How do we prove that the normal vectors to the sides of a polygon add up to 0?

Question

Given any triangle in the plane, associate to each side an outward pointing normal vector of the same length as the side. Show that the sum of these three vectors is always 0.

Answer 1

Hint: Note that cyclic vectors parallel to the sides of the triangle (and having the same length as each) sum to zero. Does this tell you anything about the sum of normals with severally equal lengths?

Elaboration: Please note that the three normals also determine a (congruent) triangle (since the lengths of the normals are equal to the lengths of the original sides). Draw a diagram to confirm that the orientation of the arrows is cyclic as in the original. Then the result follows.

Answer 2

Let $A_0, A_1, A_2, \ldots, A_n = A_0$ be the vertices of a $n$-gon oriented counterclockwisely and $N_i$ be the outward point vector for edge $A_iA_{i+1}$ for $0 \le i < n$ with length $|A_i A_{i+1}|$.

If one embed $\mathbb{R}^2$ into $\mathbb{R}^3$ by the map $$p =(x,y) \quad\mapsto\quad \vec{p} = (x,y,0)$$ convince yourself that

$$\vec{N}_i = (\vec{A}_{i+1}-\vec{A}_i) \times \hat{z}$$

This implies

$$\sum_{i=0}^{n-1} \vec{N}_i = \sum_{i=0}^{n-1}(\vec{A}_{i+1}-\vec{A}_i) \times \hat{z} = (\vec{A}_{n}-\vec{A}_0) \times \hat{z} = \vec{0}$$

This phenomenon is not limited to $2$-d. For example, for any polyhedron in $\mathbb{R}^3$, if you associate an outward pointing normal to each face with length proportional to the face area, these outward pointing normals also sum to zero.

Answer 3

It is convenient to prove it explicitly with vector cross products.

Let u be a unit vector normal to the plane of polygon. Then, the n-th normal vector N_n for the vertexes A_n and A_n+1 is

$$N_n = u\times(A_{n}-A_{n+1})$$

where $A_{n}-A_{n+1}$ is the vector of the n-th side.

The sum is then,

$$N_1+N_2+ \>... \>+N_n=u\times(A_{1}-A_2)+u\times(A_{2}-A_3)+\>...\>+u\times(A_{n}-A_1)$$ $$=u\times(A_{1}-A_2 + A_{2}-A_3+...+A_{n}-A_1)=u\times 0=0$$

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For a particular triangle below

Along the horizontal direction, you need to show

$$b\sin\gamma = c\sin\beta$$

which is guaranteed by the sine rule.

And, along the vertical direction, you need to show

$$ a = c\cos\beta + b\cos\gamma$$

which is evident from the diagram, i.e. the RHS matches the base length $a$.

Answer 4

You can prove it very simply using complex numbers: in the Argand-Cauchy plane, let $z-1, z_2, \dots,z_n$ be the affixes of the vertices $A_1,A_2,\dots ,A_n$ and consider the vectors determined by the sides of the polygon: $\overrightarrow{A_1A_2}$, with affix $z_2-z_1$, $\overrightarrow{A_2A_32}$, with affix $z_3-z_2$, $\dots$ ,$\overrightarrow{A{n}A_1}$, with affix $z_n-z_1$.

The vector directly orthogonal to vector $\overrightarrow{A_iA_{i+1}}$ has affix $\mathrm e^{\tfrac{i\pi}2}\!(z_{i+1}-z_i)$. Thus the sum of these vectors has affix $$e^{\tfrac{i\pi}2}\bigl((z_2-z_1)+(z_3-z_2)+\dots+(z_n-z_{n-1})+(z_1-z_n)\bigr)=e^{\tfrac{i\pi}2}\cdot 0.$$

Answer 5

Draw a triangle or polygon. Choose a point(vertex) and start traversing clockwise (for example) on the sides. The vectors you get in this process are the vectors whose magnitudes are the length of the sides and direction is the direction of traverse. As you complete traversing you reach to the point from where you started traversing. So you see all the vectors can not be in the same direction so that they will add up to a non zero value. Now this addition will add up to zero by definition of vector addition (as you reached to the same point) . Now vectors normal(outward) to the sides with equal magnitudes are just these vectors rotated 90 degree counterclockwise. As all the vectors rotated by same angle in the same direction ( here it is 90 degree counterclockwise, but could have been any angle in the same direction); the resultant vectors magnitude will remain same. Here the resultant magnitude is zero because earlier also it was zero.

Edit:

(1) For a polygon you need to apply triangle law of vector addition repeatedly to get the resultant vector. If sides of a polygon is A,B,C,D (in order) then you see $\overrightarrow{\rm A} + \overrightarrow{\rm B} + \overrightarrow{\rm C} + \overrightarrow{\rm D} = (\overrightarrow{\rm A} + \overrightarrow{\rm B}) + \overrightarrow{\rm C} + \overrightarrow{\rm D}$. (here $\overrightarrow{\rm A} + \overrightarrow{\rm B}$ is a vector). Which can be be further written as $( (\overrightarrow{\rm A} + \overrightarrow{\rm B}) + \overrightarrow{\rm C} ) + \overrightarrow{\rm D}$.

Now we see that vector $( (\overrightarrow{\rm A} + \overrightarrow{\rm B}) + \overrightarrow{\rm C} )$ is equal in magnitude but opposite in direction of vector D. So they cancel out.

(2) Now consider vectors normal to these vectors and outward from the enclosed area. The outward is defined by direction which is rotated 90 degree anticlockwise if you traverse clockwise (or rotated 90 degree clockwise if you traverse anticlockwise). If you consider two adjacent vectors of the polygon say A,B and their normal vectors $A^{'},B^{'}$ then vector $A^{'}+B^{'}$ will be normal to vector $A+B$ and equal in magnitude to $A+B$. Similarly $A^{'}+B^{'}+C^{'}$ will be normal to vector $A+B+C$ but equal in magnitude to $A+B+C$. So vector $A^{'}+B^{'}+C^{'}$ will be equal in magnitude but opposite in direction of vector $D^{'}$ which is normal to vector D. So they will cancel out each other.