# Difference between revisions of "Shoelace Theorem"

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− | The '''Shoelace Theorem''' is a nifty formula for finding the [[area]] of a [[polygon]] given the [[Cartesian coordinate system | coordinates]] of | + | The '''Shoelace Theorem''' is a nifty formula for finding the [[area]] of a [[polygon]] given the [[Cartesian coordinate system | coordinates]] of its [[vertex|vertices]]. |

==Theorem== | ==Theorem== | ||

− | Suppose the polygon <math>P</math> has vertices <math>(a_1, b_1)</math>, <math>(a_2, b_2)</math>, ... , <math>(a_n, b_n)</math>, listed in clockwise order. Then area of <math>P</math> is | + | Suppose the polygon <math>P</math> has vertices <math>(a_1, b_1)</math>, <math>(a_2, b_2)</math>, ... , <math>(a_n, b_n)</math>, listed in clockwise order. Then the area (<math>A</math>) of <math>P</math> is |

− | <cmath>\dfrac{1}{2} |a_1b_2+a_2b_3+\cdots +a_nb_1-b_1a_2 | + | <cmath>A = \dfrac{1}{2} \left|(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1) \right|</cmath> |

− | The Shoelace Theorem gets its name because if one lists | + | 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, | ||

<cmath>\begin{align*} | <cmath>\begin{align*} | ||

(a_1 &, b_1) \\ | (a_1 &, b_1) \\ | ||

Line 12: | Line 14: | ||

& \vdots \\ | & \vdots \\ | ||

(a_n &, b_n) \\ | (a_n &, b_n) \\ | ||

− | (a_1 &, b_1) | + | (a_1 &, b_1) \\ |

− | \end{align*} | + | \end{align*}</cmath> |

− | and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes. | + | and marks the pairs of coordinates to be multiplied, |

+ | <asy> | ||

+ | unitsize(1cm); | ||

+ | string[] subscripts={"$1$","$2$"," ","$n$","$1$"}; | ||

+ | for(int i=1; i < 6; ++i) | ||

+ | { | ||

+ | label(i==3 ? "$\vdots$" : "$a$",(0,-i*.7)); | ||

+ | label(i==3 ? "$\vdots$" : "$b$",(1.2,-i*.7)); | ||

+ | label(subscripts[i-1],(0,-i*.7),SE,fontsize(9pt)); | ||

+ | label(subscripts[i-1],(1.2,-i*.7),SE,fontsize(9pt)); | ||

+ | } | ||

+ | for(int i=1; i<5; ++i) | ||

+ | draw((0.3,-i*.7)--(1,-(i+1)*.7)); | ||

+ | |||

+ | pair c=(1.2,0); | ||

+ | label("$-$",shift(c)*(1.2,-2.1)); | ||

+ | label("$A=\frac12$",shift(-c)*(0,-2.1)); | ||

+ | draw(shift(-1/3*c)*((0,-.5)--(0,-3.9))); | ||

+ | draw(shift(13/3*c)*((0,-.5)--(0,-3.9))); | ||

+ | |||

+ | for(int i=1; i < 6; ++i) | ||

+ | { | ||

+ | label(i==3 ? "$\vdots$" : "$a$",shift(3*c)*(0,-i*.7)); | ||

+ | label(i==3 ? "$\vdots$" : "$b$",shift(3*c)*(1.2,-i*.7)); | ||

+ | label(subscripts[i-1],shift(3*c)*(0,-i*.7),SE,fontsize(9pt)); | ||

+ | label(subscripts[i-1],shift(3*c)*(1.2,-i*.7),SE,fontsize(9pt)); | ||

+ | } | ||

+ | for(int i=1; i<5; ++i) | ||

+ | draw(shift(3*c)*(0.3,-(i+1)*.7)--shift(3*c)*(1,-i*.7)); | ||

+ | </asy> | ||

+ | the resulting image looks like laced-up shoes. | ||

+ | |||

+ | ==Other Forms== | ||

+ | This can also be written in form of a summation <cmath>A = \dfrac{1}{2} \left|\sum_{i=1}^n{(x_{i+1}+x_i)(y_{i+1}-y_i)}\right|</cmath> | ||

+ | or in terms of determinants as <cmath>A = \dfrac{1}{2} \left|\sum_{i=1}^n{\det\begin{pmatrix}x_i&x_{i+1}\\y_i&y_{i+1}\end{pmatrix}}\right|</cmath> | ||

+ | which is useful in the <math>3D</math> variant of the Shoelace theorem. Note here that <math>x_{n+1} = x_1</math> and <math>y_{n+1} = y_1</math>. | ||

+ | |||

+ | The formula may also be considered a special case of Green's Theorem | ||

+ | <center> | ||

+ | <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> | ||

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

+ | |||

+ | ==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== | ||

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

+ | We have that | ||

+ | <cmath> | ||

+ | A=\int_{\Omega}\alpha, | ||

+ | </cmath> | ||

+ | where <math>\alpha=dx\wedge dy</math>. | ||

+ | The volume form <math>\alpha</math> is an exact form since <math>d\omega=\alpha</math>, where | ||

+ | <cmath> | ||

+ | \omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega} | ||

+ | </cmath> | ||

+ | Using this substitution, we have | ||

+ | <cmath> | ||

+ | \int_{\Omega}\alpha=\int_{\Omega}d\omega. | ||

+ | </cmath> | ||

+ | Next, we use the Theorem of Stokes to obtain | ||

+ | <cmath> | ||

+ | \int_{\Omega}d\omega=\int_{\partial\Omega}\omega. | ||

+ | </cmath> | ||

+ | We can write <math>\partial \Omega=\bigcup A(i)</math>, where <math>A(i)</math> is the line | ||

+ | segment from <math>(x_i,y_i)</math> to <math>(x_{i+1},y_{i+1})</math>. With this notation, | ||

+ | we may write | ||

+ | <cmath> | ||

+ | \int_{\partial\Omega}\omega=\sum_{i=1}^n\int_{A(i)}\omega. | ||

+ | </cmath> | ||

+ | If we substitute for <math>\omega</math>, we obtain | ||

+ | <cmath> | ||

+ | \sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}. | ||

+ | </cmath> | ||

+ | If we parameterize, we get | ||

+ | <cmath> | ||

+ | \frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}. | ||

+ | </cmath> | ||

+ | Performing the integration, we get | ||

+ | <cmath> | ||

+ | \frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- | ||

+ | (y_{i}+y_{i+1})(x_{i+1}-x_i)]. | ||

+ | </cmath> | ||

+ | More algebra yields the result | ||

+ | <cmath> | ||

+ | \frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i). | ||

+ | </cmath> | ||

+ | |||

+ | ==Proof 3== | ||

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

+ | |||

+ | The proof is in this book: | ||

+ | https://cses.fi/book/book.pdf#page=281 | ||

+ | |||

+ | (The only thing that needs to be slightly modified 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 <math>ABC</math>, we have <math>\angle ACB=90^{\circ}</math>, <math>AC=2</math>, and <math>BC=3</math>. [[Median of a triangle|Median]]s <math>AD</math> and <math>BE</math> are drawn to sides <math>BC</math> and <math>AC</math>, respectively. <math>AD</math> and <math>BE</math> intersect at point <math>F</math>. Find the area of <math>\triangle ABF</math>. | ||

+ | |||

+ | === Exploratory === | ||

+ | Observe that <cmath>\frac12\left|\det\begin{pmatrix} | ||

+ | x_1 & y_1\\ | ||

+ | x_2 & y_2 | ||

+ | \end{pmatrix}\right|</cmath> is the area of a triangle with vertices <math>(x_1,y_1),(x_2,y_2),(0,0)</math> and <cmath>\frac16\left|\det\begin{pmatrix} | ||

+ | x_1 & y_1 & z_1\\ | ||

+ | x_2 & y_2 & z_2\\ | ||

+ | x_3 & y_3 & z_3 | ||

+ | \end{pmatrix}\right|</cmath> is the volume of a tetrahedron with vertices <math>(x_1, y_1, z_1), (x_2, y_2, z_2),(x_3, y_3, z_3),(0,0,0)</math>. Does a similar formula hold for <math>n</math>Dimensional triangles for any <math>n</math>? If so how can we use this to derive the <math>n</math>D Shoelace Formula? | ||

+ | |||

+ | == External Links== | ||

+ | A good explanation and exploration into why the theorem works by James Tanton: | ||

+ | [http://www.jamestanton.com/wp-content/uploads/2012/03/Cool-Math-Essay_June-2014_SHOELACE-FORMULA.pdf] | ||

+ | |||

+ | Nice geometric approach and discussion for proving the 3D Shoelace Theorem by Nicholas Patrick and Nadya Pramita: [http://media.icys2018.com/2018/04/IndonesiaPatrickNicholas105190.pdf] | ||

+ | |||

+ | Nice integral approach for proving the 3D Shoelace Theorem (ignoring sign of volume) by @george2079: [https://mathematica.stackexchange.com/a/26015] | ||

+ | |||

+ | |||

− | |||

− | |||

− | |||

[[Category:Geometry]] | [[Category:Geometry]] | ||

[[Category:Theorems]] | [[Category:Theorems]] | ||

+ | AOPS |

## Latest revision as of 04:28, 30 July 2022

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. Note here that and .

The formula may also be considered a 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.

The proof is in this book: https://cses.fi/book/book.pdf#page=281

(The only thing that needs to be slightly modified 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 .

### Exploratory

Observe that is the area of a triangle with vertices and is the volume of a tetrahedron with vertices . Does a similar formula hold for Dimensional triangles for any ? If so how can we use this to derive the D Shoelace Formula?

## External Links

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

Nice geometric approach and discussion for proving the 3D Shoelace Theorem by Nicholas Patrick and Nadya Pramita: [2]

Nice integral approach for proving the 3D Shoelace Theorem (ignoring sign of volume) by @george2079: [3] AOPS