Difference between revisions of "Shoelace Theorem"

Revision as of 13:17, 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

@@ Line 17: / Line 17: @@
 ==Proof 1==
-Typing...
+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>.
+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>.
+We will proceed with induction. We start by proving it is true for a triangle:
+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.
 ==Proof 2==
 Let <math>\Omega</math> be the set of points belonging to the polygon.

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