Product of Diagonals of Regular Polygon?

In summary: Now, let's say we have a regular polygon with n vertices inscribed in a unit circle. We can label the vertices v0, v1, v2, ..., v(n-1), with v0 always being equal to 1. We want to find the product of the lengths of diagonals drawn from v0 to each of the other vertices, which can be written as |v1-1||v2-1|...|v(n-1)-1|. Using the properties of the nth roots of unity, we can show that the sum of the nth roots of unity is equal to 0. This means that when we expand the polynomial (z-1)(z^{n-1} + z^{
  • #1
nfljets
3
0
So any help would be really appreciated! I really have no idea where to start, and I can use any help.

So essentially the problem is we have a regular polygon P inscribed in a unit circle. This regular polygon has n vertices. Fix one vertex and take the product of the lengths of diagonals drawn from the one vertex to each of the other vertices.

Show that the product of these diagonals is equal to n.

...so I guess what I have so far (which is quite minimal) is this:

So I suppose I figure as much that the vertices of the regular polygon are nth roots of unity. And so I also suppose that one vertex is always 1. (1 to the nth power is always 1)

So let's label the vertices of the polygon v0, v1, v2, ... v(n-1), where v0=1.

So essentially we want to find the product |v1-1||v2-1|...|v(n-1)-1| -- and we want to show that equals n.

I guess this is what I have so far, but I'm not really sure where to go from there.

Again, any help will be greatly appreciated!
 
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  • #2
It might be easier to show that the square of the product of the diagonals = ##n^2##. That gets rid of all the modulus functions.

Think about the properties of the n'th roots of unity. They are the roots of a polynomial equation, so it is easy to show that the sum of the n'th roots = 0. There are similar results for the sums of products and powers of the n'th roots.
 
  • #3
hmmm so I'm a little conrfused...so I guess okay, the nth roots of unity are the roots of a polynomial equation..(first of all I don't think I really get that)

And, I don't seem to understand how the sum of them equal 0...
 
  • #4
The nth roots of unity, ##\cos 2 \pi k/n + i \sin 2 \pi k/n## for k = 0, 1, ... n-1 are the roots of the equation ##z^n - 1 = 0##.

If ##z## is a root , then ##z^2##, ##z^3##, ... are also roots.

##z^n -1 = (z-1)(z^{n-1} + z^{n-2} \cdots + 1)##

So if ##z \ne 1## is a root of ##z^n - 1 = 0##, ##z^{n-1} + z^{n-2} \cdots + 1 = 0##.

See http://en.wikipedia.org/wiki/Root_of_unity (or Google for "roots of unity").
 
  • #5
AlephZero said:
The nth roots of unity, ##\cos 2 \pi k/n + i \sin 2 \pi k/n## for k = 0, 1, ... n-1 are the roots of the equation ##z^n - 1 = 0##.

If ##z## is a root , then ##z^2##, ##z^3##, ... are also roots.

##z^n -1 = (z-1)(z^{n-1} + z^{n-2} \cdots + 1)##

So if ##z \ne 1## is a root of ##z^n - 1 = 0##, ##z^{n-1} + z^{n-2} \cdots + 1 = 0##.

See http://en.wikipedia.org/wiki/Root_of_unity (or Google for "roots of unity").
I don't really get that...can you restate?
 
  • #6
##z^n - 1 = (z-1)(z^{n-1} + z^{n-2} \cdots + 1)##. (To show that, just multiply the two factors).

If ##z## is a root of ##z^n - 1##, then ##(z^n - 1) = 0##.

So either ##(z-1) = 0## or ##(z^{n-1} + z^{n-2} \cdots + 1) = 0##.

If ##z \ne 1##, then ##(z-1) \ne 0##.

So ##(z^{n-1} + z^{n-2} \cdots + 1) = 0##.
 

1. What is a regular polygon?

A regular polygon is a two-dimensional shape with all sides and angles equal in measure. Some examples of regular polygons are triangles, squares, and hexagons.

2. How do you find the product of diagonals of a regular polygon?

To find the product of diagonals of a regular polygon, you need to first know the number of sides of the polygon. Then, you can use the formula n(n-3)/2, where n is the number of sides, to calculate the number of diagonals. Once you have the number of diagonals, you can multiply them by each other to find the product.

3. Why is the product of diagonals of a regular polygon important?

The product of diagonals of a regular polygon is important in geometry because it can help us calculate the area and perimeter of the polygon. It is also useful in solving problems related to symmetry and angles in regular polygons.

4. Can the product of diagonals of a regular polygon be negative?

No, the product of diagonals of a regular polygon cannot be negative. The number of diagonals is always a positive integer, and multiplying two positive numbers will always result in a positive product.

5. How can the product of diagonals of a regular polygon be used in real life?

The product of diagonals of a regular polygon can be used in real life in various applications such as architecture, engineering, and design. For example, it can help in calculating the dimensions of a regular polygon-shaped window or in designing a symmetrical building with a regular polygon floor plan.

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