO+32–4ipO–16–4) VO+1 = ⅛(beO–8+4(ieO+8)–4(ipO–4)–4) VN = ⅛(beN+4ieN–4ipN–4)

Note: the following combinations are disregarded since holes or hollowed parts are restricted: 2 opposite faces (because a hole would be formed), 4 adjacent faces in a row (because it would start with a hole), and 6 faces (because it would start with a hollow part).

Since we have shown that this variation of Pick’s Theorem for three dimensions is true for a unit block, and have shown that this variation of Pick’s Theorem is preserved for combining a unit block to another eligible polyhedron made up of unit blocks, this variation of Pick’s Theorem must be true for all polyhedrons as well.

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.

The Garden Border Problem

The following problem is in the McDougal Littell Algebra 2 textbook and is a typical word problem for a section on solving quadratic equations: 

You have just planted a rectangular flower bed of red roses in a park near your home.  You want to plant a border of yellow roses around the flower bed as shown.  Since you bought the same number of red and yellow roses, the areas of the border and inner flower bed will be equal.  What should the width x of the border be? 

To solve this problem, you must first write an area equation.  The length of the whole garden is 12 feet plus the unknown widths of the left and right borders, which can be expressed as 2x + 12.  The width of the whole garden is 8 feet plus the unknown widths of the top and bottom borders, which can be expressed as 2x + 8.  The area of the whole garden is the area of the red rose garden (which is 8 feet times 12 feet or 96 feet squared) plus the area of the yellow rose garden (which is the same as the red rose garden or also 96 feet squared), which added together is 192 feet squared.  Since the area is length times width, the equation to solve is (2x + 12)(2x + 8) = 192.

The next step is to solve this area equation.  Multiplying (2x + 12)(2x + 8) gives us 4x² + 40x + 96, so 4x² + 40x + 96 = 192, and subtracting 192 to the left side gives us 4x² + 40x – 96 = 0, and dividing everything by the common factor 4 gives us x² + 10x – 24 = 0.  At this point, there are several methods for solving this quadratic (such as factoring, completing the square, quadratic equation, and graphing) but we will use the quadratic equation x = ^{-b ± √(b^2 – 4ac)} / _2a, where a = 1, b = 10, and c = -24.  Therefore, x = ^{-10 ± √(10^2 – 4·1·-24)} / _2·1 = ^{-10 ± √(100 + 96)} / ₂ = ^{-10 ± √196} / ₂ = ^{-10 ± 14} / ₂, which means x = -12 or x = 2.  Since x represents a geometrical dimension, it cannot be negative, and therefore the border width x must be 2 feet long.

You will notice that this answer conveniently comes out as an integer, and not as a decimal.  But what would happen if the problem started out with different dimensions for the inner garden?  Would the border width still be an integer?  Let us examine the same problem but with a starting inner garden of 8 feet by 8 feet, as pictured below:

  

This time both the length and the width of the whole garden can be expressed as 2x + 8, and the area of the whole garden is 2 times 8 feet by 8 feet, or 128 feet squared, giving us the equation (2x + 8)(2x + 8) = 128.  Multiplying (2x + 8)(2x + 8) gives us 4x² + 32x + 64 = 128, subtracting 128 to the left side gives us 4x² + 32x – 64 = 0, and dividing everything by the common factor 4 gives us x² + 8x – 16 = 0.  Using the quadratic equation x = ^{-b ± √(b^2 – 4ac)} / _2a, where a = 1, b = 8, and c = -16 gives us x = ^{-8 ± √(8^2 – 4·1·-16)} / _2·1 = ^{-8 ± √(64 + 64)} / ₂ = ^{-8 ± √128} / ₂ = ^{-8 ± 8√2} / ₂ = -4 ± 4√2.  Since x cannot be negative, the border width must be -4 + 4√2 feet long, which is not an integer answer.

Can we come up with different dimensions for the inner garden such that the border width solution comes out as an integer?  We already know one solution set is (8, 12, 2) from the original problem, and using the properties of proportions and dividing each number by two we can also include (4, 6, 1).  In fact, using the same argument we can include all solution sets in the form of (4k, 6k, k) where k is any positive integer.  To simplify things, we will say that (4, 6, 1) is a “garden border triple” that includes all solutions sets in the form (4k, 6k, k), so the garden border triple (4, 6, 1) includes (4, 6, 1), (8, 12, 2), (12, 18, 3), and so on (just like the Pythagorean triple (3, 4, 5) includes all solution sets in the form (3k, 4k, 5k)).

Are there other garden border triples other than (4, 6, 1)?  Just as there are different Pythagorean triple solutions to the formula a² + b² = c² ((3, 4, 5), (5, 12, 13), etc.), there are also different garden border triples.  And just as there are different Pythagorean triple generators (see here), there are different garden border triple generators.  To make one, we must generalize the garden border problem by calling the length of the inner garden b and the width of the inner garden h, as pictured below:

The length of the whole garden can then be expressed as 2x + b, the width of the whole garden can be expressed as 2x + h, and the area of the whole garden can be expressed as 2bh, giving us the equation (2x + b)(2x + h) = 2bh.  This time, however, we are going to solve this equation for b.  Multiplying (2x + b)(2x + h) gives us 4x² + 2bx + 2hx + bh = 2bh, subtracting bh on both sides gives us 4x² + 2bx + 2hx = bh, subtracting 2bx on both sides gives us 4x² + 2hx = bh – 2bx, factoring 2x from the left side and b from the right side gives us 2x(2x + h) = b(h – 2x), and dividing both sides by h – 2x gives us b = ^{2x(h + 2x)}/_{h – 2x}.

We can now use the formula b = ^{2x(h + 2x)}/_{h – 2x} to generate garden border triples.  If we let x = 1, then b = ^{2(h + 2)}/_{h – 2}.  Then if h = 3, b = ^{2(3 + 2)}/_{3 – 2} = 10, then the garden border triple is (3, 10, 1).  If h = 4, b = ^{2(4 + 2)}/_{4 – 2} = 6, then the garden border triple is (4, 6, 1) (which is a repeat of a triple we already knew).  If h = 5, b = ^{2(5 + 2)}/_{5 – 2} = ¹⁴/₃, then the garden border triple is (5, ¹⁴/₃, 1), and to eliminate the fraction we can multiply each number by 3 to get (15, 14, 3).  Continuing on in this fashion, we also arrive at (6, 4, 1) (a repeat), (7, ¹⁸/₅, 1) ≡ (35, 18, 5), (8, ¹⁰/₃, 1) ≡ (24, 10, 3), (9, ²²/₇, 1) ≡ (63, 22, 7), (10, 3, 1) (a repeat) and so on.  If we let x = 2, then b = ^{4(h + 4)}/_{h – 4}, and the resulting garden border triples are (5, 36, 2), (6, 20, 2) ≡ (3, 10, 1) (a repeat), (7, ⁴⁴/₃, 2) ≡ (21, 44, 6), (8, 12, 2) ≡ (4, 6, 1) (a repeat), (9, ⁵²/₅, 2) ≡ (45, 52, 10), (10, ²⁸/₃, 2) ≡ (15, 14, 3) (a repeat), and so on.  If we let x = 3, then b = ^{6(h + 6)}/_{h – 6}, and the resulting garden border triples are (7, 78, 3), (8, 42, 3), (9, 30, 3) ≡ (3, 10, 1) (a repeat), (10, 24, 3), and so on. 

The garden border problem is a common word problem given to students to practice solving quadratic equations.  Most solutions come out as a decimal answer, but there are a few scenarios in which the width, length, and border width are all integers, which we called garden border triples.  Generalizing the problem in terms of b and h and solving for b gave us a garden border triple generator b = ^{2x(h + 2x)}/_{h – 2x}.  The unique garden border triples we generated in this article were (3, 10, 1), (4, 6, 1), (15, 14, 3), (35, 18, 5), (24, 10, 3), (63, 22, 7), (5, 36, 2), (21, 44, 6), (45, 52, 10), (7, 78, 3), (8, 42, 3), and (10, 24, 3); but there are many, many more.