Money, Baseball, and Apple Orchards Dan Biebighauser December 7, - - PowerPoint PPT Presentation

Money, Baseball, and Apple Orchards

Dan Biebighauser December 7, 2010

1

Some Questions

How many ways are there to make change for a dollar using quarters, dimes, and nickels?

2

Some Questions

How many ways are there to make change for a dollar using quarters, dimes, and nickels? In how many ways can a baseball player achieve the batting average .273 with at most 2,000 at-bats?

2

Some Questions

How many ways are there to make change for a dollar using quarters, dimes, and nickels? In how many ways can a baseball player achieve the batting average .273 with at most 2,000 at-bats? What is the area of the following apple orchard if the rows and columns of trees are one unit apart?

2

Some Questions

How many ways are there to make change for a dollar using quarters, dimes, and nickels? In how many ways can a baseball player achieve the batting average .273 with at most 2,000 at-bats? What is the area of the following apple orchard if the rows and columns of trees are one unit apart? How are the previous three questions related?

2

Outline

1

Apple Orchards Approximation Area

2

Money Make a List Coins on the Table Geometry

3

The Main Theorem Statement Rectangles and Right Triangles More Money Adding and Subtracting Polygons Finishing the Proof

4

Baseball

5

Conclusion

3

Outline

1

Apple Orchards Approximation Area

2

Money Make a List Coins on the Table Geometry

3

The Main Theorem Statement Rectangles and Right Triangles More Money Adding and Subtracting Polygons Finishing the Proof

4

Baseball

5

Conclusion

4

Approximation

5

Approximation

One way to approximate this area is to count the number of trees in the

• rchard.

5

Approximation

6

Approximation

Area ≈ Number of Trees = 23

6

Approximation

A better approximation only counts 1/2 for each tree on the boundary.

7

Approximation

A better approximation only counts 1/2 for each tree on the boundary. Area ≈ Number of Trees − 1 2

• (Number of Boundary Trees)

= 23 − 1 2

• (8) = 19

7

Approximation

In fact, the actual area is 18, so this better approximation is exactly 1 too big.

8

Approximation

In fact, the actual area is 18, so this better approximation is exactly 1 too big. Surprising Fact: This better approximation is always 1 too big.

8

Area using Triangles

(0, 0) (3, 6) (4, 4) (6, 6) (5, 1) A1 A2 A3

9

Area using Triangles

(0, 0) (3, 6) (4, 4) (6, 6) (5, 1) A1 A2 A3

For each triangle, we could use the formula A = 1

2bh, but . . .

9

Area using Triangles

(0, 0) (3, 6) (4, 4) (6, 6) (5, 1) A1 A2 A3

For each triangle, we could use the formula A = 1

2bh, but . . .

We could also use Heron’s formula, but . . .

9

Area using Triangles

Instead, let’s “frame” each triangle.

10

Area using Triangles

Instead, let’s “frame” each triangle.

(0, 0) (3, 6) (4, 4) A1

10

Area using Triangles

Instead, let’s “frame” each triangle.

(0, 0) (3, 6) (4, 4) A1 (0, 0) (3, 6) (4, 4) A1 3 1 2 4 4 6

10

Area using Triangles

Instead, let’s “frame” each triangle.

(0, 0) (3, 6) (4, 4) A1 (0, 0) (3, 6) (4, 4) A1 3 1 2 4 4 6

A1 = 6 · 4 − 1 2(3 · 6) − 1 2(1 · 2) − 1 2(4 · 4) = 6

10

Area using Triangles

Similarly,

(0, 0) (4, 4) (5, 1) A2 4 4 1 3 1 5 (4, 4) (6, 6) (5, 1) A3 3 2 1 1 2 5

A2 = 5 · 4 − 1 2(4 · 4) − 1 2(1 · 3) − 1 2(1 · 5) = 8 and A3 = 5 · 2 − 1 2(2 · 2) − 1 2(5 · 1) − 1 2(1 · 3) = 4

11

Area using Triangles

Similarly,

(0, 0) (4, 4) (5, 1) A2 4 4 1 3 1 5 (4, 4) (6, 6) (5, 1) A3 3 2 1 1 2 5

A2 = 5 · 4 − 1 2(4 · 4) − 1 2(1 · 3) − 1 2(1 · 5) = 8 and A3 = 5 · 2 − 1 2(2 · 2) − 1 2(5 · 1) − 1 2(1 · 3) = 4 So the total area is A1 + A2 + A3 = 6 + 8 + 4 = 18.

11

Outline

1

Apple Orchards Approximation Area

2

Money Make a List Coins on the Table Geometry

3

The Main Theorem Statement Rectangles and Right Triangles More Money Adding and Subtracting Polygons Finishing the Proof

4

Baseball

5

Conclusion

12

Money

How many ways are there to make change for a dollar using quarters, dimes, and nickels?

13

Money

• 1. Make a list

14

Money

• 1. Make a list

Let q be the number of quarters, d be the number of dimes, and n be the number of nickels. We need all triples (q, d, n) such that 25q + 10d + 5n = 100.

14

Money

d = 10 (0, 10, 0) d = 9 (0, 9, 2) d = 8 (0, 8, 4) d = 7 (0, 7, 6) (1, 7, 1) d = 6 (0, 6, 8) (1, 6, 3) d = 5 (0, 5, 10) (1, 5, 5) (2, 5, 0) d = 4 (0, 4, 12) (1, 4, 7) (2, 4, 2) d = 3 (0, 3, 14) (1, 3, 9) (2, 3, 4) d = 2 (0, 2, 16) (1, 2, 11) (2, 2, 6) (3, 2, 1) d = 1 (0, 1, 18) (1, 1, 13) (2, 1, 8) (3, 1, 3) d = 0 (0, 0, 20) (1, 0 15) (2, 0, 10) (3, 0, 5) (4, 0, 0) q = 0 q = 1 q = 2 q = 3 q = 4

15

Money

d = 10 (0, 10, 0) d = 9 (0, 9, 2) d = 8 (0, 8, 4) d = 7 (0, 7, 6) (1, 7, 1) d = 6 (0, 6, 8) (1, 6, 3) d = 5 (0, 5, 10) (1, 5, 5) (2, 5, 0) d = 4 (0, 4, 12) (1, 4, 7) (2, 4, 2) d = 3 (0, 3, 14) (1, 3, 9) (2, 3, 4) d = 2 (0, 2, 16) (1, 2, 11) (2, 2, 6) (3, 2, 1) d = 1 (0, 1, 18) (1, 1, 13) (2, 1, 8) (3, 1, 3) d = 0 (0, 0, 20) (1, 0 15) (2, 0, 10) (3, 0, 5) (4, 0, 0) q = 0 q = 1 q = 2 q = 3 q = 4 So there are 29 possibilities.

15

Money

• 2. Coins on the Table

16

Money

• 2. Coins on the Table

It suffices to find q and d such that 25q + 10d ≤ 100.

16

Money

• 2. Coins on the Table

It suffices to find q and d such that 25q + 10d ≤ 100. Suppose we put 4 quarters and 10 dimes on a table. This gives 2 dollars and enough coins to make any feasible amount up to 1 dollar.

16

Money

• 2. Coins on the Table

It suffices to find q and d such that 25q + 10d ≤ 100. Suppose we put 4 quarters and 10 dimes on a table. This gives 2 dollars and enough coins to make any feasible amount up to 1 dollar. Suppose we select q quarters and d dimes. There are 5 choices for q (0, 1, 2, 3, 4) and 11 choices for d (0, 1, . . . , 10).

16

Money

• 2. Coins on the Table

It suffices to find q and d such that 25q + 10d ≤ 100. Suppose we put 4 quarters and 10 dimes on a table. This gives 2 dollars and enough coins to make any feasible amount up to 1 dollar. Suppose we select q quarters and d dimes. There are 5 choices for q (0, 1, 2, 3, 4) and 11 choices for d (0, 1, . . . , 10). So there are 5 · 11 = 55 total choices.

16

Money

• 2. Coins on the Table

The coins we have selected give 25q + 10d cents, and the remaining 4 − q quarters and 10 − d dimes on the table give the complementary 200 − 25q − 10d cents.

17

Money

• 2. Coins on the Table

The coins we have selected give 25q + 10d cents, and the remaining 4 − q quarters and 10 − d dimes on the table give the complementary 200 − 25q − 10d cents. Exactly three of these choices for q and d give exactly one dollar:

17

Money

• 2. Coins on the Table

The coins we have selected give 25q + 10d cents, and the remaining 4 − q quarters and 10 − d dimes on the table give the complementary 200 − 25q − 10d cents. Exactly three of these choices for q and d give exactly one dollar: (q, d) = (4, 0), (2, 5), and (0, 10)

17

Money

• 2. Coins on the Table

The coins we have selected give 25q + 10d cents, and the remaining 4 − q quarters and 10 − d dimes on the table give the complementary 200 − 25q − 10d cents. Exactly three of these choices for q and d give exactly one dollar: (q, d) = (4, 0), (2, 5), and (0, 10) The other 52 coin combinations occur in complementary pairs, where one combination in each pair is strictly less than 1 dollar and one combination is strictly more than 1 dollar.

17

Money

• 2. Coins on the Table

The coins we have selected give 25q + 10d cents, and the remaining 4 − q quarters and 10 − d dimes on the table give the complementary 200 − 25q − 10d cents. Exactly three of these choices for q and d give exactly one dollar: (q, d) = (4, 0), (2, 5), and (0, 10) The other 52 coin combinations occur in complementary pairs, where one combination in each pair is strictly less than 1 dollar and one combination is strictly more than 1 dollar. So there are 3 + 26 = 29 ways to make an amount not exceeding 1 dollar using quarters and dimes, and hence with quarters, dimes, and nickels.

17

Money

• 3. Geometry

18

Money

• 3. Geometry

As before, it suffices to find q and d such that 25q + 10d ≤ 100.

18

Money

• 3. Geometry

As before, it suffices to find q and d such that 25q + 10d ≤ 100. Let’s draw this inequality in the plane. Of course, we need q ≥ 0 and d ≥ 0 too.

18

Money

(0, 0) (4, 0) (0, 10) d q

25q + 10d = 100

19

Money

(0, 0) (4, 0) (0, 10) d q

25q + 10d = 100

There are 29 integer points (q, d) that satisfy these inequalities.

19

Money

• 1. Make a List

d = 10 (0, 10, 0) d = 9 (0, 9, 2) d = 8 (0, 8, 4) d = 7 (0, 7, 6) (1, 7, 1) d = 6 (0, 6, 8) (1, 6, 3) d = 5 (0, 5, 10) (1, 5, 5) (2, 5, 0) d = 4 (0, 4, 12) (1, 4, 7) (2, 4, 2) d = 3 (0, 3, 14) (1, 3, 9) (2, 3, 4) d = 2 (0, 2, 16) (1, 2, 11) (2, 2, 6) (3, 2, 1) d = 1 (0, 1, 18) (1, 1, 13) (2, 1, 8) (3, 1, 3) d = 0 (0, 0, 20) (1, 0 15) (2, 0, 10) (3, 0, 5) (4, 0, 0) q = 0 q = 1 q = 2 q = 3 q = 4

20

Money

• 1. Make a List

d = 10 (0, 10, 0) d = 9 (0, 9, 2) d = 8 (0, 8, 4) d = 7 (0, 7, 6) (1, 7, 1) d = 6 (0, 6, 8) (1, 6, 3) d = 5 (0, 5, 10) (1, 5, 5) (2, 5, 0) d = 4 (0, 4, 12) (1, 4, 7) (2, 4, 2) d = 3 (0, 3, 14) (1, 3, 9) (2, 3, 4) d = 2 (0, 2, 16) (1, 2, 11) (2, 2, 6) (3, 2, 1) d = 1 (0, 1, 18) (1, 1, 13) (2, 1, 8) (3, 1, 3) d = 0 (0, 0, 20) (1, 0 15) (2, 0, 10) (3, 0, 5) (4, 0, 0) q = 0 q = 1 q = 2 q = 3 q = 4

• 3. Geometry

(0, 0) (4, 0) (0, 10) d q

25q + 10d = 100 20

Money

• 2. Coins on the Table

(0, 0) (4, 0) (0, 10) d q

25q + 10d = 100

• 3. Geometry

(0, 0) (4, 0) (0, 10) d q

25q + 10d = 100

21

Money

(0, 0) (4, 0) (0, 10) d q

22

Money

(0, 0) (4, 0) (0, 10) d q

Notice that there are 16 integer points on the boundary of this triangle, so our apple

• rchard area estimate is

22

Money

(0, 0) (4, 0) (0, 10) d q

Notice that there are 16 integer points on the boundary of this triangle, so our apple

• rchard area estimate is

Area ≈ 29 − 1 2

• (16) = 21

22

Money

(0, 0) (4, 0) (0, 10) d q

Notice that there are 16 integer points on the boundary of this triangle, so our apple

• rchard area estimate is

Area ≈ 29 − 1 2

• (16) = 21

Since the actual area is 20, this estimate is again 1 too high.

22

Outline

1

Apple Orchards Approximation Area

2

Money Make a List Coins on the Table Geometry

3

The Main Theorem Statement Rectangles and Right Triangles More Money Adding and Subtracting Polygons Finishing the Proof

4

Baseball

5

Conclusion

23

Pick’s Theorem

Suppose we have a simple lattice polygon (so every vertex has integer coordinates).

24

Pick’s Theorem

Suppose we have a simple lattice polygon (so every vertex has integer coordinates). Let T = total number of integer lattice points inside or on the boundary of the polygon, and B = number of boundary integer lattice points.

24

Pick’s Theorem

Suppose we have a simple lattice polygon (so every vertex has integer coordinates). Let T = total number of integer lattice points inside or on the boundary of the polygon, and B = number of boundary integer lattice points.

Theorem (Pick’s Theorem (1899))

Area of polygon = T − B

2 − 1.

24

Pick’s Theorem

Suppose we have a simple lattice polygon (so every vertex has integer coordinates). Let T = total number of integer lattice points inside or on the boundary of the polygon, and B = number of boundary integer lattice points.

Theorem (Pick’s Theorem (1899))

Area of polygon = T − B

2 − 1.

Notice that this means that every such area is either an integer or half

• f an odd integer.

24

Pick’s Theorem

For our apple orchard,

25

Pick’s Theorem

For our apple orchard, T = 23

25

Pick’s Theorem

For our apple orchard, T = 23 B = 8

25

Pick’s Theorem

For our apple orchard, T = 23 B = 8 And the area was indeed T − B

2 − 1 = 23 − 8 2 − 1 = 18.

25

Pick’s Theorem

Georg Pick (1859-1942)

26

Pick’s Theorem

Georg Pick (1859-1942) Early in his career, was an assistant to the physicist Ernst Mach

26

Pick’s Theorem

Georg Pick (1859-1942) Early in his career, was an assistant to the physicist Ernst Mach Headed the committee at the University

• f Prague which appointed his friend

Albert Einstein to a position in 1911

26

Pick’s Theorem

Georg Pick (1859-1942) Early in his career, was an assistant to the physicist Ernst Mach Headed the committee at the University

• f Prague which appointed his friend

Albert Einstein to a position in 1911 Died tragically in a concentration camp in 1942

26

Pick’s Theorem

Our lattice polygons must be simple:

27

Pick’s Theorem

Our lattice polygons must be simple:

27

Pick’s Theorem

Our lattice polygons must be simple: The boundary of the polygon must be a continuous loop with no crossings, no repeated vertices, and no holes.

27

Pick’s Theorem for Axis-Parallel Rectangles

Consider the following rectangle:

28

Pick’s Theorem for Axis-Parallel Rectangles

Consider the following rectangle: Here, T = 18 · 4 = 72

28

Pick’s Theorem for Axis-Parallel Rectangles

Consider the following rectangle: Here, T = 18 · 4 = 72 B = 17 + 3 + 17 + 3 = 40

28

Pick’s Theorem for Axis-Parallel Rectangles

Consider the following rectangle: Here, T = 18 · 4 = 72 B = 17 + 3 + 17 + 3 = 40 and T − B

2 − 1 = 72 − 40 2 − 1 = 51.

28

Pick’s Theorem for Axis-Parallel Rectangles

Consider the following rectangle: Here, T = 18 · 4 = 72 B = 17 + 3 + 17 + 3 = 40 and T − B

2 − 1 = 72 − 40 2 − 1 = 51.

Surprisingly, this is indeed the area of the above rectangle.

28

Pick’s Theorem for Axis-Parallel Rectangles

In general, suppose we have an axis-parallel a by b rectangle:

a b

29

Pick’s Theorem for Axis-Parallel Rectangles

In general, suppose we have an axis-parallel a by b rectangle:

a b

T = (a + 1)(b + 1)

29

Pick’s Theorem for Axis-Parallel Rectangles

In general, suppose we have an axis-parallel a by b rectangle:

a b

T = (a + 1)(b + 1) B = a + b + a + b = 2a + 2b

29

Pick’s Theorem for Axis-Parallel Rectangles

In general, suppose we have an axis-parallel a by b rectangle:

a b

T = (a + 1)(b + 1) B = a + b + a + b = 2a + 2b so T − B

2 − 1 = (a + 1)(b + 1) − 2a+2b 2

− 1 = ab.

29

Pick’s Theorem for Axis-Parallel Right Triangles

If we introduce one of the diagonals in our a by b rectangle:

a b

30

Pick’s Theorem for Axis-Parallel Right Triangles

If we introduce one of the diagonals in our a by b rectangle:

a b

Suppose there are B∗ integer points on the diagonal, including the two corners.

30

Pick’s Theorem for Axis-Parallel Right Triangles

If we introduce one of the diagonals in our a by b rectangle:

a b

Suppose there are B∗ integer points on the diagonal, including the two corners. Then each triangle has B = a + b + B∗ − 1.

30

Pick’s Theorem for Axis-Parallel Right Triangles

If we introduce one of the diagonals in our a by b rectangle:

a b

Suppose there are B∗ integer points on the diagonal, including the two corners. Then each triangle has B = a + b + B∗ − 1. Since the B∗ diagonal points are in both triangles, then, for each triangle, we have 2T = (a + 1)(b + 1) + B∗.

30

Pick’s Theorem for Axis-Parallel Right Triangles

If we introduce one of the diagonals in our a by b rectangle:

a b

Suppose there are B∗ integer points on the diagonal, including the two corners. Then each triangle has B = a + b + B∗ − 1. Since the B∗ diagonal points are in both triangles, then, for each triangle, we have 2T = (a + 1)(b + 1) + B∗. Thus T − B

2 − 1 = (a+1)(b+1)+B∗ 2

− a+b+B∗−1

2

− 1 = ab

2 .

30

Pick’s Theorem for Axis-Parallel Right Triangles

a b

We didn’t need it here, but it useful to know that B∗ = gcd(a, b) + 1.

31

Pick’s Theorem for Axis-Parallel Right Triangles

a b

We didn’t need it here, but it useful to know that B∗ = gcd(a, b) + 1. Thus each triangle satisfies T = (a + 1)(b + 1) + B∗ 2 = (a + 1)(b + 1) + gcd(a, b) + 1 2

31

More Money

Interlude: Let’s look at money again.

32

More Money

Interlude: Let’s look at money again. Suppose we want to make change for D dollars using quarters, dimes, and nickels.

32

More Money

Interlude: Let’s look at money again. Suppose we want to make change for D dollars using quarters, dimes, and nickels. Then we need 25q + 10d + 5n = 100D.

32

More Money

Interlude: Let’s look at money again. Suppose we want to make change for D dollars using quarters, dimes, and nickels. Then we need 25q + 10d + 5n = 100D. As before, we can ignore nickels and solve 25q + 10d ≤ 100D, which is 5q + 2d ≤ 20D.

32

More Money

So we have 5q + 2d ≤ 20D with q ≥ 0 and d ≥ 0.

33

More Money

So we have 5q + 2d ≤ 20D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with corners (0, 0), (4D, 0), and (0, 10D).

33

More Money

So we have 5q + 2d ≤ 20D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with corners (0, 0), (4D, 0), and (0, 10D).

(0, 0) (4D, 0) (0, 10D) d q

5q + 2d = 20D 33

More Money

So we have 5q + 2d ≤ 20D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with corners (0, 0), (4D, 0), and (0, 10D).

(0, 0) (4D, 0) (0, 10D) d q

5q + 2d = 20D

For this right triangle, we have

33

More Money

So we have 5q + 2d ≤ 20D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with corners (0, 0), (4D, 0), and (0, 10D).

(0, 0) (4D, 0) (0, 10D) d q

5q + 2d = 20D

For this right triangle, we have T = (a + 1)(b + 1) + gcd(a, b) + 1 2

33

More Money

So we have 5q + 2d ≤ 20D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with corners (0, 0), (4D, 0), and (0, 10D).

(0, 0) (4D, 0) (0, 10D) d q

5q + 2d = 20D

For this right triangle, we have T = (a + 1)(b + 1) + gcd(a, b) + 1 2 = (4D + 1)(10D + 1) + gcd(4D, 10D) + 1 2

33

More Money

So we have 5q + 2d ≤ 20D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with corners (0, 0), (4D, 0), and (0, 10D).

(0, 0) (4D, 0) (0, 10D) d q

5q + 2d = 20D

For this right triangle, we have T = (a + 1)(b + 1) + gcd(a, b) + 1 2 = (4D + 1)(10D + 1) + gcd(4D, 10D) + 1 2 = (4D + 1)(10D + 1) + 2D + 1 2

33

More Money

So we have 5q + 2d ≤ 20D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with corners (0, 0), (4D, 0), and (0, 10D).

(0, 0) (4D, 0) (0, 10D) d q

5q + 2d = 20D

For this right triangle, we have T = (a + 1)(b + 1) + gcd(a, b) + 1 2 = (4D + 1)(10D + 1) + gcd(4D, 10D) + 1 2 = (4D + 1)(10D + 1) + 2D + 1 2 = 20D2 + 8D + 1

33

More Money

Theorem

The number of ways to make change for D dollars from a supply of quarters, dimes, and nickels is 20D2 + 8D + 1.

34

Adding and Subtracting Polygons

Now we will prove Pick’s Theorem for general simple lattice polygons.

35

Adding and Subtracting Polygons

Now we will prove Pick’s Theorem for general simple lattice polygons. We first show that Pick’s Theorem is additive (and subtractive). Suppose the simple polygon P is partitioned into two simple polygons P′ and P′′.

35

Adding and Subtracting Polygons

Now we will prove Pick’s Theorem for general simple lattice polygons. We first show that Pick’s Theorem is additive (and subtractive). Suppose the simple polygon P is partitioned into two simple polygons P′ and P′′.

P ′ P ′′ P

35

Adding and Subtracting Polygons

Now we will prove Pick’s Theorem for general simple lattice polygons. We first show that Pick’s Theorem is additive (and subtractive). Suppose the simple polygon P is partitioned into two simple polygons P′ and P′′.

P ′ P ′′ P

Then definitely Area of P = Area of P′ + Area of P′′.

35

Adding and Subtracting Polygons

P ′ P ′′ P

36

Adding and Subtracting Polygons

P ′ P ′′ P

Now assume that P′ and P′′ satisfy Pick’s Theorem, so that

36

Adding and Subtracting Polygons

P ′ P ′′ P

Now assume that P′ and P′′ satisfy Pick’s Theorem, so that Area of P′ = T ′ − B′ 2 − 1 and Area of P′′ = T ′′ − B′′ 2 − 1.

36

Adding and Subtracting Polygons

P ′ P ′′ P

We will show that the big polygon P also satisfies Pick’s Theorem. Let B∗ be the number of common lattice points on the boundary between P′ and P′′, including the two endpoints.

37

Adding and Subtracting Polygons

P ′ P ′′ P

We will show that the big polygon P also satisfies Pick’s Theorem. Let B∗ be the number of common lattice points on the boundary between P′ and P′′, including the two endpoints. For P, notice that T = T ′ + T ′′ − B∗

37

Adding and Subtracting Polygons

P ′ P ′′ P

We will show that the big polygon P also satisfies Pick’s Theorem. Let B∗ be the number of common lattice points on the boundary between P′ and P′′, including the two endpoints. For P, notice that T = T ′ + T ′′ − B∗ and B = B′ + B′′ − 2B∗ + 2.

37

Adding and Subtracting Polygons

P ′ P ′′ P

Then Area of P = Area of P′ + Area of P′′

38

Adding and Subtracting Polygons

P ′ P ′′ P

Then Area of P = Area of P′ + Area of P′′ =

• T ′ − B′

2 − 1

• +
• T ′′ − B′′

2 − 1

• 38

Adding and Subtracting Polygons

P ′ P ′′ P

Then Area of P = Area of P′ + Area of P′′ =

• T ′ − B′

2 − 1

• +
• T ′′ − B′′

2 − 1

• =
• T ′ + T ′′ − B∗

− B′ + B′′ − 2B∗ + 2 2 − 1

38

Adding and Subtracting Polygons

P ′ P ′′ P

Then Area of P = Area of P′ + Area of P′′ =

• T ′ − B′

2 − 1

• +
• T ′′ − B′′

2 − 1

• =
• T ′ + T ′′ − B∗

− B′ + B′′ − 2B∗ + 2 2 − 1 = T − B 2 − 1

38

Adding and Subtracting Polygons

P ′ P ′′ P

So if P′ and P′′ satisfy Pick’s Theorem, then so does P. Thus the Pick relation T − B

2 − 1 is additive, just like area.

39

Adding and Subtracting Polygons

P ′ P ′′ P

So if P′ and P′′ satisfy Pick’s Theorem, then so does P. Thus the Pick relation T − B

2 − 1 is additive, just like area.

It is also subtractive too: If Pick’s Theorem holds for P and P′, then it holds for P′′.

39

Finishing the Proof

We now show that Pick’s Theorem holds for any lattice triangle.

40

Finishing the Proof

We now show that Pick’s Theorem holds for any lattice triangle. But this is easy — just “frame” the triangle like we were doing before.

40

Finishing the Proof

We now show that Pick’s Theorem holds for any lattice triangle. But this is easy — just “frame” the triangle like we were doing before.

40

Finishing the Proof

We now show that Pick’s Theorem holds for any lattice triangle. But this is easy — just “frame” the triangle like we were doing before. If the frame looks like this, then Pick’s Theorem is true for the big rectangle and for the three axis-parallel right triangles, so, since it is subtractive, it is true for the framed triangle too.

40

Finishing the Proof

Note that the other possibility for the frame looks like this:

41

Finishing the Proof

Note that the other possibility for the frame looks like this:

41

Finishing the Proof

Note that the other possibility for the frame looks like this: Here we use the fact that Pick’s Theorem is additive and subtractive.

41

Finishing the Proof

Finally, we use the fact that any polygon can be partitioned into triangles (and that any lattice polygon can be partitioned into lattice triangles).

42

Finishing the Proof

Finally, we use the fact that any polygon can be partitioned into triangles (and that any lattice polygon can be partitioned into lattice triangles).

42

Finishing the Proof

Finally, we use the fact that any polygon can be partitioned into triangles (and that any lattice polygon can be partitioned into lattice triangles).

42

Finishing the Proof

Finally, we use the fact that any polygon can be partitioned into triangles (and that any lattice polygon can be partitioned into lattice triangles). Since Pick’s Theorem is additive, it holds for any lattice polygon.

42

Finishing the Proof

Finally, we use the fact that any polygon can be partitioned into triangles (and that any lattice polygon can be partitioned into lattice triangles). Since Pick’s Theorem is additive, it holds for any lattice polygon. This finishes the proof.

42

Outline

1

Apple Orchards Approximation Area

2

Money Make a List Coins on the Table Geometry

3

The Main Theorem Statement Rectangles and Right Triangles More Money Adding and Subtracting Polygons Finishing the Proof

4

Baseball

5

Conclusion

43

Baseball

In how many ways can a baseball player achieve the batting average .273 with at most 2,000 at-bats?

44

Baseball

Batting Average = Number of hits Number of at-bats rounded to the nearest .001.

45

Baseball

Batting Average = Number of hits Number of at-bats rounded to the nearest .001. Examples:

45

Baseball

Batting Average = Number of hits Number of at-bats rounded to the nearest .001. Examples: 3 11 = .2727272727 . . .

45

Baseball

Batting Average = Number of hits Number of at-bats rounded to the nearest .001. Examples: 3 11 = .2727272727 . . . 41 150 = .2733333333 . . .

45

Baseball

Batting Average = Number of hits Number of at-bats rounded to the nearest .001. Examples: 3 11 = .2727272727 . . . 41 150 = .2733333333 . . . 131 480 = .2729166666 . . .

45

Baseball

Batting Average = Number of hits Number of at-bats rounded to the nearest .001. Examples: 3 11 = .2727272727 . . . 41 150 = .2733333333 . . . 131 480 = .2729166666 . . . 273 1000 = .273

45

Baseball

So we want integer solutions (x, y) such that 0.2725 ≤ y x < 0.2735 and 0 < x ≤ 2000.

46

Baseball

So we want integer solutions (x, y) such that 0.2725 ≤ y x < 0.2735 and 0 < x ≤ 2000. The first is equivalent to 545 2000 ≤ y x < 547 2000.

46

Baseball

Thus we want all integer points (x, y) in the triangle bounded by (0, 0), (2000, 545), and (2000, 547), except for the points on the segment joining (0, 0) and (2000, 547).

47

Baseball

Thus we want all integer points (x, y) in the triangle bounded by (0, 0), (2000, 545), and (2000, 547), except for the points on the segment joining (0, 0) and (2000, 547).

Slope = y/x = exact batting average y hits x at-bats (0, 0) (2000, 545) (2000, 547) (x, y)

47

Baseball

Slope = y/x = exact batting average y hits x at-bats (0, 0) (2000, 545) (2000, 547) (x, y)

This triangle has an area of 1

2 · 2000 · 2 = 2000.

48

Baseball

Slope = y/x = exact batting average y hits x at-bats (0, 0) (2000, 545) (2000, 547) (x, y)

This triangle has an area of 1

2 · 2000 · 2 = 2000.

Since gcd(2000, 545) = 5 and gcd(2000, 547) = 1, we have B = (5 + 1) + (1 + 1) + 3 − 3 = 8.

48

Baseball

Slope = y/x = exact batting average y hits x at-bats (0, 0) (2000, 545) (2000, 547) (x, y)

This triangle has an area of 1

2 · 2000 · 2 = 2000.

Since gcd(2000, 545) = 5 and gcd(2000, 547) = 1, we have B = (5 + 1) + (1 + 1) + 3 − 3 = 8. Thus, by Pick’s formula, 2000 = T − 8 2 − 1 = T − 5.

48

Baseball

Slope = y/x = exact batting average y hits x at-bats (0, 0) (2000, 545) (2000, 547) (x, y)

Since 2000 = T − 5, we have T = 2005.

49

Baseball

Slope = y/x = exact batting average y hits x at-bats (0, 0) (2000, 545) (2000, 547) (x, y)

Since 2000 = T − 5, we have T = 2005. We don’t count (0, 0) and (2000, 547), so there are 2005 − 2 = 2003 ways to achieve the batting average .273 with at most 2000 at-bats.

49

Outline

1

Apple Orchards Approximation Area

2

Money Make a List Coins on the Table Geometry

3

The Main Theorem Statement Rectangles and Right Triangles More Money Adding and Subtracting Polygons Finishing the Proof

4

Baseball

5

Conclusion

50

Conclusion

What about higher dimensions?

51

Conclusion

What about higher dimensions?

(1, 0, 0) (0, 1, 0) (0, 0, 1) (1, 1, 1)

51

Conclusion

What about higher dimensions?

(1, 0, 0) (0, 1, 0) (0, 0, 1) (1, 1, 1)

Each of the five polyhedra have four lattice points, all of which are boundary points.

51

Conclusion

What about higher dimensions?

(1, 0, 0) (0, 1, 0) (0, 0, 1) (1, 1, 1)

Each of the five polyhedra have four lattice points, all of which are boundary points. The four congruent outer pyramids have volume V = area of base·height

3

= 1

6.

51

Conclusion

What about higher dimensions?

(1, 0, 0) (0, 1, 0) (0, 0, 1) (1, 1, 1)

Each of the five polyhedra have four lattice points, all of which are boundary points. The four congruent outer pyramids have volume V = area of base·height

3

= 1

6.

So the central tetrahedron has volume 1

3.

51

Conclusion

What about higher dimensions?

(1, 0, 0) (0, 1, 0) (0, 0, 1) (1, 1, 1)

Each of the five polyhedra have four lattice points, all of which are boundary points. The four congruent outer pyramids have volume V = area of base·height

3

= 1

6.

So the central tetrahedron has volume 1

3.

A modified version of Pick’s Theorem does generalize to higher dimensions,

51

Conclusion

What about higher dimensions?

(1, 0, 0) (0, 1, 0) (0, 0, 1) (1, 1, 1)

Each of the five polyhedra have four lattice points, all of which are boundary points. The four congruent outer pyramids have volume V = area of base·height

3

= 1

6.

So the central tetrahedron has volume 1

3.

A modified version of Pick’s Theorem does generalize to higher dimensions, which would allow us to use more coins!

51

Reference

How to Guard an Art Gallery and Other Discrete Mathematical Adventures, T. S. Michael, Johns Hopkins University Press, 2009.

52