For any number greater than or equal to 3, a suitable polygon can be found.
Assuming all the vertices of the segmentation lie on the circle, then you can choose any three of them as the corners of a triangle circumscribed by the circle. The perpendicular bisectors of the sides of that triangle intersect at the center of the circle.
By definition, a regular polygon has all interior angles the same, but a concave polygon has some interior angles that are not identical. Also, it violates the axiom that all vertices lie on a circle.While it is possible to construct a polygon with equilateral sides, to be concave would require a form that is equally convex and laterally opposite. (An example is a 'solid arrow shape.')
It will always lie on a diameter.
The only general requirement for a quadrilateral on a plane (flat surface) is that the angles must add up to 360 degrees. There are special cases e.g. if the 4 vertices (corners) lie on a circle each pair of opposite angles must add up to 180 degrees. In a rectangle or a square each angle is 90 degrees.
Lots of points don't lie on the circle. In fact, there are (in a way) more points NOT on the circle, than points on the circle.
triangle
It is a regular polygon as for example an equilateral triangle
A polygon which has a circumscribed circle is called a cyclic polygon.All regular simple polygons, all triangles and all rectangles are cyclic.
false
No.
First make sure you understand that concyclic simply means the points lie on a common circle. We are not told it is a regular pentagon but we will assume it is. We could create pentagons that are not even convex and would not be concyclic.Let's start with a regular pentagon. You can split it up into 5 congruent triangles with the points meeting at the middle. Any side of one of these triangle is connected from each of the vertices of the pentagon to the center of the pentagon. Since all 5 triangles are congruent, this distance must be the same for each of the vertices. So, we see that each of the vertices is equidistant from a given point. Now if we drew a circle centered at that point with a radius equal to the distance between the point and any vertex, that circle would pass through all 5 vertices. Therefore any four ( really all 5), vertices of a regular pentagon are concyclic.A nice proof would use Ptolemy's theorem. I will place a line to an answers.com page that helps with that.Another solution:The pentagon has to be regular. Otherwise, the question is impossible to prove.Consider a regular polygon ABCDE.Take triangles BCD & ECDBC=ED (sides of a regular polygon are equal)CD=CD (common side)
Assuming all the vertices of the segmentation lie on the circle, then you can choose any three of them as the corners of a triangle circumscribed by the circle. The perpendicular bisectors of the sides of that triangle intersect at the center of the circle.
A convex polygon.
Rectangles are 2-dimensional figures- they lie in a plane- they have four vertices There really is no such thing as a 3D rectangle. If you mean a rectangular prism, it has 8 vertices, 4 on each of its two parallel faces.
No, a polygon must be 2 dimensional and lie on a single plane.
By definition, a regular polygon has all interior angles the same, but a concave polygon has some interior angles that are not identical. Also, it violates the axiom that all vertices lie on a circle.While it is possible to construct a polygon with equilateral sides, to be concave would require a form that is equally convex and laterally opposite. (An example is a 'solid arrow shape.')
It will always lie on a diameter.