\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)In this note, I will prove that the area formed by connecting every kth vertex of a regular polygon is \(\dfrac{a^2n(\cot{\frac{180k_i}{n}})(\csc^2{(\frac{180}{n})})(1-\cos{(\frac{k_i360}{n})})}{8k_i}\) when \(\textbf{k divides n without remainder}\) 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.

\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)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 \(=\boxed{\dfrac{n}{k_i}}\)

\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)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 \(\theta\). It’s quite obvious that each angle is congruent. Also, \(\boxed{\theta=\frac{360}{n}}\). 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 \(\boxed{\angle{JPK_i=k_i\theta}}\).

\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)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 \(\frac{a}{2}\). The angle at P of the right triangle is \(\dfrac{\theta}{2}\). Therefore, \(\boxed{r=(\frac{a}{2})\csc{(\frac{\theta}{2})}}\).

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

\[m^2=r^2+r^2-2(r)(r)\cos{(k_i\theta)}\]

\[m^2=2r^2(1-\cos{(k_i\theta}))\]

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

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

Substituting for \(\theta=\frac{360}{n}\)

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

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

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

Plugging in our value of \(m^2\) yeilds

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

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

\[\boxed{area=\dfrac{a^2n(\cot{\frac{180k_i}{n}})(\csc^2{(\frac{180}{n})})(1-\cos{(\frac{k_i360}{n})})}{8k_i}}\]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

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.

Log in to reply

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

Log in to reply

Log in to reply

If \(k \nmid n\) then you cannot have a regular polygon.

Log in to reply

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

Log in to reply