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[This post is under review.]
If you are aware of elementary facts of geometry, then you might know that the area of a disk with radius is .
The radius is actually the measure(length) of a line joining the center of disk and any point on the circumference of the disk or any other circular lamina. Radius for a disk is always same, irrespective of the location of point at circumference to which you are joining the center of disk. The area of disk is defined as the ‘measure of surface‘ surrounded by the round edge (circumference) of the disk.
The area of a disk can be derived by breaking it into a number of identical parts of disk as units — calculating their areas and summing them up till disk is reformed. There are many ways to imagine a unit of disk. We can imagine the disk to be made up of several concentric very thin rings increasing in radius from zero to the radius of disc. In this method we can take an arbitrary ring, calculate its area and then in similar manner, induce areas of other rings -sum them till whole disk is obtained. (more…)
Triangle inequality has its name on a geometrical fact that the length of one side of a triangle can never be greater than the sum of the lengths of other two sides of the triangle. If , and be the three sides of a triangle, then neither can be greater than , nor can be greater than , nor can be than .
Consider the triangle in the image, side shall be equal to the sum of other two sides and , only if the triangle behaves like a straight line. Thinking practically, one can say that one side is formed by joining the end points of two other sides.
In modulus form, represents the side if represents side and represents side . A modulus is nothing, but the distance of a point on the number line from point zero.
For example, the distance of and from on the initial line is . So we may write that .
Triangle inequalities are not only valid for real numbers but also for complex numbers, vectors and in Euclidean spaces. In this article, I shall discuss them separately. (more…)
A triangle is divided into smaller triangles such that any two of the smaller triangles either have no point in common, or have a vertex in common, or actually have an edge in common. Thus no two smaller triangles touch along part of an edge of them.
For an illustration let me denote the three vertices of T by 1, 2 and 3. Now number each of the vertices of the small triangles by 1, 2, 3. Do this in an arbitrary way, but such that vertices lying on an edge of T must not be numbered by the same number as the vertex of T opposite to that edge.
Show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3.
To show that among the small triangles there is always one whose vertices are numbered by 1, 2 and 3, we show that the number of small triangles whose vertices are labeled with is odd and thus actually !
We enumerate all small triangles in the picture as , and denote by the number of edges with endpoints and in each triangle . Thus, if say the vertices of are labeled by , then , and so on …
Observe now that obviously we have
where is the number of triangles whose vertices are labeled , while is the number of those triangles labeled by or . (Actually it is easily seen that for such triangles, while if the vertices of are and otherwise.) All we have to show is that is odd.
Let denote the number of -edges lying inside the original triangle and let be the number of -edges lying on the boundary of . Every interior -edge lies in two triangles and thus it is counted twice in the sum , while every boundary -edge is counted only once. In conclusion we get
Hence is odd if and only if is odd. It is therefore enough to show that is odd.
According to the hypothesis of the problem, edges labeled or can occur only on the -edge of the large triangle . We start walking along the edge of the triangle starting at the vertex toward the vertex . Now, only when we first pass an edge labeled will we arrive at the first vertex labeled . A number of vertices labeled may now follow, and only after we have passed a segment do we reach a label , and so on. Thus after an odd number of segments or we arrive at vertices labeled , and after an even number of such segments we arrive at vertices labeled . Since the last vertex we will reach is the vertex of the big triangle , it follows that the total number of segments or lying on the side of the big triangle must be odd! The same reasoning applies for each of the other edges of the big triangle , so we deduce that , the total number of or -edges lying on the boundary of , must be odd. Proved
It is obvious. As a result of this numbering we get following diagram: