Difference between revisions of "Shoelace Theorem"
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\int_{\Omega}\alpha=\int_{\Omega}d\omega. | \int_{\Omega}\alpha=\int_{\Omega}d\omega. | ||
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\int_{\Omega}d\omega=\int_{\partial\Omega}\omega. | \int_{\Omega}d\omega=\int_{\partial\Omega}\omega. | ||
Revision as of 14:58, 13 July 2016
The Shoelace Theorem is a nifty formula for finding the area of a polygon given the coordinates of its vertices.
Theorem
Suppose the polygon 
has vertices 
, 
, ... , 
, listed in clockwise order. Then the area of is
![\[\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|\]](http://latex.artofproblemsolving.com/9/f/5/9f5392e8f69d97540c2fad3437685d2a8c6c19cb.png)
The Shoelace Theorem gets its name because if one lists the coordinates in a column,
and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes.
Proof
Let 
be the set of points belonging to the polygon.
We have that
![\[A=\int_{\Omega}\alpha,\]](http://latex.artofproblemsolving.com/6/b/3/6b3c51608b877346fbf5630bfb05f1122f15a277.png)
where 
.
The volume form 
is an exact form since 
, where
![\[\omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}\]](http://latex.artofproblemsolving.com/5/9/5/595a3ca55f21e85a6304722e556ff5daf938d9d5.png)
Using this substitution, we have
![\[\int_{\Omega}\alpha=\int_{\Omega}d\omega.\]](http://latex.artofproblemsolving.com/c/5/6/c5681412cc306a0473b53738bc9ab47c3eecdda2.png)
Next, we use the theorem of Stokes to obtain
![\[\int_{\Omega}d\omega=\int_{\partial\Omega}\omega.\]](http://latex.artofproblemsolving.com/c/6/f/c6fd7bed6cda7ed4b1fba9313d992340e24a0ec5.png)
We can write 
, where 
is the line
segment from 
to 
. With this notation,
we may write
![\[\int_{\partial\Omega}\omega=\sum_{i=1}^n\int_{A(i)}\omega.\]](http://latex.artofproblemsolving.com/c/3/b/c3bcbae43cc1750ad12da7a4b7c2b62f98d7e407.png)
If we substitute for 
, we obtain
![\[\sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.\]](http://latex.artofproblemsolving.com/8/d/8/8d8e73ad43b7ca1e7d62fb8170e9608a14c35741.png)
If we parameterize, we get
![\[\frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}.\]](http://latex.artofproblemsolving.com/c/5/0/c5019a5e21206e30287b09169ffa0f2604df4838.png)
Performing the integration, we get
![\[\frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- (y_{i}+y_{i+1})(x_{i+1}-x_i)].\]](http://latex.artofproblemsolving.com/f/6/d/f6d18d118e85da2e0f52830a3ebec78cb9dc2161.png)
More algebra yields the result
![\[\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).\]](http://latex.artofproblemsolving.com/c/0/f/c0f864227f4cb4abcc26e29f7d097faddecd54fd.png)
Problems
Introductory
In right triangle 
, we have 
, 
, and 
. Medians 
and 
are drawn to sides 
and 
, respectively. and intersect at point 
. Find the area of 
.
External Links
A good explanation and exploration into why the theorem works by James Tanton: [1]
