Difference between revisions of "Shoelace Theorem"

Latest revision as of 17:09, 28 September 2024

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

Theorem

Suppose the polygon $P$ has vertices $(x_1, y_1)$ , $(x_2, y_2)$ , ... , $(x_n, y_n)$ , listed in clockwise order. Then the area ( $A$ ) of $P$ is

$A = \dfrac{1}{2} \left|(x_1y_2 + x_2y_3 + \cdots + x_ny_1) - (y_1x_2 + y_2x_3 + \cdots + y_nx_1) \right|$

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, $\begin{align*} (x_1 &, y_1) \\ (x_2 &, y_2) \\ & \vdots \\ (x_n &, y_n) \\ (x_1 &, y_1) \\ \end{align*}$ 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 ? "$\vdots$" : "$x$",(0,-i*.7)); label(i==3 ? "$\vdots$" : "$y$",(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 ? "$\vdots$" : "$x$",shift(3*c)*(0,-i*.7)); label(i==3 ? "$\vdots$" : "$y$",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 $A = \dfrac{1}{2} \left|\sum_{i=1}^n{(x_{i+1}+x_i)(y_{i+1}-y_i)}\right|$ or in terms of determinants as $A = \dfrac{1}{2} \left|\sum_{i=1}^n{\det\begin{pmatrix}x_i&x_{i+1}\\y_i&y_{i+1}\end{pmatrix}}\right|$ which is useful in the $3D$ variant of the Shoelace theorem. Note here that $x_{n+1} = x_1$ and $y_{n+1} = y_1$ .

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

$\tilde{A}=\int \int \left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy=\oint(Ldx+Mdy)$

where $L=-y$ and $M=0$ so $\tilde{A}=A$ .

Proof 1

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

Proof of claim 1:

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

)

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 $A_1A_2A_3...A_n$ then it is also true for $A_1A_2A_3...A_nA_{n+1}$ .

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

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

Hence

$[A_1A_2A_3...A_nA_{n+1}]=[A_1A_2A_3...A_n]+[A_1A_nA_{n+1}]=\frac{1}{2}\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)+\frac{1}{2}(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)$ $=\frac{1}{2}((x_2y_1+x_3y_2+...+x_{n+1}y_n+x_1y_{n+1})-(x_1y_2+x_2y_3+...+x_ny_{n+1}+x_{n+1}y_1))=\boxed{\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i)}$

as claimed.

~ShreyJ

Proof 2

Let $\Omega$ be the set of points belonging to the polygon. We have that $A=\int_{\Omega}\alpha,$ where $\alpha=dx\wedge dy$ . The volume form $\alpha$ is an exact form since $d\omega=\alpha$ , where $\omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}$ Using this substitution, we have $\int_{\Omega}\alpha=\int_{\Omega}d\omega.$ Next, we use the Theorem of Stokes to obtain $\int_{\Omega}d\omega=\int_{\partial\Omega}\omega.$ We can write $\partial \Omega=\bigcup A(i)$ , where $A(i)$ is the line segment from $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$ . With this notation, we may write $\int_{\partial\Omega}\omega=\sum_{i=1}^n\int_{A(i)}\omega.$ If we substitute for $\omega$ , we obtain $\sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.$ If we parameterize, we get $\frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}.$ Performing the integration, we get $\frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- (y_{i}+y_{i+1})(x_{i+1}-x_i)].$ More algebra yields the result $\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).$

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 $ABC$ , we have $\angle ACB=90^{\circ}$ , $AC=2$ , and $BC=3$ . Medians $AD$ and $BE$ are drawn to sides $BC$ and $AC$ , respectively. $AD$ and $BE$ intersect at point $F$ . Find the area of $\triangle ABF$ .

Exploratory

Observe that $\frac12\left|\det\begin{pmatrix} x_1 & y_1\\ x_2 & y_2 \end{pmatrix}\right|$ is the area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(0,0)$ and $\frac16\left|\det\begin{pmatrix} x_1 & y_1 & z_1\\ x_2 & y_2 & z_2\\ x_3 & y_3 & z_3 \end{pmatrix}\right|$ is the volume of a tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2),(x_3, y_3, z_3),(0,0,0)$ . Does a similar formula hold for $n$ Dimensional triangles for any $n$ ? If so how can we use this to derive the $n$ 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

@@ Line 1: / Line 1: @@
-The '''Shoelace Theorem''' is a nifty formula for finding the [[area]] of a [[polygon]] given the [[Cartesian coordinate system | coordinates]] of its [[vertex|vertices]].
+The '''Shoelace Theorem''' is a nifty formula for finding the [[area]] of a simple [[polygon]] given the [[Cartesian coordinate system|coordinates]] of its [[vertex|vertices]].
 ==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 the area (<math>A</math>) of <math>P</math> is
+Suppose the polygon <math>P</math> has vertices <math>(x_1, y_1)</math>, <math>(x_2, y_2)</math>, ... , <math>(x_n, y_n)</math>, listed in clockwise order.  Then the area (<math>A</math>) of <math>P</math> is
-<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>
+<cmath>A = \dfrac{1}{2} \left|(x_1y_2 + x_2y_3 + \cdots + x_ny_1) - (y_1x_2 + y_2x_3 + \cdots + y_nx_1) \right|</cmath>
-You can also go counterclockwise order, as long as you find the absolute value of the answer.
+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*}
-(a_1 &, b_1) \\
+(x_1 &, y_1) \\
-(a_2 &, b_2) \\
+(x_2 &, y_2) \\
 & \vdots \\
-(a_n &, b_n) \\
+(x_n &, y_n) \\
-(a_1 &, b_1) \\
+(x_1 &, y_1) \\
 \end{align*}</cmath>
 and marks the pairs of coordinates to be multiplied,
@@ Line 22: / Line 22: @@
 for(int i=1; i < 6; ++i)
 {
-   label(i==3 ? "$\vdots$" : "$a$",(0,-i*.7));
+   label(i==3 ? "$\vdots$" : "$x$",(0,-i*.7));
-   label(i==3 ? "$\vdots$" : "$b$",(1.2,-i*.7));
+   label(i==3 ? "$\vdots$" : "$y$",(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));
@@ Line 32: / Line 32: @@
 pair c=(1.2,0);
 label("$-$",shift(c)*(1.2,-2.1));
-label("$A=$",shift(-c)*(0,-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 ? "$\vdots$" : "$x$",shift(3*c)*(0,-i*.7));
-   label(i==3 ? "$\vdots$" : "$b$",shift(3*c)*(1.2,-i*.7));
+   label(i==3 ? "$\vdots$" : "$y$",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));
@@ Line 46: / Line 48: @@
 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>
-And thus we can introduce determinants to get <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>
+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>
-This is more helpful in the <math>3D</math> formula variant of the Shoelace theorem.
+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
+<cmath>\tilde{A}=\int \int \left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy=\oint(Ldx+Mdy)</cmath>
+where <math>L=-y</math> and <math>M=0</math> so <math>\tilde{A}=A</math>.
 ==Proof 1==
@@ Line 56: / Line 65: @@
 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:===
@@ Line 73: / Line 83: @@
 <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.
+as claimed.
 ~ShreyJ
@@ Line 92: / Line 102: @@
 \int_{\Omega}\alpha=\int_{\Omega}d\omega.
 </cmath>
-Next, we use the theorem of Stokes to obtain
+Next, we use the Theorem of Stokes to obtain
 <cmath>
 \int_{\Omega}d\omega=\int_{\partial\Omega}\omega.
@@ Line 123: / Line 133: @@
 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.)
+The proof is in this book:
-https://cses.fi/book/book.pdf
+https://cses.fi/book/book.pdf#page=281
-(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.)
+(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]]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>.
+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]
@@ Line 142: / Line 166: @@
 [[Category:Geometry]]
 [[Category:Theorems]]
+[[Category:Mathematics]]
 AOPS

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