GeometryDifficulty 25-7513 tagged problems

Pick's Theorem

Pick's Theorem is the habit of finding lattice polygon area from interior and boundary lattice points. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

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The Key Result

$$A = I + \frac{B}{2} - 1$$

The area of a polygon whose vertices are lattice points equals the interior lattice points plus half the boundary lattice points minus one.

Why It Works

1
Here $I$ counts lattice points strictly inside the polygon and $B$ counts lattice points on its boundary (edges and vertices).
2
Pick's theorem holds for a single primitive lattice triangle (no interior points, only $3$ boundary points): $A = 0 + \tfrac32 - 1 = \tfrac12$, the smallest possible lattice triangle area.
3
Any lattice polygon triangulates into such primitive triangles; both sides of $A = I + \tfrac{B}{2} - 1$ are additive when you glue triangles along an edge (shared boundary points get recounted exactly to make the $-1$ terms combine correctly).
4
By induction over the triangulation, the formula holds for the whole polygon.

Worked Examples

Example 1

A polygon drawn on grid paper with all vertices at lattice points has $12$ interior lattice points and $8$ lattice points on its boundary. Find its area.

1
Apply Pick's theorem with $I = 12$ and $B = 8$.
2
$A = I + \dfrac{B}{2} - 1 = 12 + \dfrac{8}{2} - 1$.
3
Compute: $A = 12 + 4 - 1 = 15$.

Answer:

Area $= 15$

Warm-up

A lattice polygon has area $20$ and $10$ lattice points on its boundary. How many lattice points lie strictly inside it?

1
Pick's theorem: $A = I + \dfrac{B}{2} - 1$, here with $A = 20$ and $B = 10$.
2
Solve for $I$: $20 = I + \dfrac{10}{2} - 1 = I + 5 - 1 = I + 4$.
3
So $I = 20 - 4 = 16$.

Answer:

$16$ interior lattice points.

Contest level

A triangle has lattice-point vertices $(0,0)$, $(6,0)$, $(0,4)$. Use Pick's theorem to count the lattice points strictly inside it.

1
Its area is $A = \tfrac12\cdot 6\cdot 4 = 12$ (right triangle with legs $6$ and $4$).
2
Count boundary lattice points $B$: a segment from $(x_1,y_1)$ to $(x_2,y_2)$ carries $\gcd(|x_2-x_1|, |y_2-y_1|)$ steps, hence that many lattice points beyond its start. Edges give $\gcd(6,0)=6$, $\gcd(0,4)=4$, $\gcd(6,4)=2$, so $B = 6 + 4 + 2 = 12$.
3
Pick's theorem: $12 = I + \dfrac{12}{2} - 1 = I + 5$, so $I = 12 - 5 = 7$.

Answer:

$7$ interior lattice points.

Olympiad / Challenge

Prove that a lattice triangle whose ONLY lattice points are its three vertices (no other boundary points and no interior points) has area exactly $\tfrac12$, and explain why this is the minimum possible area for a lattice triangle.

1
Such a triangle has $I = 0$ interior lattice points and $B = 3$ boundary lattice points (just the vertices).
2
Pick's theorem gives $A = I + \dfrac{B}{2} - 1 = 0 + \dfrac{3}{2} - 1 = \dfrac{1}{2}$.
3
This is the minimum: any lattice triangle has $I \ge 0$ and $B \ge 3$, so $A = I + \tfrac{B}{2} - 1 \ge 0 + \tfrac32 - 1 = \tfrac12$, with equality exactly for these 'primitive' (empty) triangles.
4
Equivalently, area $\tfrac12$ lattice triangles are precisely the images of the unit triangle $(0,0),(1,0),(0,1)$ under a unimodular (determinant $\pm 1$) integer transformation.

Answer:

Area $= \tfrac12$, the minimum area of any lattice triangle (a primitive triangle).

Going Deeper

Generalization: Pick's theorem is the 2D specialization of the Ehrhart theory of lattice-point counting; in 3D there is NO analogous formula (Reeve tetrahedra of arbitrary volume all have $I = 0$, $B = 4$), so 'count interior points to get volume' fails in space. The boundary count itself generalizes via $\gcd$: a segment from $P$ to $Q$ contains $\gcd(|\Delta x|, |\Delta y|) + 1$ lattice points.
Where it appears: Pick's theorem is the fast route to lattice-polygon areas on AMC/AIME, and run backward it counts interior or boundary lattice points once the area is known (as in the warm-up). It also proves the minimum-area lattice triangle result and supports lattice-point parity arguments.
Pitfall: it applies ONLY to simple polygons with ALL vertices at lattice points — not to polygons with non-lattice vertices, and not to self-intersecting outlines. Also, $B$ counts every lattice point on the boundary (interior edge points AND vertices), not just the vertices; undercounting collinear edge points is the usual error.

Spot the Signal

Look for problems where the key step is finding lattice polygon area from interior and boundary lattice points.
You can describe the hard part as finding lattice polygon area from interior and boundary lattice points, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the diagram relation that calls for pick's theorem, then rewrite the givens around it.
Name the relation that makes Pick's Theorem legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Pick's Theorem just because the surface looks familiar; verify the required condition first.
Applying Pick's Theorem because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let pick's theorem reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is finding lattice polygon area from interior and boundary lattice points.