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 3. Geometry d (0 , 10) 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) 25 q + 10 d = 100 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 (0 , 0) q (4 , 0) 20
Money 2. Coins on the Table 3. Geometry d d (0 , 10) (0 , 10) 25 q + 10 d = 100 25 q + 10 d = 100 q (0 , 0) (4 , 0) q (0 , 0) (4 , 0) 21
Money d (0 , 10) q (0 , 0) (4 , 0) 22
Money d Notice that there are 16 integer points on the (0 , 10) boundary of this triangle, so our apple orchard area estimate is q (0 , 0) (4 , 0) 22
Money d Notice that there are 16 integer points on the (0 , 10) boundary of this triangle, so our apple orchard area estimate is � 1 � Area ≈ 29 − ( 16 ) = 21 2 q (0 , 0) (4 , 0) 22
Money d Notice that there are 16 integer points on the (0 , 10) boundary of this triangle, so our apple orchard area estimate is � 1 � Area ≈ 29 − ( 16 ) = 21 2 Since the actual area is 20, this estimate is again 1 too high. q (0 , 0) (4 , 0) 22
Outline Apple Orchards 1 Approximation Area Money 2 Make a List Coins on the Table Geometry The Main Theorem 3 Statement Rectangles and Right Triangles More Money Adding and Subtracting Polygons Finishing the Proof Baseball 4 Conclusion 5 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 of 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 Early in his career, was an assistant to the physicist Ernst Mach Georg Pick (1859-1942) 26
Pick’s Theorem Early in his career, was an assistant to the physicist Ernst Mach Headed the committee at the University of Prague which appointed his friend Albert Einstein to a position in 1911 Georg Pick (1859-1942) 26
Pick’s Theorem Early in his career, was an assistant to the physicist Ernst Mach Headed the committee at the University of Prague which appointed his friend Albert Einstein to a position in 1911 Died tragically in a concentration camp in 1942 Georg Pick (1859-1942) 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: b a 29
Pick’s Theorem for Axis-Parallel Rectangles In general, suppose we have an axis-parallel a by b rectangle: b a 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: b a T = ( a + 1 )( b + 1 ) B = a + b + a + b = 2 a + 2 b 29
Pick’s Theorem for Axis-Parallel Rectangles In general, suppose we have an axis-parallel a by b rectangle: b a T = ( a + 1 )( b + 1 ) B = a + b + a + b = 2 a + 2 b so T − B 2 − 1 = ( a + 1 )( b + 1 ) − 2 a + 2 b − 1 = ab . 2 29
Pick’s Theorem for Axis-Parallel Right Triangles If we introduce one of the diagonals in our a by b rectangle: b a 30
Pick’s Theorem for Axis-Parallel Right Triangles If we introduce one of the diagonals in our a by b rectangle: b a 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: b a 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: b a 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 2 T = ( 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: b a 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 2 T = ( a + 1 )( b + 1 ) + B ∗ . 2 − 1 = ( a + 1 )( b + 1 )+ B ∗ Thus T − B − a + b + B ∗ − 1 − 1 = ab 2 . 2 2 30
Pick’s Theorem for Axis-Parallel Right Triangles b a 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 b a 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 ∗ = ( a + 1 )( b + 1 ) + gcd ( a , b ) + 1 2 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 25 q + 10 d + 5 n = 100 D . 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 25 q + 10 d + 5 n = 100 D . As before, we can ignore nickels and solve 25 q + 10 d ≤ 100 D , which is 5 q + 2 d ≤ 20 D . 32
More Money So we have 5 q + 2 d ≤ 20 D with q ≥ 0 and d ≥ 0. 33
More Money So we have 5 q + 2 d ≤ 20 D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with corners ( 0 , 0 ) , ( 4 D , 0 ) , and ( 0 , 10 D ) . 33
More Money d (0 , 10 D ) So we have 5 q + 2 d ≤ 20 D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with 5 q + 2 d = 20 D corners ( 0 , 0 ) , ( 4 D , 0 ) , and ( 0 , 10 D ) . q (0 , 0) (4 D, 0) 33
More Money d (0 , 10 D ) So we have 5 q + 2 d ≤ 20 D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with 5 q + 2 d = 20 D corners ( 0 , 0 ) , ( 4 D , 0 ) , and ( 0 , 10 D ) . q (0 , 0) (4 D, 0) For this right triangle, we have 33
More Money d (0 , 10 D ) So we have 5 q + 2 d ≤ 20 D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with 5 q + 2 d = 20 D corners ( 0 , 0 ) , ( 4 D , 0 ) , and ( 0 , 10 D ) . q (0 , 0) (4 D, 0) For this right triangle, we have ( a + 1 )( b + 1 ) + gcd ( a , b ) + 1 T = 2 33
More Money d (0 , 10 D ) So we have 5 q + 2 d ≤ 20 D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with 5 q + 2 d = 20 D corners ( 0 , 0 ) , ( 4 D , 0 ) , and ( 0 , 10 D ) . q (0 , 0) (4 D, 0) For this right triangle, we have ( a + 1 )( b + 1 ) + gcd ( a , b ) + 1 T = 2 ( 4 D + 1 )( 10 D + 1 ) + gcd ( 4 D , 10 D ) + 1 = 2 33
More Money d (0 , 10 D ) So we have 5 q + 2 d ≤ 20 D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with 5 q + 2 d = 20 D corners ( 0 , 0 ) , ( 4 D , 0 ) , and ( 0 , 10 D ) . q (0 , 0) (4 D, 0) For this right triangle, we have ( a + 1 )( b + 1 ) + gcd ( a , b ) + 1 T = 2 ( 4 D + 1 )( 10 D + 1 ) + gcd ( 4 D , 10 D ) + 1 = 2 ( 4 D + 1 )( 10 D + 1 ) + 2 D + 1 = 2 33
More Money d (0 , 10 D ) So we have 5 q + 2 d ≤ 20 D with q ≥ 0 and d ≥ 0. This is solved by all integer points in the triangle with 5 q + 2 d = 20 D corners ( 0 , 0 ) , ( 4 D , 0 ) , and ( 0 , 10 D ) . q (0 , 0) (4 D, 0) For this right triangle, we have ( a + 1 )( b + 1 ) + gcd ( a , b ) + 1 T = 2 ( 4 D + 1 )( 10 D + 1 ) + gcd ( 4 D , 10 D ) + 1 = 2 ( 4 D + 1 )( 10 D + 1 ) + 2 D + 1 = 2 20 D 2 + 8 D + 1 = 33
More Money Theorem The number of ways to make change for D dollars from a supply of quarters, dimes, and nickels is 20 D 2 + 8 D + 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
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