Difference between revisions of "Pick's Theorem"
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== Proof == | == Proof == | ||
− | If a triangle on the lattice points with exactly <math>3</math> points in its interior or on its edges, it has an area of <math>\frac{1}{2}</math>. Such triangle must contain two lattice points distance <math>1</math> from each other and one lattice point on a line parallel to the opposite edge distance <math>1</math> apart. The minimum distance between two distinct lattice points is <math>1</math>. If no two lattice points have distance <math>1</math>, then by <math>\frac{1}{2}bh</math> the area is more than 1 and similarly for the height. | + | If a triangle on the lattice [https://artofproblemsolving.com/wiki/index.php/TOTO_SLOT_:_SITUS_TOTO_SLOT_MAXWIN_TERBAIK_DAN_TERPERCAYA TOTO SLOT] points with exactly <math>3</math> points in its interior or on its edges, it has an area of <math>\frac{1}{2}</math>. Such triangle must contain two lattice points distance <math>1</math> from each other and one lattice point on a line parallel to the opposite edge distance <math>1</math> apart. The minimum distance between two distinct lattice points is <math>1</math>. If no two lattice points have distance <math>1</math>, then by <math>\frac{1}{2}bh</math> the area is more than 1 and similarly for the height. |
Removing a triangle either removes <math>1</math> boundary point or turns <math>1</math> interior point into a boundary point, accounting for the <math>I+\frac{1}{2}B</math> part. The <math>-1</math> part is accounted for by looking at the area of the unit triangle with <math>3</math> boundary points, <math>0</math> interior points, and <math>\frac{1}{2}</math> area. | Removing a triangle either removes <math>1</math> boundary point or turns <math>1</math> interior point into a boundary point, accounting for the <math>I+\frac{1}{2}B</math> part. The <math>-1</math> part is accounted for by looking at the area of the unit triangle with <math>3</math> boundary points, <math>0</math> interior points, and <math>\frac{1}{2}</math> area. | ||
Revision as of 17:03, 19 February 2024
Pick's Theorem expresses the area of a polygon, all of whose vertices are lattice points in a coordinate plane, in terms of the number of lattice points inside the polygon and the number of lattice points on the sides of the polygon. The formula is:
where is the number of lattice points in the interior and
being the number of lattice points on the boundary.
It is similar to the Shoelace Theorem, and although it is less powerful, it is a good tool to have in solving problems.
Proof
If a triangle on the lattice TOTO SLOT points with exactly points in its interior or on its edges, it has an area of
. Such triangle must contain two lattice points distance
from each other and one lattice point on a line parallel to the opposite edge distance
apart. The minimum distance between two distinct lattice points is
. If no two lattice points have distance
, then by
the area is more than 1 and similarly for the height.
Removing a triangle either removes
boundary point or turns
interior point into a boundary point, accounting for the
part. The
part is accounted for by looking at the area of the unit triangle with
boundary points,
interior points, and
area.
Solution by a1b2