Difference between revisions of "Shoelace Theorem"

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==Proof 1==
 
==Proof 1==
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Lemma 1: The area of a triangle with coordinates <math>(x_1, y_1, 0)</math>, <math>(x_2, y_2, 0)</math>, and <math>(x_3, y_3, 0)</math> is <math>x_1,y_2-x_2y_1</math>.
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Let's translate <math>(x_1, y_1, 0)</math> to the origin so that the other two points are now <math>A'(x_2-x_1, y_2-y_1, 0)</math> and <math>B'(x_3-x_1, y_3-y_1, 0)</math>.
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We will proceed with induction. We start by proving it is true for a triangle:
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Let the triangle have coordinates <math>(x_1, y_1)</math> <math>(x_2, y_2)</math> and <math>(x_3, y_3)</math>. To simplify calculations let's translate <math>(x_1, y_1)</math> to the origin.
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==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 14:17, 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 $P$ has vertices $(a_1, b_1)$, $(a_2, b_2)$, ... , $(a_n, b_n)$, listed in clockwise order. Then the area of $P$ is

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

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

Lemma 1: The area of a triangle with coordinates $(x_1, y_1, 0)$, $(x_2, y_2, 0)$, and $(x_3, y_3, 0)$ is $x_1,y_2-x_2y_1$.

Let's translate $(x_1, y_1, 0)$ to the origin so that the other two points are now $A'(x_2-x_1, y_2-y_1, 0)$ and $B'(x_3-x_1, y_3-y_1, 0)$.

We will proceed with induction. We start by proving it is true for a triangle: Let the triangle have coordinates $(x_1, y_1)$ $(x_2, y_2)$ and $(x_3, y_3)$. To simplify calculations let's translate $(x_1, y_1)$ to the origin.

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

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