# Difference between revisions of "Shoelace Theorem"

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

− | + | Claim 1: The area of a triangle with coordinates <math>A(x_1, y_1)</math>, <math>B(x_2, y_2)</math>, and <math>C(x_3, y_3)</math> is <math>\frac{x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2}{2}</math>. | |

+ | |||

+ | Proof of claim 1: | ||

+ | |||

+ | Writing the coordinates in 3D and translating <math>\triangle ABC</math> so that <math>A=(0, 0, 0)</math> we get the new coordinates <math>A'(0, 0, 0)</math>, <math>B(x_2-x_1, y_2-y_1, 0)</math>, and <math>C(x_3-x_1, y_3-y_1, 0)</math>. Now if we let <math>\vec{b}=(x_2-x_1 \quad y_2-y_1 \quad 0)</math> and <math>\vec{c}=(x_3-x_1 \quad y_3-y_1 \quad 0)</math> then by definition of the cross product <math>[ABC]=\frac{||\vec{b} \times \vec{c}||}{2}=\frac{1}{2}||(0 \quad 0 \quad x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)||=\frac{x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2}{2}</math>. | ||

+ | |||

+ | 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 <math>A_1A_2A_3...A_n</math> then it is also true for <math>A_1A_2A_3...A_nA_{n+1}</math>. | ||

+ | |||

+ | We cut <math>A_1A_2A_3...A_nA_{n+1}</math> into two polygons, <math>A_1A_2A_3...A_n</math> and <math>A_1A_nA_{n+1}</math>. Let the coordinates of point <math>A_i</math> be <math>(x_i, y_i)</math>. Then, applying the shoelace theorem on <math>A_1A_2A_3...A_n</math> and <math>A_1A_nA_{n+1}</math> we get | ||

+ | |||

+ | <cmath>[A_1A_2A_3...A_n]=\frac{1}{2}\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)</cmath> | ||

+ | <cmath>[A_1A_nA_{n+1}]=\frac{1}{2}(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)</cmath> | ||

+ | |||

+ | Hence | ||

+ | |||

+ | <cmath>[A_1A_2A_3...A_nA_{n+1}]=[A_1A_2A_3...A_n]+[A_1A_nA_{n+1}]=\frac{1}{2}\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)+\frac{1}{2}(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)</cmath> | ||

+ | <cmath>=\frac{1}{2}((x_2y_1+x_3y_2+...+x_{n+1}y_n+x_1y_{n+1})-(x_1y_2+x_2y_3+...+x_ny_{n+1}+x_{n+1}y_1))=\boxed{\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i)}</cmath> | ||

+ | |||

+ | As claimed | ||

+ | ~ShreyJ | ||

+ | |||

==Proof 2== | ==Proof 2== | ||

Let <math>\Omega</math> be the set of points belonging to the polygon. | Let <math>\Omega</math> be the set of points belonging to the polygon. |

## Revision as of 16:14, 6 July 2018

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

## Theorem

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

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.

## 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

## 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