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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 \(\dfrac{a^2n(\cot{\frac{180k_i}{n}})(\csc^2{(\frac{180}{n})})(1-\cos{(\frac{k_i360}{n})})}{8k_i}\) 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.

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

Step two: determining the length of each side



\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)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).



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


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



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 \(\dfrac{ \left(\dfrac{n}{k_i}\right)\cot(\dfrac{180k_i}{n})m^2}{4}\). (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


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


Note by Trevor Arashiro
1 year, 10 months ago

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Is this a new theorem? And can anyone build upon it such that k doesn't necessarily have to divide n? Trevor Arashiro · 1 year, 10 months ago

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@Trevor Arashiro If \(k \nmid n\) then you cannot have a regular polygon. Raphael Nasif · 1 year, 7 months ago

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@Trevor Arashiro 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 · 1 year, 10 months ago

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@Jake Lai 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 Trevor Arashiro · 1 year, 10 months ago

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@Trevor Arashiro I didn't check the formula, so my formula is probably invalid :< Jake Lai · 1 year, 10 months ago

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Thanks...But I am not getting it. Is there anyone who can explain me in simpler words..please? Ubaidullah Khan · 1 year, 7 months ago

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