Difference between revisions of "Shoelace Theorem"

(Theorem)
Line 17: Line 17:
  
 
==Proof==
 
==Proof==
{{incomplete|proof}}
+
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 Green 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>
  
 
== Problems ==
 
== Problems ==

Revision as of 18:39, 28 February 2010

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 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

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 Green 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 $\triangle ABF$.


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