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The Area of a Disk

Friday, January 27th, 2012 16:22 / 1 Comment

~~[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 $R$ is $\pi R^2$ .

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…)

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Triangle Inequality

Friday, January 20th, 2012 10:57 / 5 Comments

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 $a$ , $b$ and $c$ be the three sides of a triangle, then neither $a$ can be greater than $b+c$ , nor $b$ can be greater than $c+a$ , nor $c$ can be than $a+b$ .

: Triangle

Consider the triangle in the image, side $a$ shall be equal to the sum of other two sides $b$ and $c$ , 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, $|x+y|$ represents the side $a$ if $|x|$ represents side $b$ and $|y|$ represents side $c$ . A modulus is nothing, but the distance of a point on the number line from point zero.

: Visual representation of Triangle inequality

For example, the distance of $5$ and $-5$ from $0$ on the initial line is $5$ . So we may write that $|5|=|-5|=5$ .

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

Saturday, August 13th, 2011 14:51 / 2 Comments

A triangle $T$ 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 $1,2,3$ is odd and thus actually $>0$ !

We enumerate all small triangles in the picture as $T_1$ , $T_2, \ldots, T_n$ and denote by $a_i$ the number of edges with endpoints $1$ and $2$ in each triangle $T_i$ . Thus, if say the vertices of $T_i$ are labeled by $1,1,2$ , then $a_i=2$ , and so on …

Observe now that obviously we have

$\displaystyle a_1+a_2+a_3+\cdots +a_n= A+2B,$

where $A$ is the number of triangles whose vertices are labeled $1,2,3$ , while $B$ is the number of those triangles labeled by $1,1,2$ or $1,2,2$ . (Actually it is easily seen that $a_i=2$ for such triangles, while $a_i=1$ if the vertices of $T_i$ are $1,2,3$ and $a_i=0$ otherwise.) All we have to show is that $A$ is odd.

Let $C$ denote the number of $12$ -edges lying inside the original triangle $T$ and let $D$ be the number of $12$ -edges lying on the boundary of $T$ . Every interior $12$ -edge lies in two triangles $T_i$ and thus it is counted twice in the sum $a_1+a_2+a_3+\cdots +a_n$ , while every boundary $12$ -edge is counted only once. In conclusion we get

$\displaystyle a_1+a_2+a_3+\cdots +a_n= 2C+D,$

which yields

$\displaystyle A+2B=2C+D.$

Hence $A$ is odd if and only if $D$ is odd. It is therefore enough to show that $D$ is odd.

According to the hypothesis of the problem, edges labeled $12$ or $21$ can occur only on the $12$ -edge of the large triangle $T$ . We start walking along the edge $12$ of the triangle $T$ starting at the vertex $1$ toward the vertex $2$ . Now, only when we first pass an edge labeled $12$ will we arrive at the first vertex labeled $2$ . A number of vertices labeled $2$ may now follow, and only after we have passed a segment $21$ do we reach a label $1$ , and so on. Thus after an odd number of segments $12$ or $21$ we arrive at vertices labeled $2$ , and after an even number of such segments we arrive at vertices labeled $1$ . Since the last vertex we will reach is the vertex $2$ of the big triangle $T$ , it follows that the total number of segments $12$ or $21$ lying on the side $12$ of the big triangle $T$ must be odd! The same reasoning applies for each of the other edges of the big triangle $T$ , so we deduce that $D$ , the total number of $12$ or $21$ -edges lying on the boundary of $T$ , must be odd. Proved

Graphical Proof

It is obvious. As a result of this numbering we get following diagram:

: Problem Image

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Tag Archives: Triangle

The Area of a Disk

Triangle Inequality

A Problem On Several Triangles

Solution

Graphical Proof

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MY DIGITAL NOTEBOOK

Tag Archives: Triangle

The Area of a Disk

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Triangle Inequality

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A Problem On Several Triangles

Solution

Graphical Proof

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