Ratio of the areas of the circumcircle and the incircle of an n-sided regular polygon

Question:

To find the ratio of the areas of the circumcircle and the incircle of an n-sided regular polygon.

Solution:

Result:

The ratio of the areas of the circumcircle and the incircle of an n sided regular polygon is square of the secant ratio of (pi divided by the number of sides of the regular ploygon).

Note:

The ratio of the areas of the circumcircle and the incircle of an n-sided regular polygon is independent of size of the side. It depends only on the number of sides.

© Rishabh Bidya

Difference between the areas of the circumcircle and the incircle of an n-sided regular polygon

Question:

To find the difference between the areas of the circumcircle and the incircle of an n-sided regular polygon.

Solution:

Result:

The difference between the areas of the circumcircle and the incircle of an n sided regular polygon is pi times square of the length of the side divided by 4.

Note:

The difference between the areas of the circumcircle and the incircle of an n-sided regular polygon is independent of the number of sides. It depends only on the length of the side.

© Rishabh Bidya

Surprise your buddies: A wonderful number trick

The trick:

We tell our friend to think of any random number. Then perform some random operations on it and .. Surprise! We tell him the final answer.

Note: Before we start the trick, we must think of two numbers, a and m. a is the number that we want as our answer and m is an arbitrary multiplier. Preferably a and m should be small for our convenience during arithmetic operations. Of course, we never disclose a and m to our friend before or during the trick !!

Let us begin !!

i) Think of a and m.
ii) Multiply a and m. P=a*m.

iii) Tell your friend to think of any number, say x.
iv) Tell him to multiply x with m.
v) Tell him to subtract P from the obtained product of x and m.
vi) Now, tell your buddy to divide the obtained difference by m.
vii) Now, you tell your friend to subtract the obtained quotient from x.

TADA !! Tell your friend - The answer is a.

Let us try the trick (An example) :

i) I think of two numbers:
   a=3. (3 should be my final answer.)
   m=2.
ii) P=6.

iii) You think of a number, say 7.         (x=7)
iv) Multiply 7 with 2. Result is 14.
v) Subtract 6 from 14. Result is 8.       (P=6)
vi) Divide obtained difference by 2, i.e. 8 by 2. Result is 4      (Here our m=2. Therefore, we divide difference by 2).
vii) Subtract obtained result from the number you had thought first, i.e. Subtract 4 from 7.

The answer is 3 !!!



© Rishabh Bidya

Total number of triangles in an inverted triangle system

Question:

To find total number of triangles in an inverted triangular system which has n triangles, each of which is inverted and contained inside its previous triangle.

Example: 

Solution : 

If the total number of triangles is n, 

Except for the innermost triangle, every triangle made results in three smaller triangles, so the number of such smaller triangles is 3*(n-1)

Apart from the smaller formed triangles, there are n triangles that form the figure.

Thus, the total number of such formed triangles will be the sum of both these numbers, i.e. 3*(n-1)+n

Therefore, we obtain the total number of triangles in an inverted triangle system is 4n-3

© Rishabh Bidya

Magic Squares 2 : Creating a 3x3 magic square using 9 consecutive integers given the sum of a row or a column

Problem Background : A 3x3 magic square is an arrangement of 9 squares arranged in 3 rows and 3 columns. We are given a sum S which is the sum of any row or column. We need to arrange 9 consecutive integers in the square.

Solution : If the sum of a row is S, S also needs to be the sum of any other row or any other column.

Derived corollary : It can be noted that for the numbers in individual squares to be integers, S must be divisible by 3, i.e. S must be 0(mod3) and S must be greater than 12.

The above result can be derived from  Magic Squares 1 :

After this, we merely substitute for a and simplify.

Q.E.D

© Rishabh Bidya

Magic Squares 1 : To create a 3x3 magic square given 9 consecutive integers

Problem Background : A 3x3 magic square is an arrangement of 9 squares arranged in 3 rows and 3 columns. We are given 9 consecutive integers to place in each of the squares such that the sum of each of the rows and columns is equal.

Solution : If the given integers begin from a, rest if the integers are a+1, a+2, a+3, ..., a+7, a+8.

In the above figure, Sum of every row as also sum of every column is 3a+12.

Q.E.D

© Rishabh Bidya