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Q: Prove that a vertex is a boundary point but not all boundary points are vertices?

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The circumference of a circle is its boundary - it is a perimeter and therefore is a linear measure. Whether it is a smooth curve, as in the case of a circle, or a set of line segments meeting at vertices is irrelevant to its being linear.

To prove that three points are colinear, pick two points and form the equation of the line they describe, and then see if the third point lies on that line.

I'm not sure whether you are asking what convex means, or to prove that all circles are convex. A shape is convex if all pairs of points within it (i.e., the interior, not just the boundary) have all points between them within the shape. That is, if you draw the straight line segment between 2 points on or within a circle, all points of that line segment are within the circle. To prove that, you could use Euclid's axioms of geometry, or analytic geometry, which deals with the coordinates of points and the equations they satisfy. I will leave the proof as an exercise, since it is rather involved (but not necessarily advanced) and I don't know if that is what you are asking for anyway!

Here is one way to prove it. It is easier if the decagon is convex but the proof is valid even if not.Take any point P in the plane. Again, it is easier if this point is inside the decagon but the proof is valid even if it is not.Join all the vertices of the decagon to this point. So now you have ten triangles with a common vertex P.The angles of these ten triangles form the internal angles of the decagon plus all the angles around P.So the sum of the internal angles of the decagon is equal to the sum of the internal angles of all ten triangles with P as their common vertex minus the 360 degrees worth of angles at P itself.The sum of the interior angles of each triangle is 180 degrees so the sum of ten triangles with common vertex P is 180*10 = 1800 degrees.So the internal angles of the decagon add to 1800 - 360 = 1440 degrees.

This problem can be solved with the Calculus of Variations. Seehttp://en.wikipedia.org/wiki/Calculus_of_variations#The_Euler.E2.80.93Lagrange_equation

Related questions

Prove that the maximum vertex connectivity one can achieve with a graph G on n. 01. Define a bipartite graph. Prove that a graph is bipartite if and only if it contains no circuit of odd lengths. Define a cut-vertex. Prove that every connected graph with three or more vertices has at least two vertices that are not cut vertices. Prove that a connected planar graph with n vertices and e edges has e - n + 2 regions. 02. 03. 04. Define Euler graph. Prove that a connected graph G is an Euler graph if and only if all vertices of G are of even degree. Prove that every tree with two or more vertices is 2-chromatic. 05. 06. 07. Draw the two Kuratowski's graphs and state the properties common to these graphs. Define a Tree and prove that there is a unique path between every pair of vertices in a tree. If B is a circuit matrix of a connected graph G with e edge arid n vertices, prove that rank of B=e-n+1. 08. 09.

The circumference of a circle is its boundary - it is a perimeter and therefore is a linear measure. Whether it is a smooth curve, as in the case of a circle, or a set of line segments meeting at vertices is irrelevant to its being linear.

yes

To prove that three points are colinear, pick two points and form the equation of the line they describe, and then see if the third point lies on that line.

There are 4

I'm not sure whether you are asking what convex means, or to prove that all circles are convex. A shape is convex if all pairs of points within it (i.e., the interior, not just the boundary) have all points between them within the shape. That is, if you draw the straight line segment between 2 points on or within a circle, all points of that line segment are within the circle. To prove that, you could use Euclid's axioms of geometry, or analytic geometry, which deals with the coordinates of points and the equations they satisfy. I will leave the proof as an exercise, since it is rather involved (but not necessarily advanced) and I don't know if that is what you are asking for anyway!

That is the definition of a closed set.

The prove is the determination of these melting points.

The Articles of Confederation did not work

Suppose that the parallelogram is a rhombus (a parallelogram with equal sides). If we draw the diagonals, isosceles triangles are formed (where the median is also an angle bisector and perpendicular to the base). Since the diagonals of a parallelogram bisect each other, and the diagonals don't bisect the vertex angles where they are drawn, then the parallelogram is not a rhombus.

No. To prove this, you can actually construct the plane.

There is no such shape. If anyone can prove otherwise, I'd be grateful if they'd let me know.

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