# Pick's theorem

**Pick's theorem**provides a simple formula for calculating the area

*A*of this polygon in terms of the number

*i*of

*interior points*of the polygon and the number

*b*of

*boundary points*on the polygon:

*A*=*i*+ ½*b*− 1.

*i*= 9 and

*b*= 14, so the area is

*A*= 9 + ½(14) − 1 = 9 + 7 − 1 = 15 (square units).

This formula so simple that it has been correctly used by first-grade children, drawing figures on square tiles on floor or wall, or stretching strings from pegs in pegboard. They learn how to add, along with subtraction as "take away". They learn to "halve" by a one-to-one correspondence between counters.

Note that the theorem as stated above is only valid for *simple* polygons, i.e. ones that consist of a single piece and do not contain "holes". For more general polygons, the "− 1" of the formula has to be replaced with "− χ(*P*)", where χ(*P*) is the Euler characteristic of *P*.

The result was first described by George Pick in 1899. It can be generalized to three dimensions and higher by Ehrhart polynomials. The formula also generalizes to surfaces of polyhedra.

### Proof

Since we are assuming the theorem for *P* and for *T* separately,

*A*=_{PT}*A*+_{P}*A*_{T}- =
*i*+ ½_{P}*b*− 1 +_{P}*i*+ ½_{T}*b*− 1'''_{T} - = (
*i*+_{P}*i*) + ½(_{T}*b*+_{P}*b*) − 2_{T} - =
*i*− (_{PT}*c*− 2) + ½(*b*+ 2(_{PT}*c*− 2) + 2) − 2 - =
*i*+ ½_{PT}*b*− 1._{PT}

*n*triangles, the theorem is also true for polygons constructed from

*n*+1 triangles. To finish the proof by mathematical induction, it remains to show that the theorem is true for triangles. The verification for this case can be done in these short steps:

- directly check that the formula is correct for any rectangle with sides parallel to the axes;
- verify from that case that it works for right-angled triangles obtained by cutting such rectangles along a diagonal;
- now any triangle can be turned into a rectangle by attaching (at most three) such right triangles; since the formula is correct for the right triangles and for the rectangle, it also follows for the original triangle.

*PT*and for the triangle

*T*, then it's also true for

*P*; this can be seen by a calculation very much similar to the one shown above.