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The '''Shoelace Theorem''' is a nifty formula for finding the [[area]] of a [[polygon]] given the [[Cartesian coordinate system | coordinates]] of its [[vertex|vertices]]. ==Theorem== Suppose the polygon <math>P</math> has vertices <math>(a_1, b_1)</math>, <math>(a_2, b_2)</math>, ... , <math>(a_n, b_n)</math>, listed in clockwise order. Then the area (<math>A</math>) of <math>P</math> is <cmath>A = \dfrac{1}{2} \left|(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1) \right|</cmath> You can also go counterclockwise order, as long as you find the absolute value of the answer. The Shoelace Theorem gets its name because if one lists the coordinates in a column, <cmath>\begin{align*} (a_1 &, b_1) \\ (a_2 &, b_2) \\ & \vdots \\ (a_n &, b_n) \\ (a_1 &, b_1) \\ \end{align*}</cmath> and marks the pairs of coordinates to be multiplied, <asy> unitsize(1cm); string[] subscripts={"$1$","$2$"," ","$n$","$1$"}; for(int i=1; i < 6; ++i) { label(i==3 ? "$\vdots$" : "$a$",(0,-i*.7)); label(i==3 ? "$\vdots$" : "$b$",(1.2,-i*.7)); label(subscripts[i-1],(0,-i*.7),SE,fontsize(9pt)); label(subscripts[i-1],(1.2,-i*.7),SE,fontsize(9pt)); } for(int i=1; i<5; ++i) draw((0.3,-i*.7)--(1,-(i+1)*.7)); pair c=(1.2,0); label("$-$",shift(c)*(1.2,-2.1)); label("$A=\frac12$",shift(-c)*(0,-2.1)); draw(shift(-1/3*c)*((0,-.5)--(0,-3.9))); draw(shift(13/3*c)*((0,-.5)--(0,-3.9))); for(int i=1; i < 6; ++i) { label(i==3 ? "$\vdots$" : "$a$",shift(3*c)*(0,-i*.7)); label(i==3 ? "$\vdots$" : "$b$",shift(3*c)*(1.2,-i*.7)); label(subscripts[i-1],shift(3*c)*(0,-i*.7),SE,fontsize(9pt)); label(subscripts[i-1],shift(3*c)*(1.2,-i*.7),SE,fontsize(9pt)); } for(int i=1; i<5; ++i) draw(shift(3*c)*(0.3,-(i+1)*.7)--shift(3*c)*(1,-i*.7)); </asy> the resulting image looks like laced-up shoes. ==Other Forms== This can also be written in form of a summation <cmath>A = \dfrac{1}{2} \left|\sum_{i=1}^n{(x_{i+1}+x_i)(y_{i+1}-y_i)}\right|</cmath> or in terms of determinants as <cmath>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|</cmath> which is useful in the <math>3D</math> variant of the Shoelace theorem. Note here that <math>x_{n+1} = x_1</math> and <math>y_{n+1} = y_1</math>. The formula may also be considered a special case of Green's Theorem <cmath>\tilde{A}=\int \int \left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)dxdy=\oint(Ldx+Mdy)</cmath> where <math>L=-y</math> and <math>M=0</math> so <math>\tilde{A}=A</math>. ==Proof 1== Claim 1: The area of a triangle with coordinates <math>A(x_1, y_1)</math>, <math>B(x_2, y_2)</math>, and <math>C(x_3, y_3)</math> is <math>\frac{|x_1y_2+x_2y_3+x_3y_1-x_1y_3-x_2y_1-x_3y_2|}{2}</math>. ===Proof of claim 1:=== Writing the coordinates in 3D and translating <math>\triangle ABC</math> so that <math>A=(0, 0, 0)</math> we get the new coordinates <math>A'(0, 0, 0)</math>, <math>B(x_2-x_1, y_2-y_1, 0)</math>, and <math>C(x_3-x_1, y_3-y_1, 0)</math>. Now if we let <math>\vec{b}=(x_2-x_1 \quad y_2-y_1 \quad 0)</math> and <math>\vec{c}=(x_3-x_1 \quad y_3-y_1 \quad 0)</math> then by definition of the cross product <math>[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}</math>. ===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 <math>A_1A_2A_3...A_n</math> then it is also true for <math>A_1A_2A_3...A_nA_{n+1}</math>. We cut <math>A_1A_2A_3...A_nA_{n+1}</math> into two polygons, <math>A_1A_2A_3...A_n</math> and <math>A_1A_nA_{n+1}</math>. Let the coordinates of point <math>A_i</math> be <math>(x_i, y_i)</math>. Then, applying the shoelace theorem on <math>A_1A_2A_3...A_n</math> and <math>A_1A_nA_{n+1}</math> we get <cmath>[A_1A_2A_3...A_n]=\frac{1}{2}\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)</cmath> <cmath>[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)</cmath> Hence <cmath>[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)</cmath> <cmath>=\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)}</cmath> as claimed. ~ShreyJ ==Proof 2== 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 Stokes 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> ==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 <math>ABC</math>, we have <math>\angle ACB=90^{\circ}</math>, <math>AC=2</math>, and <math>BC=3</math>. [[Median of a triangle|Median]]s <math>AD</math> and <math>BE</math> are drawn to sides <math>BC</math> and <math>AC</math>, respectively. <math>AD</math> and <math>BE</math> intersect at point <math>F</math>. Find the area of <math>\triangle ABF</math>. === Exploratory === Observe that <cmath>\frac12\left|\det\begin{pmatrix} x_1 & y_1\\ x_2 & y_2 \end{pmatrix}\right|</cmath> is the area of a triangle with vertices <math>(x_1,y_1),(x_2,y_2),(0,0)</math> and <cmath>\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|</cmath> is the volume of a tetrahedron with vertices <math>(x_1, y_1, z_1), (x_2, y_2, z_2),(x_3, y_3, z_3),(0,0,0)</math>. Does a similar formula hold for <math>n</math>Dimensional triangles for any <math>n</math>? If so how can we use this to derive the <math>n</math>D Shoelace Formula? == External Links== A good explanation and exploration into why the theorem works by James Tanton: [http://www.jamestanton.com/wp-content/uploads/2012/03/Cool-Math-Essay_June-2014_SHOELACE-FORMULA.pdf] Nice geometric approach and discussion for proving the 3D Shoelace Theorem by Nicholas Patrick and Nadya Pramita: [http://media.icys2018.com/2018/04/IndonesiaPatrickNicholas105190.pdf] Nice integral approach for proving the 3D Shoelace Theorem (ignoring sign of volume) by @george2079: [https://mathematica.stackexchange.com/a/26015] [[Category:Geometry]] [[Category:Theorems]] AOPS
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