A cell is a polygon
Four sites stand in a square region from 0 to 12 on both axes: A at (2, 3), B at (2, 9), C at (10, 3) and D at (7, 8). The cell of A is every point of the region nearer to A than to B, C or D.
Every edge of the cell is straight, either a piece of a perpendicular bisector or a piece of the border, so the cell is a polygon. Its area comes from its corners, like the area of any other polygon.
The cell of A, in gold, in the region from 0 to 12.
The corners
Three bisectors bound the cell. B is directly above A, so their bisector is the horizontal line through the midpoint (2, 6): y = 6. C is level with A, so their bisector is the vertical line through (6, 3): x = 6. For A and D, the midpoint is (4.5, 5.5) and AD has gradient 1, so the bisector has gradient −1: y = −x + 10, or x + y = 10.
The line y = 6 meets x + y = 10 at (4, 6), which is from A, B and D: a vertex. The line x = 6 meets x + y = 10 at (6, 4), which is from A, C and D. The cell’s corners, in order around it, are (0, 0), (6, 0), (6, 4), (4, 6) and (0, 6).
Two of the five edges lie on the border of the region, from (0, 6) down to (0, 0) and from (0, 0) across to (6, 0). The other three are pieces of the bisectors with C, D and B.
A square with a corner cut off
The cell sits inside the 6 by 6 square with corners (0, 0), (6, 0), (6, 6) and (0, 6), whose area is 36. The corner (6, 6) is not in the cell: its squared distance is 25 to A and only 5 to D, so it is in the cell of D. The edge x + y = 10 slices that corner off.
The slice is the right triangle with corners (6, 4), (6, 6) and (4, 6). Both legs are 2, so its area is ½ × 2 × 2 = 2. The cell covers 36 − 2 = 34 square units.
D’s cell, in gold, holds the corner (6, 6) of the square, and the triangle it takes from the square has legs 2 and 2.
The shoelace formula
The shoelace formula gives the area of any polygon from its corners listed in order around it. For each corner and the next, multiply the first x by the second y and subtract the second x times the first y. Add the results and halve the total.
For (0, 0), (6, 0), (6, 4), (4, 6), (0, 6) and back to (0, 0), the products are 0 × 0 − 6 × 0 = 0, 6 × 4 − 6 × 0 = 24, 6 × 6 − 4 × 4 = 20, 4 × 6 − 0 × 6 = 24 and 0 × 0 − 0 × 6 = 0. They total 68, and half of 68 is 34, the same area.
All four cells
The edge between B and D is y = 5x − 14, which reaches the top of the region at (5.2, 12). The edge between C and D is 3x − 5y = −2, which reaches the right side at (12, 7.6).
B’s cell has corners (0, 6), (4, 6), (5.2, 12) and (0, 12): a trapezoid with parallel sides 4 and 5.2 that are 6 apart, so its area is ½ × (4 + 5.2) × 6 = 27.6. C’s cell has corners (6, 0), (12, 0), (12, 7.6) and (6, 4): a trapezoid with parallel sides 4 and 7.6 that are 6 apart, so its area is ½ × (4 + 7.6) × 6 = 34.8.
The four cells fill the region, whose area is 12 × 12 = 144, so D’s cell is what is left: 144 − 34 − 27.6 − 34.8 = 47.6.
B’s cell, in gold, is a trapezoid of area 27.6, with its slanting edge on y = 5x − 14.
Closed by the border, or by bisectors
A’s cell has two edges on the border. Without the border it would run on forever down and to the left, with no area at all, so its area depends on the region it is drawn in. Every cell of a site on the outside of the group is like this.
A site surrounded by others can have a cell bounded by bisectors alone. Put O at (6, 6) among P at (2, 2), Q at (10, 2), R at (10, 10) and S at (2, 10). The bisectors of O with the four outer sites are x + y = 8, x − y = 4, x + y = 16 and x − y = −4, and they enclose a diamond with corners (6, 2), (10, 6), (6, 10) and (2, 6). Its diagonals are both 8, so its area is 8 × 8 ÷ 2 = 32, and it stays 32 in any larger region.
O’s cell, in gold, is a diamond of area 32. None of its edges touches the border.
Adding a site
A new site takes every point that is nearer to it than to its old site. Adding it changes no distance to the old sites, so no point moves into an old cell, and an old cell can only lose area.
Add E at (3, 2), close to A. The bisector of A and E is the line y = x, and A keeps the side above it. A’s cell becomes (0, 0), (5, 5), (4, 6) and (0, 6), where (5, 5) is on both y = x and x + y = 10. The shoelace products are 0, 5 × 6 − 4 × 5 = 10, 4 × 6 − 0 × 6 = 24 and 0, which total 34, so the new area is 17: half of what it was.
With E added, A’s cell, in gold, is cut along y = x, and its area falls from 34 to 17.
The usual mistakes
Adding the sides instead of multiplying. For a square cell of side 6, 4 × 6 = 24 is the perimeter; the area is 6 × 6 = 36.
Using the whole region. The region holds four cells, so its area, 144, is not the area of one of them.
Forgetting the slice. The 6 by 6 square gives 36, but the corner beyond x + y = 10 belongs to D.
Taking off a whole 2 by 2 square. The piece cut off is the triangle inside it, of area 2, not 4.
Listing the corners out of order for the shoelace formula. They must go around the cell in turn, or the sum no longer measures its area.
Stores and rain gauges
In the first application below, a town is split between three stores, and each store’s catchment is a cell whose area gives its share of the town. In the second, the cells of three rain gauges weight their readings, and the weighted average rainfall is compared with the plain mean.
Worked example: The Catchment Area of a Store in a Town with Three Stores, and the Share of the Town Each Store Serves
Question A town is the rectangle 0 ≤ x ≤ 12, 0 ≤ y ≤ 10, in kilometers, with three stores of one chain: A at (1, 2), B at (5, 8) and C at (11, 2). The chain takes each store's catchment to be the part of the town nearer to it than to the other two stores. (a) Find the area of B's catchment. (b) What percentage of the town's area does each store's catchment cover?
1.The edge between A and C is x = 6. The edge between A and B: the midpoint is (3, 5) and AB has gradient 64 = 32, so the edge has gradient −23: y − 5 = −23(x − 3), which is 2x + 3y = 21. The edge between B and C: the midpoint is (8, 5) and BC has gradient −1, so the edge is y = x − 3.
The three bisectors meet at (6, 3). 2.The three edges meet at (6, 3): 2 × 6 + 3 × 3 = 21 and 3 = 6 − 3. From there, 2x + 3y = 21 runs to the west boundary at (0, 7), and y = x − 3 runs to the east boundary at (12, 9). B's catchment has corners (0, 7), (6, 3), (12, 9), (12, 10) and (0, 10).
B's catchment has corners (0, 7), (6, 3), (12, 9), (12, 10) and (0, 10). 3.(a) With the corners in that order, the shoelace products are 0 × 3 − 6 × 7 = −42, 6 × 9 − 12 × 3 = 18, 12 × 10 − 12 × 9 = 12, 12 × 10 − 0 × 10 = 120 and 0 × 7 − 0 × 10 = 0. The area is 12(−42 + 18 + 12 + 120 + 0) = 54 km².
(a) The shoelace formula gives 12 × 108 = 54 km². 4.A's catchment is the trapezoid (0, 0), (6, 0), (6, 3), (0, 7), with parallel sides 7 and 3 that are 6 apart: 12(7 + 3) × 6 = 30 km². C's is the trapezoid (6, 0), (12, 0), (12, 9), (6, 3): 12(3 + 9) × 6 = 36 km². Check: 54 + 30 + 36 = 120 = 12 × 10.
A's catchment is 30 km² and C's is 36 km², and the three make 120 km². 5.(b) As shares of the town's 120 km²: A covers 30120 = 25%, B covers 54120 = 45% and C covers 36120 = 30%.
(b) A covers 25%, B 45% and C 30% of the town.
Answer: (a) 54 km². (b) A 25%, B 45%, C 30%
Common mistakes
- Drawing the edge between A and B through (3, 5) with the gradient of AB, 32. The edge is perpendicular to AB, with gradient −23; with the wrong gradient the three edges no longer meet at one point.
- Taking the shoelace products with the corners out of order. The corners must be listed in order round the cell, or the sum no longer measures its area; the check 54 + 30 + 36 = 120 catches the slip.
Worked example: Rainfall over a River Catchment from Three Rain Gauges, Weighted by the Area Nearest Each Gauge
Question A river's catchment is modeled as the rectangle 0 ≤ x ≤ 16, 0 ≤ y ≤ 10, in kilometers, with three rain gauges: A at (1, 3), B at (7, 3) and C at (7, 9). After a storm they record 30 mm, 20 mm and 40 mm. Hydrologists estimate the average rainfall over the catchment by giving each gauge's reading to the part of the catchment nearest that gauge, which is the Thiessen polygon method. (a) Find the area of each gauge's cell. (b) Find the estimated average rainfall over the catchment, and compare it with the plain mean of the three readings.
1.The edge between A and B is x = 4, and the edge between B and C is y = 6. For A and C, the midpoint is (4, 6) and AC has gradient 1, so the edge is x + y = 10. All three edges meet at (4, 6), which is √18 km from each gauge.
The edges x = 4, y = 6 and x + y = 10 meet at (4, 6). 2.B's cell is the rectangle with corners (4, 0), (16, 0), (16, 6) and (4, 6), so its area is 12 × 6 = 72 km².
B's cell is a 12 km by 6 km rectangle: 72 km². 3.A's cell has corners (0, 0), (4, 0), (4, 6) and (0, 10): a trapezoid with parallel sides 10 and 6 that are 4 apart, so its area is 12(10 + 6) × 4 = 32 km². C's cell has corners (4, 6), (16, 6), (16, 10) and (0, 10): a trapezoid with parallel sides 12 and 16 that are 4 apart, so its area is 12(12 + 16) × 4 = 56 km².
A's cell is a trapezoid of 32 km² and C's a trapezoid of 56 km². 4.(a) The cells are 32 km² for A, 72 km² for B and 56 km² for C. Check: 32 + 72 + 56 = 160 = 16 × 10.
(a) The cells are 32, 72 and 56 km², which make up the 160 km² catchment. 5.Weight each reading by its cell's area and divide by the whole area: 32 × 30 + 72 × 20 + 56 × 40160 = 960 + 1440 + 2240160 = 4640160 = 29 mm.
Each reading is weighted by its cell's area: 4640160 = 29 mm. 6.(b) The estimate is 29 mm, which is 1 mm less than the plain mean, 30 + 20 + 403 = 30 mm. It is lower because B, with the lowest reading, stands for the largest cell.
(b) 29 mm, 1 mm less than the plain mean of 30 mm.
Answer: (a) A 32 km², B 72 km², C 56 km². (b) 29 mm, which is 1 mm less than the plain mean of 30 mm
Common mistakes
- Averaging the three readings to get 30 mm. That counts every gauge as standing for the same area, but B's cell is 72 km² and A's is only 32 km².
- Dividing the weighted total, 4640, by 3 instead of by the whole area, 160 km². The weights are areas, so the total is divided by the sum of the areas.