# Pick's Theorem

**Pick's theorem** gives a way to find the area of polygons in a plane whose endpoints have integer vertices.

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## Lattice Polygons

**Lattice points** are points whose coordinates are both integers, such as \((1,2), (-4, 11)\), and \((0,5)\). The set of all lattice points forms a grid. A *lattice polygon* is a shape made of straight lines whose vertices are all lattice points and Pick's theorem gives a formula for the area of a lattice polygon.

First, observe that for any lattice polygon \(P\), the polygon contains some lattice points on its boundary edges \((\)including the lattice points that are the vertices of \(P)\) and may contain some lattice points in its interior (not counting the points on the boundary). Let

\[\begin{align} B(P) &= \mbox{ number of points on the boundary of the polygon}\\ I(P) &= \mbox{ number of points in the interior of the polygon}. \end{align}\]

Pick's TheoremLet \(P\) be a lattice polygon, let \(B(P)\) be the points on the boundary of the polygon, and let \(I(P)\) be the number of points in the interior of the polygon. Then

\[\mbox{Area of the polygon } P = I(P) + \frac{1}{2} B(P) -1.\]

Notice that Pick's theorem applies to *any* polygon, not only convex polygons.

For a rectangle \(R\) of length \(l\) and height \(h,\) there are \(B(R) = 2l + 2h\) points along the boundary of the rectangle, and \(I(R) = (l-1)(h-1) = lh - l - h + 1\) points in the interior of the rectangle. Applying Pick's theorem gives

\[\begin{align} \mbox{Area}(R) &= I(R) + \frac{1}{2} B(R) - 1\\ &= lh - l - h + 1 + \frac{1}{2} (2l + 2h) - 1\\ &= lh - l - h + 1 + l + h - 1\\ &= lh. \end{align}\]

Without Pick's theorem, we might calculate the area of a lattice polygon by decomposing the lattice polygon into triangles, computing the area of each triangle using the sine rule, and then summing the resulting triangle areas to obtain the area of the lattice polygon. Pick's theorem gives a way to find the area of a lattice polygon without performing all of these calculations. Pick's theorem also implies the following interesting corollaries:

The area of a lattice polygon is always an integer or half an integer.

Proof:By Pick's theorem, the area of a lattice polygon \(P\) is given by \( \mbox{Area}(P) = I(P) + \frac{1}{2} B(P) - 1.\) In a lattice polygon, the number of points in the interior of \(P\) and the number of points on the boundary of \(P\) are both integers. Then \( \mbox{Area}(P) = I(P) + \frac{1}{2} B(P) - 1\) is an integer if \(B(P)\) is even and is half an integer if \(B(P)\) is odd. \(_\square\)

It is impossible to draw an equilateral triangle as a lattice polygon.

Proof: Suppose we could draw an equilateral triangle as a lattice polygon with lattice vertices \(A\), \(B\) and \(C\). The the area of \(\triangle ABC\) must be an integer or half an integer by the previous theorem. Now, given an equilateral triangle with side length \(s,\) the area of the triangle is \(\frac{s^2 \sqrt{3}}{4}\). However, by considering the midpoint \(D\) of side \(BC\), we have that the length of \(AD\) and \(BD\) are both integers (since \(ABC\) is a lattice polygon), which implies that \(s^2 = \lvert AD \rvert^2 + \lvert BD \rvert^2\) is an integer. Since \(s^2\) and \(\frac{s^2 \sqrt{3}}{4}\) are both integers, this implies \(\sqrt{3}\) is an integer, a contradiction. Therefore, it is impossible to draw an equilateral triangle as a lattice polygon. \(_\square\)

## Example Problems

An integer lattice point is a point with coordinates \( (n, m) \), where \(n\) and \(m\) are integers. As \(N\) ranges from 1 to 905, what is the maximum number of integer lattice points in the interior of a triangle with vertices \( (0,0),\) \((N, 907-N),\) and \((N+1, 907-N-1) \)?

Note: The point \( (0,0) \) is not in the interior of any of the triangles described above.