Area Created by Joining Every Kth Vertex in a Polygon

 ~ ~ ~ ~ ~ ~ ~ ~ ~In this note, I will prove that the area formed by connecting every kth vertex of a regular polygon is a2n(cot180kin)(csc2(180n))(1cos(ki360n))8ki\dfrac{a^2n(\cot{\frac{180k_i}{n}})(\csc^2{(\frac{180}{n})})(1-\cos{(\frac{k_i360}{n})})}{8k_i} when k divides n without remainder\textbf{k divides n without remainder} and n=number of sides of the original polygon and a represents the side length of the original polygon.

 ~ ~ ~ ~ ~ ~ ~ ~ ~ Can someone link me to a website so I can read up on it? I haven't been able to find one.


Step one: determining the number of sides

 ~ ~ ~ ~ ~ ~ ~ ~ ~This step is rather simple. The number of sides of the original polygon R divided by k will yield the new polygon’s number of sides.

 ~ ~ ~ ~ ~ ~ ~ ~ ~This is because #vertices=#edges. And since we’re dividing the number of vertices by k, the number of edges will be divided by k as well.

 ~ ~ ~ ~ ~ ~ ~ ~ ~Therefore, the number of sides =nki=\boxed{\dfrac{n}{k_i}}


Step two: determining the length of each side

image image


 ~ ~ ~ ~ ~ ~ ~ ~ ~Ok, this is the hard part. For this part, I will refer to the second picture (also, NOTE 1,2,3,4 represent possible values of k). Begin by looking at the angled formed by θ\theta. It’s quite obvious that each angle is congruent. Also, θ=360n\boxed{\theta=\frac{360}{n}}. Now, if we’re going to connect every kth vertex starting with vertex J, the angle (NOTE: point k varies in position) formed will be equal to JPKi=kiθ\boxed{\angle{JPK_i=k_i\theta}}.

 ~ ~ ~ ~ ~ ~ ~ ~ ~Now, the one thing that remains constant no matter what vertex we chose is the length from the center of the original polygon to each vertex, AKA: the radius. The radius r is the hypotenuse of a right triangle with one leg as its apothem and the other leg being a2\frac{a}{2}. The angle at P of the right triangle is θ2\dfrac{\theta}{2}. Therefore, r=(a2)csc(θ2)\boxed{r=(\frac{a}{2})\csc{(\frac{\theta}{2})}}.

 ~ ~ ~ ~ ~ ~ ~ ~ ~Now, using Law of Co-sines, we can find M (length of segment Jki\overline{Jk_i}) in terms of kiθk_i\theta and r (r is in terms of a and csc\csc here).

m2=r2+r22(r)(r)cos(kiθ)m^2=r^2+r^2-2(r)(r)\cos{(k_i\theta)}

m2=2r2(1cos(kiθ))m^2=2r^2(1-\cos{(k_i\theta}))

Substituting for r=(a2)csc(θ2)r=(\frac{a}{2})\csc{(\frac{\theta}{2})}

m2=2[(a2)csc(θ2)](1cos(kiθ))m^2=2\left[(\frac{a}{2})\csc{(\frac{\theta}{2})}\right](1-\cos{(k_i\theta)})

Substituting for θ=360n\theta=\frac{360}{n}

m2=2[a2csc2((360n)2)4](1cos(ki(360n))m^2=2\left[\frac{a^2\csc^2{(\frac{(\frac{360}{n})}{2})}}{4}\right](1-\cos{(k_i(\frac{360}{n})})

m2=a2csc2(180n)(1cos(360kn))2\boxed{m^2=\dfrac{a^2\csc^2{(\frac{180}{n})}(1-\cos{(\frac{360k}{n})})}{2}}


Finding the area

 ~ ~ ~ ~ ~ ~ ~ ~ ~This step isn't so hard, in this note, I prove that the area of any regular polygon can be represented as (nki)cot(180kin)m24\dfrac{ \left(\dfrac{n}{k_i}\right)\cot(\dfrac{180k_i}{n})m^2}{4}. (NOTE: that I substituted a=m from the other note and the number of sides is nki\dfrac{n}{k_i} rather than n because in this note I’m using different variables than in the other note)

Plugging in our value of m2m^2 yeilds

area=(nki)cot(180kin)(a2csc2(180n)(1cos(360kin))2)4area=\dfrac{\left(\dfrac{n}{k_i}\right)\cot{(\dfrac{180k_i}{n})}\left(\dfrac{a^2\csc^2{(\frac{180}{n})}(1-\cos{(\frac{360k_i}{n})})}{2}\right)}{4}.

area=ncot(180kin)(a2csc2(180n)(1cos(360kin)))8kiarea=\dfrac{ n\cot{(\dfrac{180k_i}{n})}\left(a^2\csc^2{(\frac{180}{n})}(1-\cos{(\frac{360k_i}{n})})\right)}{8k_i}.

area=a2n(cot180kin)(csc2(180n))(1cos(ki360n))8ki\boxed{area=\dfrac{a^2n(\cot{\frac{180k_i}{n}})(\csc^2{(\frac{180}{n})})(1-\cos{(\frac{k_i360}{n})})}{8k_i}}

Note by Trevor Arashiro
4 years, 8 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Is this a new theorem? And can anyone build upon it such that k doesn't necessarily have to divide n?

Trevor Arashiro - 4 years, 8 months ago

Log in to reply

Probably not. You can probably use the same application of cosine rule to prove the case where k doesn't divide n; I did that when doing proving the case k = 2 for all n.

Jake Lai - 4 years, 8 months ago

Log in to reply

How does your formula work? I'm plugging in x=cos60x=\cos{60} for a hexagon and I'm getting 25/4. But I think that it should be 1/4

Trevor Arashiro - 4 years, 8 months ago

Log in to reply

@Trevor Arashiro I didn't check the formula, so my formula is probably invalid :<

Jake Lai - 4 years, 8 months ago

Log in to reply

If knk \nmid n then you cannot have a regular polygon.

Raphael Nasif - 4 years, 4 months ago

Log in to reply

Thanks...But I am not getting it. Is there anyone who can explain me in simpler words..please?

Ubaidullah Khan - 4 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...