Shoelace Theorem

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 $\triangle ABF$.