Math Detours · Grade 6 · Sharpen · 12

Area by counting dots

School teaches you to chop a shape into rectangles and triangles. That works. But for shapes whose corners sit on grid points there is a formula so simple it feels like cheating.

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Tonight's idea

You know two honest ways to find the area of a polygon drawn on a grid. Cut it up into rectangles and right triangles, and add. Or box it in: draw the smallest rectangle around it, then subtract the pieces in the corners. Both work for any polygon. Both are fiddly when the shape has five or six corners.

In 1899 Georg Pick noticed something odd. If every corner of the polygon sits exactly on a grid point (a lattice polygon), then the area depends only on two counts:

\(I\) = number of grid points strictly inside the shape
\(B\) = number of grid points on the boundary, corners included (a point on a side counts even if it is not a corner)
Area \(= I + \dfrac{B}{2} - 1\)

No lengths, no heights, no cutting. Just count dots. It works for any lattice polygon that does not cross itself, however spiky.

Roughly why: each inside point owns a unit square of area around it, and each boundary point about half of one. The full proof is a page long. Checking the formula takes thirty seconds, and you will do that now.

Watch one

Filled dots are on the boundary (\(B = 9\)). Hollow dots are strictly inside (\(I = 10\)). Faint dots are outside.

Pick: \(A = 10 + \tfrac{9}{2} - 1 = 13.5\).

Check by boxing in. The bounding box runs from \(x = 1\) to \(6\) and \(y = 1\) to \(5\), area \(5 \times 4 = 20\). Subtract the four outside pieces: bottom right triangle \(\tfrac{1}{2} \times 1 \times 2 = 1\); top right triangle \(\tfrac{1}{2} \times 2 \times 2 = 2\); top left triangle \(\tfrac{1}{2} \times 2 \times 1 = 1\); left piece, a \(1 \times 1\) square plus a triangle \(\tfrac{1}{2} \times 1 \times 3\), total \(2.5\). \(20 - 1 - 2 - 1 - 2.5 = 13.5\). Same answer, five times the work.

Your turn

  1. Draw a \(5 \times 3\) rectangle with corners on grid points. Count \(I\) and \(B\), apply Pick's formula, and check it gives 15.
  2. Draw a right triangle with corners at (0,0), (4,0) and (0,3). You know its area is 6. Count \(I\) and \(B\) and check. Be careful: does the slanted side pass through any grid points?
  3. Find the area of this shape twice: once by boxing in and subtracting, once by Pick's formula. One boundary dot is not a corner: it sits on the slanted side between the top corner and the inner corner. Find it.
  4. A lattice polygon has \(B = 12\) and area 20. How many interior points does it have?
  5. On the grid below, draw a lattice triangle with \(I = 0\) and \(B = 3\), so the only grid points touching it are its three corners. What does Pick say its area is? Draw a second, very different-looking one. Same area?
  6. Explain why Pick's formula can never give a lattice polygon an area like 2.3 or 7.1. What kind of numbers can it give?

Stretch: a shape with a hole

Draw a \(5 \times 5\) square and cut a \(1 \times 1\) square hole out of its middle, with the hole's corners on grid points. The true area is \(25 - 1 = 24\). Count \(I\) and \(B\) for the whole figure, treating the hole's edge as boundary too. What does Pick give? By how much is it wrong? Try a \(2 \times 2\) hole instead. Guess the corrected formula for a polygon with \(h\) holes.

Answers

  1. \(B = 16\) (count round the edge: 5 + 3 + 5 + 3, each corner once), \(I = 4 \times 2 = 8\). \(8 + 8 - 1 = 15\).
  2. The slanted side from (4,0) to (0,3) passes through no grid points, because 4 and 3 have no common factor. \(B = 8\): the five points along the bottom and the three more up the left side. \(I = 3\): (1,1), (2,1), (1,2). \(3 + 4 - 1 = 6\).
  3. \(B = 6\): the five corners plus (2,4), which sits on the edge from (3,3) to (1,5). \(I = 12\). Pick: \(12 + 3 - 1 = 14\). Boxing in: the box is \(6 \times 5 = 30\). Three pieces lie outside the shape. Lower right, the quadrilateral with corners (0,0), (6,0), (6,4), (4,1): split it along the line \(y = 1\) into a trapezoid below (area 4) and a triangle above (area 3), total 7. Upper, the quadrilateral (6,4), (6,5), (1,5), (3,3): area 6.5. Left, the triangle (0,0), (0,5), (1,5): area 2.5. \(30 - 7 - 6.5 - 2.5 = 14\). Pick's formula had it in one line.
  4. \(20 = I + 6 - 1\), so \(I = 15\).
  5. \(0 + \tfrac{3}{2} - 1 = \tfrac{1}{2}\). Every such triangle has area exactly one half, however long and thin. Try corners (0,0), (3,1), (5,2) for a surprising one.
  6. \(I\) and \(B\) are whole numbers, so \(I + B/2 - 1\) is either a whole number or a whole number plus a half. Lattice polygon areas are always multiples of \(\tfrac{1}{2}\).

Stretch. \(5 \times 5\) with a \(1 \times 1\) hole in the middle: \(B = 20 + 4 = 24\), \(I = 16 - 4 = 12\). Pick gives \(12 + 12 - 1 = 23\), one too small. With a \(2 \times 2\) hole: \(B = 20 + 8 = 28\) and \(I = 7\) (the 16 inner points lose 8 on the hole's edge and 1 inside the hole). Pick gives \(7 + 14 - 1 = 20\) against a true area of 21, again one too small. Each hole costs exactly one: \(A = I + B/2 - 1 + h\).

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