Tile Patterns of Regular Polygons

A regular polygon is a shape with all congruent sides and all congruent angles.  For example, an equilateral triangle is a regular polygon because it has 3 congruent sides and 3 congruent angles, and a square is a regular polygon because it has 4 congruent sides and 4 congruent angles.

Regular Tessellations

Some regular polygons can be used to make a tile pattern (or a tessellation), and others cannot.  In order for a regular polygon to be tileable, its interior angle must divide evenly into 360°, and the interior angle of a regular polygon with n sides is θ = ¹/_n(n – 2)180°.  A regular triangle, which has n = 3 sides, has an interior angle θ = ¹/₃(3 – 2)180° = 60°, which divides evenly into 360°, which means it can be used to make a tile pattern.  A square, which has n = 4 sides, has an interior angle θ = ¹/₄(4 – 2)180° = 90°, which divides evenly into 360°, which means it also can be used to make a tile pattern.  However, a regular pentagon, which has n = 5 sides, has an interior angle θ = ¹/₅(5 – 2)180° = 108°, which does not divide evenly into 360°, which means it cannot be used to make a tile pattern.  A regular hexagon, which has n = 6 sides, has an interior angle θ = ¹/₆(6 – 2)180° = 120°, which divides evenly into 360°, which means it can be used to make a tile pattern. 

Interior Angles of Regular Polygons

Since the next biggest factor of 360° after 120° is 180°, which is too big to be an interior angle of a regular polygon, there are no other regular polygons that can be used to make a tile pattern.  Therefore, the equilateral triangle, the square, and the hexagon are the only regular polygons that can be used to make a tile pattern.

Regular Polygons That Can Be Tiled

Semi-Regular Tessellations

If the requirement that all the regular polygons in the tile pattern must have the same number of sides is removed, then there are a few more tile patterns that can be obtained called semi-regular tessellations.  Semi-regular tessellations (or Archimedean tessellations) are tile patterns that contain two or more regular polygons with the same order around each vertex.  For example, at my house growing up we had a brick driveway consisting of octagons and squares.  This was a semi-regular tessellation because it was a tile pattern containing regular polygons in which each vertex was surrounded by 2 regular octagons and 1 square.

Brick Driveway Pattern Consisting of

Octagons and Squares

One way to find all the possible semi-regular tessellations is to examine which combinations of regular polygons fit snugly around a single vertex.  In order for this to happen, each interior angle, which for a regular polygon with n sides is θ = ¹/_n(n – 2)180°, must add up to 360°.  If there are 3 regular polygons around one vertex with sides n₁, n₂, and n₃; then ¹/_n1(n₁ – 2)180° + ¹/_n2(n₂ – 2)180° + ¹/_n3(n₃ – 2)180° = 360°.  After distributing, the equation becomes 180° – ^360°/_n1 + 180° – ^360°/_n2 + 180° – ^360°/_n3 = 360°; after combining like terms, the equation becomes -^360°/_n1 – ^360°/_n2 – ^360°/_n3 = -180°; and after dividing by -360°, the equation simplifies to

¹/_n1 + ¹/_n2 + ¹/_n3 = ¹/₂

This holds true for the known three hexagons example (¹/₆ + ¹/₆ + ¹/₆ = ¹/₂ or (6, 6, 6)), and for the brick driveway example consisting of 1 square and 2 octagons (¹/₄ + ¹/₈ + ¹/₈ = ¹/₂ or (4, 8, 8)).  Using trial and error and the fact that n is an integer and n ≥ 3, the only possible solutions are (3, 7, 42), (3, 8, 24), (3, 9, 18), (3, 10, 15), (3, 12, 12), (4, 5, 20), (4, 6, 12), (4, 8, 8), (5, 5, 10), and (6, 6, 6).

All Combinations of 3 Polygons Fitting

Snugly around a Single Vertex

However, even though all of these solutions represent all the possible ways for 3 regular polygons to fit snugly around a single vertex, not all of them can be used to make a tessellation.  One final requirement for a tessellation is that the number of different sides must divide evenly into the third side.  For example, the solution (5, 5, 10), which represents a pentagon, pentagon, and decagon, fits snugly around a single vertex but cannot be tessellated.  The first pentagon, with 5 sides, would need to alternately share sides with the other pentagon and decagon, 2 different shapes, but 2 does not divide evenly into 5, and so these 3 shapes cannot be tessellated.  (See here for more details.)

Attempt at a (5, 5, 10) Tessellation Results in Some Gaps

This requirement eliminates all the above solutions except for (3, 12, 12), (4, 6, 12), (4, 8, 8), and (6, 6, 6).

(3, 12, 12)

(4, 6, 12)

(4, 8, 8)

The same logic can be applied to find semi-regular tessellations with 4 or more regular polygons.  In general, if there are k regular polygons around one vertex with sides with sides n₁, n₂, … n_k; then ∑ ^k _p=1 ¹/_np(n_p – 2)180° = 360°.  After distributing, the equation becomes ∑ ^k _p=1180° – ∑ ^k _p=1^360°/_np = 360°, which is 180°k – ∑ ^k _p=1^360°/_np = 360°; after some rearranging, the equation becomes -∑ ^k _p=1^360°/_np = -180k + 360°; and after dividing by -360°, the equation simplifies to ∑ ^k _p=1¹/_np = ¹/₂(k – 2) or

¹/_n1 + ¹/_n2 + … +  ¹/_nk= ¹/₂(k – 2)

For k = 3, ¹/_n1 + ¹/_n2 + ¹/_n3 = ¹/₂ (as proved above); for k = 4,

¹/_n1 + ¹/_n2 + ¹/_n3 + ¹/_n4 = 1

for k = 5,

¹/_n1 + ¹/_n2 + ¹/_n3 + ¹/_n4 + ¹/_n5 = ³/₂

and for k = 6,

¹/_n1 + ¹/_n2 + ¹/_n3 + ¹/_n4 + ¹/_n5 + ¹/_n6 = 2

Since n ≥ 3, there are no integer solutions when k ≥ 7 (geometrically, 7 or more equilateral triangles cannot fit around a single vertex). 

Once again using trial and error and the fact that n is an integer and n ≥ 3, the only possible solutions for k ≥ 4 are (3, 3, 4, 12), (3, 3, 6, 6), (3, 4, 4, 6), (4, 4, 4, 4), (3, 3, 3, 3, 6), (3, 3, 3, 4, 4), and (3, 3, 3, 3, 3, 3).  For some of these solutions, a new ordering produces a new unique pattern, and so (3, 4, 3, 12), (3, 6, 3, 6), (3, 4, 6, 4), and (3, 3, 4, 3, 4) can be added to the list. 

All Combinations of More Than 3 Polygons

Fitting Snugly around a Single Vertex

However, the requirement for semi-regular tessellations that there is the same order around each vertex eliminates (3, 3, 4, 12), (3, 3, 6, 6), (3, 4, 4, 6), and (3, 4, 3, 12); leaving only the solutions (3, 4, 6, 4), (3, 6, 3, 6), (4, 4, 4, 4), (3, 3, 3, 3, 6), (3, 3, 3, 4, 4), (3, 3, 4, 3, 4), and (3, 3, 3, 3, 3, 3).   

(3, 4, 6, 4)

(3, 6, 3, 6)

(3, 3, 3, 3, 6)

(3, 3, 3, 4, 4)

(3, 3, 4, 3, 4)

Since (6, 6, 6), (4, 4, 4, 4), and (3, 3, 3, 3, 3, 3) are regular tessellations, that leaves 8 possible semi-regular tessellation patterns: 1 triangle and 2 dodecagons (3, 12, 12); 1 square, 1 hexagon, and 1 dodecagon (4, 6, 12); 1 square and 2 octagons (4, 8, 8); 1 triangle, 2 squares, and 1 hexagon (3, 4, 6, 4); 2 triangles and 2 hexagons (3, 6, 3, 6); 4 triangles and 1 hexagon (3, 3, 3, 3, 6); and 3 triangles and 2 squares (3, 3, 3, 4, 4) and (3, 3, 4, 3, 4).  (See here for more details.)

The Eight Semi-Regular Tessellations

Quasi-Regular Tessellations

If the semi-regular tessellation requirement that there is the same order around each vertex is removed, then there are countless other variations of tile patterns that can be obtained called quasi-regular tessellations.  Many quasi-regular tessellations can be formed by modifying an existing semi-regular tessellation.  For example, since 1 regular hexagon can be formed from 6 equilateral triangles, the semi-regular tessellation of (6, 6, 6) can be transformed to a quasi-regular tessellation in which some of the vertices are (3, 3, 6, 6) and others are (3, 3, 3, 3, 3, 3); and the semi-regular tessellation of (4, 6, 12) can be transformed to a quasi-regular tessellation in which some of the vertices are (3, 3, 4, 12) and others are (3, 3, 3, 3, 3, 3). 

A regular hexagon formed by

6 equilateral triangles

A regular dodecagon formed by

1 regular hexagon, 6 squares,

and 6 equilateral triangles

In addition, 1 regular dodecagon can be formed from 1 regular hexagon, 6 squares, and 6 equilateral triangles, so the semi-regular tessellation of (3, 12, 12) can be transformed to a quasi-regular tessellation in which some of the vertices are (3, 3, 3, 4, 4) and some are (3, 4, 6, 4).  There are countless other variations of quasi-regular tessellations, but all of them use the same vertex combinations found in regular tessellations and semi-regular tessellations.

Some (of Many) Quasi-Regular Tessellations

Conclusion

There are many ways to tile an area with shapes all consisting of segments of the same length.  Three of those ways are by regular tessellations, in which all the shapes are the same.  These include 3 hexagons (6, 6, 6), 4 squares (4, 4, 4, 4), and 6 triangles (3, 3, 3, 3, 3, 3).  Eight other ways are by semi-regular tessellations, in which some of the shapes are different, but all of the shapes follow the same order around each vertex used in the tessellation.  These include 1 triangle and 2 dodecagons (3, 12, 12); 1 square, 1 hexagon, and 1 dodecagon (4, 6, 12); 1 square and 2 octagons (4, 8, 8); 1 triangle, 2 squares, and 1 hexagon (3, 4, 6, 4); 2 triangles and 2 hexagons (3, 6, 3, 6); 4 triangles and 1 hexagon (3, 3, 3, 3, 6); and 3 triangles and 2 squares (3, 3, 3, 4, 4) and (3, 3, 4, 3, 4).  Finally, there are several more ways to tile an area with shapes all consisting of segments of the same length using quasi-regular tessellations, in which some of the shapes are different and in which some of the vertices have different orders of shapes around them.  

Triangles Inscribed in Rectangles

When a triangle is inscribed in a rectangle, the vertices of that triangle are on the sides or vertices of that rectangle.  Sometimes when additional conditions are added, some interesting properties occur.

For example, if the inscribed triangle is an equilateral triangle, and it shares a common vertex with the rectangle that it is inscribed in, then the sum of the areas of the two outer triangles (that share the same vertex) is equal to the area of the third outer triangle.  This theorem was described by a mathematician named Honsberger in the late 20^th century.

If the Inscribed Triangle is an Equilateral Triangle,

then Area X + Area of Y = Area of Z

The amazing thing about this theorem is that it holds true for any angle θ between the bottom of the equilateral triangle and the bottom of the rectangle.  As long as the equilateral triangle can be inscribed in the rectangle, the areas of the two outer triangles add up to the area of the third outer triangle.

To prove this, first find the value of each of the angles of the outer triangles in terms of θ, the angle between the bottom of the equilateral triangle and the bottom of the rectangle, using the fact that the angle sum of a triangle is 180°, and that the sum of the angles forming a line is 180°, and that the sum of the angles forming a right angle is 90°.  This means that ∠ADE = 30° – θ, ∠AED = 60° + θ, ∠ADE = 30° – θ, ∠BEF = 60° – θ, ∠BFE = 30° + θ, and ∠CFD = 90° – θ. 

Second, use A = ½bh =  ½(c sin θ)(c cos θ) = ½ c² sin θ cos θ = ¼ c²(2 sin θ cos θ) = ¼ c² sin 2θ for the area of each right triangle, where c is the hypotenuse and θ is one of the acute angles.  This means that the area of X = ¼ s² sin 2(30° – θ), the area of Y = ¼ s² sin 2θ, and the area of Z = ¼ s² sin 2(30° + θ).  Therefore:

01	X + Y	the sum of the areas
02	= ¼ s2 sin 2(30° – θ) + ¼ s2 sin 2θ	substitution
03	= ¼ s2 (sin 2(30° – θ) + sin 2θ)	factor out ¼ s2
04	= ¼ s2 (sin (60° – 2θ) + sin 2θ)	distribute 2
05	= ¼ s2 (sin 60° cos 2θ – cos 60° sin 2θ + sin 2θ)	sin (a – b) = sin a cos b – cos a sin b
06	= ¼ s2 (√3/2 cos 2θ – ½ sin 2θ + sin 2θ)	sin 60° = √3/2, cos 60° = ½
07	= ¼ s2 (√3/2 cos 2θ + ½ sin 2θ)	combine like terms
08	= ¼ s2 (sin 60° cos 2θ + cos 60° sin 2θ)	sin 60° = √3/2, cos 60° = ½
09	= ¼ s2 (sin (60° + 2θ))	sin (a + b) = sin a cos b + cos a sin b
10	= ¼ s2 sin 2(30° + θ)	factor out 2
11	= Z	substitution

Another two fascinating theorems concerning a triangle inscribed in a rectangle occur when two sides of the rectangle are cut in the same ratio.  If a triangle is inscribed in a rectangle so that it shares a common vertex and cuts two consecutive sides of the rectangle in the same ratio, then the sum of the areas of the two outer triangles (that also share the same vertex) is equal to area of half the rectangle it is inscribed in, and the product of the areas of the two outer triangles (that also share the same vertex) is equal to the product of the areas of the third outer triangle and half the rectangle it is inscribed in.

If E and F Cut the Sides of AB and BC in the Same Ratio,

then Area of X + Area Y = Area of Half the Rectangle

and Area of X · Area of Y = Area of Z · Area of Half the Rectangle

The amazing thing about this theorem is that it holds true no matter where the side is cut.  As long as the second side is cut in the same ratio, the sum of the areas of the two outer triangles is equal to the area of the half the rectangle, and the product of the areas of the two outer triangles is equal to the product of the areas of the third outer triangle and half the rectangle.

To prove this, label the sides of the rectangle b and h, and the ratio k, so that the cut segments are kb and b – kb and kh and h – kh.  Then using A = ½bh, the area of X = ½kbh, the area of Y = ½b(h – kh), the area of Z = ½kh(b – kb), and the area of half the rectangle is ½bh.  Therefore:

01	X + Y	the sum of the areas
02	= ½kbh + ½b(h – kh)	substitution
03	= ½kbh + ½bh – ½kbh	distribute ½b
04	= ½bh	combine like terms
05	= Area of Half the Rectangle	substitution

And also:

01	XY	the product of the areas
02	= ½kbh ∙ ½b(h – kh)	substitution
03	= ¼b2hk(h – kh)	commutative
04	= ¼b2h2k(1 – k)	factor out h
05	= ¼bh2k(b – kb)	distribute b
06	= ½kh(b – kb) ∙ ½bh	commutative
07	= Z ∙ Area of Half the Rectangle	substitution

There are several amazing theorems concerning triangles inscribed in rectangles and the outer triangles that are formed.  If the triangle is equilateral, then no matter the angle between the side of the triangle and the side of the rectangle, the area of two of the outer triangles add up to the area of the third triangle.  If the triangle cuts two consecutive sides of the rectangle in the same ratio, then no matter what the ratio the sides are cut at, the sum of the areas of two of the outer triangles is equal to the area of half the rectangle it is inscribed in, and the product of the areas of two of the outer triangles is equal to the product of the areas of the third outer triangle and half the rectangle it is inscribed in.  All of these theorems show just how wonderful and amazing geometry is.

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.