Pick’s Theorem — Can You Find an Area Just by Counting Dots?

Forget base times height for a moment. Could you calculate the exact area of a complicated shape simply by counting dots?

Draw an irregular polygon on squared paper. Make sure every corner sits exactly on a grid intersection.

Now imagine asking a student to calculate its area.

They might divide it into rectangles and triangles. They might draw a larger rectangle around it and subtract the unwanted pieces. A more advanced student might start working with coordinates.

All perfectly sensible approaches.

But there is another method that initially seems rather unlikely.

Count the grid points inside the shape. Count the grid points on its boundary. Put those two numbers into a short formula.

Out comes the exact area.

This is Pick’s theorem, a lovely example of mathematics beyond the usual GCSE calculation exercises: accessible enough to investigate with a pencil, but rich enough to lead towards conjecture, proof and computer programming.

Two Numbers That Reveal the Area

Pick’s theorem states:

A = I + B/2 - 1

where:

• A is the area in square grid units.

• I is the number of grid points strictly inside the polygon.

• B is the number of grid points on its boundary.

Suppose we count 12 interior points and 10 boundary points.

Then:

A = 12 + 10/2 - 1
A = 12 + 5 - 1
A = 16 square units

No measuring angles. No calculating diagonal lengths.

For a suitable polygon, those two counts are enough.

There are conditions, however. We need a closed polygon made from straight sides, with every vertex on a point of a regular square lattice. Its sides must not cross, and the version of the formula we are using assumes there are no holes.

An inward dent is perfectly acceptable. A curved edge is not.

First, Avoid the Most Common Counting Mistake

Boundary points are not the same as corners.

Consider a horizontal line from (0,0) to (4,0).

It passes through:

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

That is five boundary points along this side, although only its endpoints are corners.

Diagonal sides can also pass through grid points.

The line from (0,0) to (4,4) passes through:

(1,1), (2,2), (3,3)

Those intermediate points count towards B too.

When counting the whole boundary, count each point once. A corner belongs to two sides, but it is still only one point.

A useful practical method is to mark:

• interior points in blue;

• boundary points in orange;

• exterior points in neither colour.

The colours make mistakes much easier to spot.

Try It: An Irregular Shape with an Exact Answer

On squared paper, plot these coordinates:

(0,0), (5,0), (5,2), (3,2), (3,4), (0,4)

Join them in that order, then join the last point back to the first.

You should have an L-shaped polygon.

Calculate its area conventionally

Imagine a rectangle 5 units wide and 4 units high.

Its area is:

5 x 4 = 20 square units

The missing top-right rectangle measures 2 units by 2 units.

Its area is:

2 x 2 = 4 square units

Therefore:

A = 20 - 4
A = 16 square units

Now count the dots

The interior points occur in three horizontal rows:

• At y = 1: four interior points.

• At y = 2: two interior points.

• At y = 3: two interior points.

Therefore:

I = 8

Count around the boundary carefully:

B = 18

Now apply Pick’s theorem:

A = 8 + 18/2 - 1
A = 8 + 9 - 1
A = 16 square units

The two methods agree exactly.

Notice that the points at (3,2) and (4,2) are on the boundary. Being part-way up the drawing does not automatically make a point an interior point.

Does It Still Work with Sloping Sides?

Yes — provided the corners remain on grid points.

Draw the parallelogram with vertices:

(0,0), (4,0), (5,4), (1,4)

Its base is 4 units and its perpendicular height is 4 units.

A = 4 x 4
A = 16 square units

Now count its lattice points.

There are four interior points on each of the rows y = 1, y = 2 and y = 3.

I = 12

The bottom and top edges each contain five grid points. The sloping sides contain no additional grid points between their endpoints.

B = 10

Therefore:

A = 12 + 10/2 - 1
A = 16 square units

This also reveals something important: shapes can have the same area but different values of I and B.

Our L-shape had I = 8 and B = 18.

Our parallelogram has I = 12 and B = 10.

Both have area 16.

A Better Investigation: Hide the Formula First

For teaching, I would resist revealing the rule immediately.

The discovery is part of the attraction.

Give students several polygons and ask them to record three things:

Interior points, boundary points and area.

These examples make a useful starting set. Join each list of vertices in order and close the shape.

Shape	Vertices	I	B	Area
Rectangle	(0,0), (3,0), (3,2), (0,2)	2	10	6
Right-angled triangle	(0,0), (4,0), (0,3)	3	8	6
L-shape	(0,0), (5,0), (5,2), (3,2), (3,4), (0,4)	8	18	16
Parallelogram	(0,0), (4,0), (5,4), (1,4)	12	10	16

For the student version, leave the final three columns blank.

Ask:

• Does counting only the interior points work?

• Does counting only the boundary points work?

• What happens if we use half the boundary count?

• Can you find one rule that fits every row?

If students become stuck, suggest calculating:

Area - interior points

Then compare that result with B.

This gives them a useful direction while leaving the final discovery to them.

Why I Like This Kind of Mathematics

One of the things I value in teaching is the moment when a student realises that mathematics contains ideas they would never have thought to look for.

Counting dots seems elementary. Calculating an irregular area seems more demanding.

Pick’s theorem connects the two.

It also gives students several ways into the same problem. One might enjoy drawing. Another might spot patterns in the table. Another might immediately ask why the rule works.

For a confident GCSE student, the calculation is only the beginning. The more interesting questions are:

Why does each interior point contribute one unit? Why does the boundary count come with a half? And where does the minus one come from?

Those questions turn a surprising trick into mathematics.

Why Does It Work? Start with a Rectangle

Take a rectangle whose sides follow the grid, with whole-number width w and height h.

Its area is:

A = w x h

Inside the rectangle there are:

I = (w - 1)(h - 1)

There are w - 1 interior columns and h - 1 interior rows.

The boundary count is:

B = 2w + 2h

Substitute into Pick’s formula:

I + B/2 - 1
= (w - 1)(h - 1) + w + h - 1
= wh - w - h + 1 + w + h - 1
= wh

So the formula gives exactly the rectangle’s area.

That proves the rectangle case. It does not yet prove the theorem for every polygon.

The wider proof needs another idea: showing that the formula behaves correctly when suitable polygons are joined or divided. Points along a shared edge can become interior points, and careful counting keeps the area formula consistent.

This allows the argument to extend through triangles to general simple lattice polygons.

Finding a Pattern Is Not the Same as Proving It

Suppose the rule works for five shapes.

That is encouraging.

Suppose it works for fifty.

That is stronger evidence.

But neither establishes that it works for every allowed polygon.

A conjecture is a proposed mathematical statement. A proof explains why the statement must hold throughout its stated scope.

This distinction is worth discussing explicitly.

A student who asks, “Could there be a shape where it fails?” is asking exactly the right kind of question.

Before answering, check the conditions. A polygon with a corner between grid points is outside the theorem’s scope. So is a shape with crossing sides.

A theorem is a precise promise, and its conditions tell us when that promise applies.

The Reverse Challenge: Same Counts, Different Shapes

Now reverse the problem:

Can you design three genuinely different polygons with I = 12 and B = 10?

Each must have area:

A = 12 + 10/2 - 1
A = 16 square units

Moving the same shape across the paper is too easy. Try changing its geometry.

Here are three solutions to investigate:

Shape	Vertices in order
Triangle	(0,0), (8,0), (1,4)
Parallelogram	(0,0), (4,0), (5,4), (1,4)
Irregular quadrilateral	(0,0), (5,0), (4,4), (0,3)

All three have:

I = 12, B = 10 and A = 16

Ask students to verify the counts and then check the areas independently.

The triangle is particularly instructive: most of its boundary points lie along its long horizontal base.

The challenge shows that these two counts determine area, but do not determine the shape.

Three Questions to Take It Further

1. Can the area be 7.3 square units?

Not for a polygon satisfying these conditions.

I and B are whole numbers, so:

A = I + B/2 - 1

must produce either a whole number or a number ending in .5.

An exact area of 7.3 square grid units would tell us that at least one condition or calculation needs checking.

2. Can an area of 0.5 have no interior points?

Yes.

Draw the triangle:

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

It has:

I = 0
B = 3

Therefore:

A = 0 + 3/2 - 1
A = 0.5 square units

An empty interior does not mean zero area.

3. What if the dots are 2 cm apart?

The formula gives area in square grid units.

If adjacent dots are 2 cm apart horizontally and vertically, one grid square has area:

2 x 2 = 4 square centimetres

An answer of 16 square grid units therefore represents:

16 x 4 = 64 square centimetres

Counting remains the same. The physical area represented by each grid square changes.

A Link to Computing

For students interested in programming, Pick’s theorem offers a useful extension.

A program can take a list of integer-coordinate vertices, calculate the polygon’s area using coordinates, and count its boundary points.

For a side with horizontal change dx and vertical change dy, the number of equal lattice steps along it is:

gcd(abs(dx), abs(dy))

Here, gcd means greatest common divisor.

Adding this value over every side gives B, with each boundary point counted once around the polygon.

Rearranging Pick’s theorem then gives:

I = A - B/2 + 1

The program can work out how many lattice points lie inside a polygon without checking each possible interior point individually.

A pencil-and-paper investigation has become an algorithm.

The Real Discovery Is Bigger Than the Formula

Pick’s theorem gives us an exact area from two simple counts.

But its educational value goes further.

Students draw, calculate, classify, look for patterns, test a conjecture and begin asking what would constitute a proof.

They also discover that an apparently complicated problem may become much simpler when we choose a different way to describe it.

This week’s challenge: draw an awkward-looking polygon on dot paper. Find its area conventionally, then count its interior and boundary points.

When the two answers agree, ask the question that opens the next stage of the investigation:

“Why must that happen?”