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The Shoelace Theorem is a nifty formula for finding the area of a polygon given the coordinates of its vertices.

Theorem

Let a simple polygon on the plane be defined by its vertices: $\left(x_1,y_1\right),\left(x_2,y_2\right),...\left(x_n,y_n\right)$ . Then, the area of the polygon is $A=\frac{1}{2}\left|\left(x_1y_2+x_1y_2+...x_{n-1}y_n+x_ny_1\right)-\left(y_1x_2+y_1x_2+...y_{n-1}x_n+y_nx_1\right)\right|$

The name of the theorem comes from the fact that when the coordinates are written out in a matrix, as shown:

$\left|\begin{matrix}x_1&y_1\\ x_2&y_2\\ \vdots&\vdots\\ x_n&y_n\\ x_1&y_1\end{matrix}\right|$

Other Forms

This can also be written in form of a summation $A = \dfrac{1}{2} \left|\sum_{i=1}^n{(x_{i+1}+x_i)(y_{i+1}-y_i)}\right|$ or in terms of determinants as $A = \dfrac{1}{2} \left|\sum_{i=1}^n{\det\begin{pmatrix}x_i&x_{i+1}\\y_i&y_{i+1}\end{pmatrix}}\right|$ which is useful in the $3D$ variant of the Shoelace theorem. Note here that $x_{n+1} = x_1$ and $y_{n+1} = y_1$ .

The formula may also be considered a special case of Green's Theorem

$\tilde{A}=\int \int \left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy=\oint(Ldx+Mdy)$

where $L=-y$ and $M=0$ so $\tilde{A}=A$ .

Proof 1

Claim 1: The area of a triangle with coordinates $A(x_1, y_1)$ , $B(x_2, y_2)$ , and $C(x_3, y_3)$ is $\frac{|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|}{2}$ .

Proof of claim 1:

Writing the coordinates in 3D and translating $\triangle ABC$ so that $A=(0, 0, 0)$ we get the new coordinates $A'(0, 0, 0)$ , $B(x_2-x_1, y_2-y_1, 0)$ , and $C(x_3-x_1, y_3-y_1, 0)$ . Now if we let $\vec{b}=(x_2-x_1 \quad y_2-y_1 \quad 0)$ and $\vec{c}=(x_3-x_1 \quad y_3-y_1 \quad 0)$ then by definition of the cross product $[ABC]=\frac{||\vec{b} \times \vec{c}||}{2}=\frac{1}{2}||(0 \quad 0 \quad x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)||=\frac{x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2}{2}$ .

Proof:

We will proceed with induction.

By claim 1, the shoelace theorem holds for any triangle. We will show that if it is true for some polygon $A_1A_2A_3...A_n$ then it is also true for $A_1A_2A_3...A_nA_{n+1}$ .

We cut $A_1A_2A_3...A_nA_{n+1}$ into two polygons, $A_1A_2A_3...A_n$ and $A_1A_nA_{n+1}$ . Let the coordinates of point $A_i$ be $(x_i, y_i)$ . Then, applying the shoelace theorem on $A_1A_2A_3...A_n$ and $A_1A_nA_{n+1}$ we get

$[A_1A_2A_3...A_n]=\frac{1}{2}\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)$ $[A_1A_nA_{n+1}]=\frac{1}{2}(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)$

Hence

$[A_1A_2A_3...A_nA_{n+1}]=[A_1A_2A_3...A_n]+[A_1A_nA_{n+1}]=\frac{1}{2}\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)+\frac{1}{2}(x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2)$ $=\frac{1}{2}((x_2y_1+x_3y_2+...+x_{n+1}y_n+x_1y_{n+1})-(x_1y_2+x_2y_3+...+x_ny_{n+1}+x_{n+1}y_1))=\boxed{\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i)}$

as claimed.

~ShreyJ

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

Proof 3

This is a very nice approach that directly helps in understanding the sum as terms which are areas of trapezoids.

The proof is in this book: https://cses.fi/book/book.pdf#page=281

(The only thing that needs to be slightly modified is that one must shift the entire polygon up by k, until all the y coordinates are positive, but this term gets canceled in the resulting sum.)

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 $\triangle ABF$ .

Exploratory

Observe that $\frac12\left|\det\begin{pmatrix} x_1 & y_1\\ x_2 & y_2 \end{pmatrix}\right|$ is the area of a triangle with vertices $(x_1,y_1),(x_2,y_2),(0,0)$ and $\frac16\left|\det\begin{pmatrix} x_1 & y_1 & z_1\\ x_2 & y_2 & z_2\\ x_3 & y_3 & z_3 \end{pmatrix}\right|$ is the volume of a tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2),(x_3, y_3, z_3),(0,0,0)$ . Does a similar formula hold for $n$ Dimensional triangles for any $n$ ? If so how can we use this to derive the $n$ D Shoelace Formula?

External Links

A good explanation and exploration into why the theorem works by James Tanton: [1]

Nice geometric approach and discussion for proving the 3D Shoelace Theorem by Nicholas Patrick and Nadya Pramita: [2]

Nice integral approach for proving the 3D Shoelace Theorem (ignoring sign of volume) by @george2079: [3] AOPS

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