\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)\(~\)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
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 . It’s quite obvious that each angle is congruent. Also, . 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 .
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 . The angle at P of the right triangle is . Therefore, .
Now, using Law of Co-sines, we can find M (length of segment ) in terms of and r (r is in terms of a and here).
Substituting for
Substituting for
This step isn't so hard, in this note, I prove that the area of any regular polygon can be represented as . (NOTE: that I substituted a=m from the other note and the number of sides is rather than n because in this note I’m using different variables than in the other note)
Plugging in our value of yeilds
.
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Top NewestIs this a new theorem? And can anyone build upon it such that k doesn't necessarily have to divide n?
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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.
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How does your formula work? I'm plugging in x=cos60 for a hexagon and I'm getting 25/4. But I think that it should be 1/4
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If k∤n then you cannot have a regular polygon.
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Thanks...But I am not getting it. Is there anyone who can explain me in simpler words..please?
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