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The Area of a Disk
[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
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
.

- Triangle
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 Problem On Several Triangles
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.
Solution
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
which yields
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
Graphical Proof
It is obvious. As a result of this numbering we get following diagram:


