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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

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 $$\theta$$. 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 $$\theta=\frac{360}{n}$$

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

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

## 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

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

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

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

Note by Trevor Arashiro
2 years, 3 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? · 2 years, 3 months ago

If $$k \nmid n$$ then you cannot have a regular polygon. · 1 year, 12 months ago

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. · 2 years, 3 months ago

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 · 2 years, 3 months ago

I didn't check the formula, so my formula is probably invalid :< · 2 years, 3 months ago