Difference between revisions of "Shoelace Theorem"

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'''Shoelace Theorem''' is a nifty formula for finding the [[area]] of a [[polygon]] given the coordinates of it's [[vertex|vertices]].
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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 its [[vertex|vertices]].
  
 
==Theorem==
 
==Theorem==
Let the coordinates, in "clockwise" order, be <math>(a_1, b_1)</math>, <math>(a_2, b_2)</math>, ... , <math>(a_n, b_n)</math>. The area of the polygon 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 of <math>P</math> is
  
<cmath>\dfrac{1}{2} |a_1b_2+a_2b_3+\cdots +a_nb_1-b_1a_2-b_2a_3-\cdots -b_na_1|.</cmath>
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<cmath>\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|</cmath>
 +
 
 +
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) \\
 +
(a_2 &, b_2) \\
 +
& \vdots \\
 +
(a_n &, b_n) \\
 +
(a_1 &, b_1) \\
 +
\end{align*}</cmath>
 +
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 <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>
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If we substitute for <math>\omega</math>, we obtain
 +
<cmath>
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\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)-
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(y_{i}+y_{i+1})(x_{i+1}-x_i)].
 +
</cmath>
 +
More algebra yields the result
 +
<cmath>
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\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.
 +
 
 +
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 <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>.
 +
 
 +
== 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]
  
Shoelace Theorem gets it's name by listing the coordinates like so:
 
  
<cmath>(a_1, b_1)</cmath>
 
<cmath>(a_2, b_2)</cmath>
 
<cmath>\vdots</cmath>
 
<cmath>(a_n, b_n)</cmath>
 
<cmath>(a_1, b_1)</cmath>
 
  
==Proof==
 
{{incomplete|proof}}
 
  
{{stub}}
 
  
 
[[Category:Geometry]]
 
[[Category:Geometry]]
 
[[Category:Theorems]]
 
[[Category:Theorems]]
 +
AOPS

Revision as of 17:18, 23 June 2020

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

Theorem

Suppose the polygon $P$ has vertices $(a_1, b_1)$, $(a_2, b_2)$, ... , $(a_n, b_n)$, listed in clockwise order. Then the area of $P$ is

\[\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|\]

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*} (a_1 &, b_1) \\ (a_2 &, b_2) \\ & \vdots \\ (a_n &, b_n) \\ (a_1 &, b_1) \\ \end{align*} 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 $A(x_1, y_1)$, $B(x_2, y_2)$, 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.

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 $ABC$, 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 $\triangle ABF$.

External Links

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

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