Pick's theorem

What just happened?

You changed the figure — from a rectangle to a non-convex polygon. Each time the number of lattice points inside ($i$) and on the boundary ($b$) was different, yet $i + \tfrac{b}{2} - 1$ always gave exactly the area of the figure. You found the area by counting dots.

Pick’s theorem:

Formulated by the Austrian mathematician Georg Pick in 1899, it gives a surprisingly simple formula for the area of a polygon whose vertices all lie on lattice points (with integer coordinates): $i$ is the number of lattice points inside the polygon, and $b$ is the number of lattice points on its boundary. This means the area can be found simply by COUNTING DOTS — no integration, no complicated formulas!

Check for yourself:

Rectangle: $8 + \tfrac{16}{2} - 1 = 15$. Triangle: $6 + \tfrac{10}{2} - 1 = 10$. L-polygon: $5 + \tfrac{16}{2} - 1 = 12$. Each time the result matches the area computed with the shoelace formula.

How to count b (boundary points)?

On each boundary segment the number of lattice points (not counting one endpoint) is the greatest common divisor of the coordinate differences: for a segment from $(x_1,y_1)$ to $(x_2,y_2)$ it is $\gcd(|x_2-x_1|,\,|y_2-y_1|)$. Summing over all sides gives $b$.

For any simple polygon:

The formula works for any (including non-convex) simple polygon with lattice vertices. Pick’s theorem is a favourite example of “discrete mathematics meeting geometry” and is used, among other things, in computational geometry.

Why this matters for the exam:

When a figure has vertices at lattice points, you can compute the area as $P = i + \tfrac{b}{2} - 1$ without splitting it into triangles and trapezia — just carefully count the dots inside and on the boundary. It is also an elegant example of the link between geometry and discrete mathematics.

Remember: for a polygon with lattice vertices $P = i + \tfrac{b}{2} - 1$ — the interior points plus half the boundary points minus one give the area.