# Difference between revisions of "Shoelace Theorem"

(Tag: Undo) |
(A tilde) |
||

Line 53: | Line 53: | ||

or as the special case of Green's Theorem | or as the special case of Green's Theorem | ||

<center> | <center> | ||

− | <math>A=\iint\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy=</math><font size="6">∳</font><math>(Ldx+Mdy)</math> | + | <math>\tilde{A}=\iint\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy=</math><font size="6">∳</font><math>(Ldx+Mdy)</math> |

</center> | </center> | ||

− | where <math>L=-y</math> and <math>M=0</math>. | + | where <math>L=-y</math> and <math>M=0</math> so <math>\tilde{A}=A</math>. |

==Proof 1== | ==Proof 1== |

## Revision as of 06:40, 24 December 2020

The **Shoelace Theorem** is a nifty formula for finding the area of a polygon given the coordinates of its vertices.

## Contents

## Theorem

Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area () of is

You can also go counterclockwise order, as long as you find the absolute value of the answer.

The Shoelace Theorem gets its name because if one lists the coordinates in a column, and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes.

## Other Forms

This can also be written in form of a summation or in terms of determinants as which is useful in the variant of the Shoelace theorem,

or as the special case of Green's Theorem

∳

where and so .

## Proof 1

Claim 1: The area of a triangle with coordinates , , and is .

### Proof of claim 1:

Writing the coordinates in 3D and translating so that we get the new coordinates , , and . Now if we let and then by definition of the cross product .

### Proof:

We will proceed with induction.

By claim 1, the shoelace theorem holds for any triangle. We will show that if it is true for some polygon then it is also true for .

We cut into two polygons, and . Let the coordinates of point be . Then, applying the shoelace theorem on and we get

Hence

As claimed.

~ShreyJ

## Proof 2

Let be the set of points belonging to the polygon. We have that where . The volume form is an exact form since , where Using this substitution, we have Next, we use the theorem of Stokes to obtain We can write , where is the line segment from to . With this notation, we may write If we substitute for , we obtain If we parameterize, we get Performing the integration, we get More algebra yields the result

## Proof 3

This is a very nice approach that directly helps in understanding the sum as terms which are areas of trapezoids.

See page 281 in this book (in the Polygon Area section.) https://cses.fi/book/book.pdf

(The only thing that needs to be modified in this proof is that one must shift the entire polygon up by k, until all the y coordinates are positive, but this term gets canceled in the resulting sum.)

## Problems

### Introductory

In right triangle , we have , , and . Medians and are drawn to sides and , respectively. and intersect at point . Find the area of .

## External Links

A good explanation and exploration into why the theorem works by James Tanton: [1] AOPS