Categories

## Concept – Division Algorithm

Let’s discuss the concept of Division Algorithm.

Given any integers a and b with ?≠0 there exist unique integers q and r such that ?=??+?,?≤?<|?| . If a does not divide b then r satisfies the stronger inequality 0<?<|?|.

Categories

## Beautiful problems from Coordinate Geometry

The following problems are collected from a variety of Math Olympiads and mathematics contests like I.S.I. and C.M.I. Entrances. They can be solved using elementary coordinate geometry and a bit of ingenuity.

1. The equation $x^2 y – 3xy + 2y = 3$ represents:
• (A) a straight line;
• (B) a circle;
• (C) a hyperbola
• (D) none of the foregoing curves;
2. The equation $r = 2a \cos \theta + 2b \sin \theta$ in polar coordinates represents:
• (A) a circle passing through the origin;
• (B) a circle with the origin lying outside it;
• (C) a circle with radius $2 \sqrt {a^2 + b^2 }$ ;
• (D) a circle with the center at the origin;
3. The curve whose equation in polar coordinates is $r \sin^2 \theta – \sin \theta – r = 0$, is
• (A) an ellipse;
• (B) a parabola;
• (C) a hyperbola;
• (D) none of the foregoing curves;
4. A point P on the line 3x + 5y = 15 is equidistant from the coordinate axes can lie in
5. The set of all points (x, y) in the plane satisfying the equation $5x^2 y – xy + y = 0$ forms:
• (A) A straight line;
• (B) a parabola;
• (C) a circle;
• (D) none of the foregoing curves;
6. The equation of the line through the intersection of the lines $$2x + 3y + 4 = 0 \textrm{and} 3x + 4y – 5 = 0$$ and perpendicular to $7x – 5y+ 8 = 0$ is:
• (A) 5x + 7y – 1 = 0;
• (B) 7x + 5y + 1 = 0;
• (C) 5x – 7y + 1 = 0;
• (D) 7x – 5y – 1 = 0;
7. The two equal sides of an isosceles triangle are given by the equations y = 7x and y = -x and its third side passes through (1, -10). Then the equation of the third side is
• (A) 3x + y + 7 = 0 or x – 3y – 31 = 0
• (B) x + 3 y + 29 = 0 or -3x + y + 13 = 0
• (C) 3x + y + 7 = 0 or x + 3y + 29 = 0
• (D) x – 3y – 31 = 0 or – 3x + y + 13 = 0
8. The equations of two adjacent sides of a rhombus are given by y = -x and y = 7x. The diagonals of the rhombus intersect each other at the point (1, 2). The area of the rhombus is:
• (A) $\frac{10}{3}$
• (B) $\frac{20}{3}$
• (C) $\frac{50}{3}$
• (D) none of the foregoing quantities.

More problems are in the Cheenta student portal. You may send answers to support@cheenta.com.

We will keep on adding more problems in this list as well.

Categories

## Clocky Rotato Arithmetic

Do you know that CLOCKS add numbers in a different way than we do? Do you know that ROTATIONS can also behave as numbers and they have their own arithmetic? Well, this post is about how clock adds numbers and rotations behave like numbers. Let’s learn about clock rotation today

Consider the clock on earth.

So, there are 12 numbers {1,2, …, 12 } are written on the clock. But let’s see how clocks add them.

What is 3+ 10 ?

Well, to the clock it is nothing else than 1. Why?

Say, it is 3 am and the clock shows 3 on the clock. Now you add 10 hours to 3 am. You get a 13th hour of the day. But to the clock, it is 1 pm.

So, 3 + 10 = 1.

If you take any other addition, say 9 + 21 = 6 to the clock ( 9 am + 21 hours = 6 pm ).

Now, you can write any other Clocky addition. But you will essentially see that the main idea is :

The clock counts 12 = 0.

Isn’t it easy? 0 comes as an integer just before 1, but on the clock, it is 12 written. So 12 must be equal to 0. Yes, it is that easy.

#### Cayley’s Table

This is a handsome and sober way to write the arithmetic of a set. It is useful if the set is finite like the numbers of the CLOCK Arithmetic.

Let me show you by an example.

Consider the planet Cheenta. A day on Cheenta consists of 6 earth hours.

So, how will the clock on Cheenta look like?

Let’s us construct the Cayley Table for Cheenta’s Clocky Arithmetic. Check it really works as you wish. Here for Cheenta Clock, 3 = 0.

Exercise: Draw the Cayley Table for the Earth (24 hours a day) and Jupiter (10 hours a day).

Nice, let’s move on to the Rotato part. I mean the arithmetic of Rotation part.

Let’s go through the following image.

Well, let’s measure the symmetry of the figure. But how?

Well, which is more symmetric : The Triskelion or the Square (Imagine).

Well, Square seems more right? But what is the thing that is catching our eyes?

It is the set of all the symmetric positions, that capture the overall symmetry of a figure.

For the Triskelion, observe that there are three symmetric operations that are possible but that doesn’t alter the picture:

• Rotation by 120 degrees. $r_1$
• Rotation by 240 degrees. $r_2$
• Rotation by 360 degrees. $r_3$

For the Square, the symmetries are:

• Rotation by 90 degrees.
• Rotation by 180 degrees.
• Rotation by 270 degrees.
• Rotation by 360 degrees.
• Four Reflections along the Four axes

For, a square there are symmetries, hence the eyes feel that too.

So, what about the arithmetic of these? Let’s consider the Triskelion.

Just like 1 interact (+) 3 to give 4.

We say $r_1$ interacts with $r_2$ if $r_1$ acts on the figure after $r_2$ i.e ( 240 + 120 = 360 degrees rotation = $r_3$ ).

Hence, this is the arithmetic of the rotations. To give a sober look to this arithmetic, we draw a Cayley Table for this arithmetic.

Well, check it out.

Exercise: Can you see any similarity of this table with that of anything before?

Challenge Problem: Can you draw the Cayley Table for the Square?

You may explore this link:- https://www.cheenta.com/tag/level-2/

Don’t stop investigating.

All the best.

Hope, you enjoyed. 🙂

Passion for Mathematics.

Categories

## A Proof from my Book

This is proof from my book – my proof of my all-time favorite true result of nature – Pick’s Theorem. This is the simplest proof I have seen without using any high pieces of machinery like Euler number as used in The Proofs from the Book.

Given a simple polygon constructed on a grid of equal-distanced points (i.e., points with integer coordinates) such that all the polygon vertices are grid points, Pick’s theorem provides a simple formula for calculating the area A of this polygon in terms of the number i of lattice points in the interior located in the polygon and the number b of lattice points on the boundary placed on the polygon’s perimeter:

In the language of pictures,

Play around with this in Geogebra.

Kudos to some of the Students of Cheenta Ganit Kendra like Shahbaz Khan and ACHUTHKRISHNAN, who have successfully completed the proof with the help of the following stepstones:

## Steps:

Step 1:

Consider the two-dimensional plane. What is the minimum area of a triangle whose vertices are points with integer coordinates?

Hint: Use Determinant form to compute the area of such a triangle.

Such triangles are called Fundamental Triangles.

Step 2

Prove that the triangle with minimum area and with coordinates as integers cannot contain ANY integer point on or inside the triangle except the three vertices.

Observe that the above picture is one such example.

Step 3

Find the area of a triangle with integer vertices with m points inside it and n points on its boundary edges. (in terms of m and n ) if possible otherwise prove that it is not possible.

Hint: Just draw pictures and use the previous steps.

Step 4

Find the area of any polygon with integer vertices with m points inside it and n points on its boundary edges. (in terms of m and n ) if possible otherwise prove that it is not possible.

Hint: Just try to follow the ideas in Step 3. Try to observe by introducing one point and by the method of induction.

I would love if you try the steps and discover the Pick’s Theorem in such simple steps yourself. Don’t forget to try out the interactive Geogebra version as given.

Enjoy. 🙂

Categories

## A Math Conversation – I

Inspired by the book of Precalculus written in a dialogue format by L.V.Tarasov, I also wanted to express myself in a similar fashion when I found that the process of teaching and sharing knowledge in an easy way is nothing but the output of a lucid conversation between a student and a teacher inside the person only.

Instead of presenting in paragraphs, I will express that discussion in the raw format of conversation to you.

Student: Do you think problem – solving is so important in learning math?

Teacher: Yes, surely without problem-solving, you will not at all enjoy the process of learning of mathematics. If you learn math without problem-solving is like charging the phone without the battery.

S: But why is it so important?

T: Math is nothing but layers of answers to a large number of Whys, Hows, Can We, etc. All the thoughts, questions and answers are refined into something called theorems and lemmas to make them look concise. Theory develops in this way only. Often someone gives a new way of looking into things or connecting two different fields of study, that becomes the definitions, but that was actually done to solve a new problem or question that occurred in their minds or make the things precise. Problems and Theory are the two sides of the same coin.

S: Interesting! So, a student how can we build our own new theory?

T: That’s a good question. As I told you, it is about not developing a new theory, it is about solving a new problem always. So find your own problems or existing ones and start spending time with them.

S: What if I couldn’t solve a problem?

T: That is the best part when you solve problems. You try to understand what doesn’t work. You have to understand why it doesn’t work. Then only you can engineer an alternate pathway, and that is eventually called a new theory if it is too influential. But whatever you have thought you can call it your own theory. Even if you cannot develop a full alternated pathway, something new you have thought will help you solve another problem you were pondering upon for sometime. This actually happenned with me.

S: Wow! This problem solving path seems to be quite adventurous. I never thought it this way.

T: Yes, it is. Infact human beings think and behave in this way only, when their rational and logical part of the brain work. Only they have to conscious of the steps they are taking and they have to take the opportunity to learn at every failure.

S: So, are there any problem solving tips for me?

T: I really don’t believe in tricks while learning, tricks are important for quick problem solving that is only required in the exam. But those are temporary. Once you develop your own strategies, first of all we will get to know something new from you and also you will never forget those strategies. But yes, from my experience I can help you in approaching problems.

S: Surely, what are they?

T: First of all, you need to understand the working principle of problem-solving. It works like whenever you think or see some problem, you are either blank or you have some ideas to approach based on your previous experiences and knowledge.

S: Yes, then?

T: Well, if you have some ideas, you approach along that path till you can’t proceed further. You have to find everytime where is the problem occurring and how to bypass it. This is why different people have different solutions to a problem because they create different bypasses.

S: What if I am blank?

T: You have to learn how to bring in ideas into you.

S: But how?

T: First of all, without knowledge or past experience ideas will rarely come, because that is how humans think using past data. Do you that our visual memory is the maximum memory among all the senses?

S: I felt that was true, but I knew that. But how is it related to math and problem-solving?

T: well, try to draw a picture in your mind or your copy. Try to draw a mindmap, some flowcharts to get some visual aspect of the theory or the problems. Now your visual memory comes into play automatically. That’s why, the important step when you learn something new or try to solve something, ALWAYS TRY TO DRAW DIAGRAMS AND PICTURES WHENEVER IT IS POSSIBLE.

S: That’s seems quite satisfying. I will start doing it from today only. What else?

T: Actually this is the first and important step. Also, intuition is also gained by the method of HIT AND TRIAL. When you have a problem, you play with the numbers in the problem or the expressions and often it happened to me that something interesting came out. So, also whenever you are blank, you can follow this HIT AND TRIAL methodology to start your idea engines and this helps you in recognizing patterns.

S: Hit and Trial seems exciting and adventurous too. What did you mean by pattern recognition?

T: Pattern recognition is the most important part of math. You have to observe a pattern and justify it with logic in a new problem. The first step to problem-solving is to find the pattern. Then comes the next part of logically justifying the pattern. To help you see the pattern and proceeding through you are guided by DRAWING PICTURES and HIT AND TRIAL methodologies.

S: What about logically justifying a problem?

T: This is the most difficult part and rewarding part of mathematics. Often people are not being properly trained in this domain that’s why they find it uninteresting. The method of proofs depends on the pattern you see and observe and that varies from person to person. But proving that the pattern actually holds requires some serious training. Isn’t it so beautiful, when you show that what you observe actually works? It is like painting your patterns by the brush of logic.

S: How do I develop that?

T: That is solely practice and you must enjoy that process. That process requires some knowledge and how people approached on similar ideas and patterns. Once you develop that habit, you will find it super rewarding. It is also a continuously learning path.

S: How do we learn like that that interesting way?

T: Yes, that is a really nice question and those who can impart this knowledge into you very fluidly and interestingly are good teachers. If you want to be a good teacher, you have to start practicing that too. You have to understand and question why every step works and are there any other ways to do it? Then only you will learn out of a problem, or a solution you have learned. Even if you are learning, you have to question WHY THIS WAY AND NOT THAT WAY. Develop the habit of questioning at every step of your learning process, then only you will enjoy.

S: Nice! I learnt a lot of new things today. Let me put that into action. See you later. Have a nice day.

T: Yes, you must put them into action. Yes see ya later. Have a nice day. Bye.

S: Bye.

Categories

## The 3n+1 Problem | Learn Collatz Conjecture

The 3n+1 Problem is known as Collatz Conjecture.

Consider the following operation on an arbitrary positive integer:

• If the number is even, divide it by two.
• If the number is odd, triple it and add one.

The conjecture is that no matter what value of the starting number, the sequence will always reach 1.

Observe that once it reaches 1, it will do like the following oscillation

1 -> 4 -> 2 -> 1-.> 4 ->2 ….

So, we see when it converges to 1, or does it?

We will investigate the problem in certain details as much as possible with occasional exercises and some computer problems ( python ).

### Let’s start!

Let’s start by playing with some numbers and observe what is happening actually.

We will need this frequently. So let’s make a little piece of code to do this.

n = 12
print(n)
while n > 1:
if n % 2 == 0 :
n = n//2
print (n)
else:
n = 3*n+1
print (n)

This will give the output :

12 6 3 10 5 16 8 4 2 1

Now depending on the input of “n” you can get different sequences.

Exercise: Please copy this code and changing the input value of “n”, play around with the sequences here. (Warning: Change it to python before implementing the code.) Eg: Do it for 7.

If you observe that different numbers requires different time to converge and that is the unpredictable which makes the problem so hard.

There are in total two possibilities:

• It may occur a particular value occurs repeatedly i.e. after some time the sequence starts repeating. We define the last non-repeating digit to be the Terminating Number. The total number of numbers in the sequence until the terminating number is called the Length of the Sequence. For eg: For 5 the Terminating Number is 1 and the Length of the sequence is 6.
• We may never end the sequence. In that case, the sequence never terminates.

Exercise: Find out the length of the sequence for all the numbers from 1 to 100 by a computer program. I will provide you the code. Your job is to find a pattern among the numbers.

c = 1
for n in range(2,101):
i = n
while i > 1:
if (i % 2 == 0):
i = i//2
c = c + 1
if i == 1 :
print( "The length of the sequence of {} is {}".format(n, c) )
c = 1
else:
i = 3*i+1
c = c + 1 

Exercise: Consider the length of a sequence corresponding to a starting number “n’ as L(n). Consider numbers of the form n = (8k+4) and n+1 = (8k+5), then find the relationship between L(n) and L(n+1). Try to observe the sequence of lengths for the numbers till 100 and predict the conjecture..

Exercise: Prove the conjecture that you discovered.

Hint:

8k+4 -> 2k+1 -> 6k+4 -> 3k+2

8k+5 -> 24k+ 16 -> 3k+2

Exercise: Prove that if n = 128k + 28, then L(n) = L(n+1) =L(n+2)

Hint:

128k+28 -> 48k+11 -> 81k+20

128k+29 -> 48k+11 -> 81k+20

128k+30 -> 81k+20

Exercise: Try to generalize the result for general k consecutive elements. Maybe you can take help of the computer to observe the pattern. Modify the code as per your choice.

You can try to do an easier problem and get the intuition as an exercise.

Exercise: Try to understand the behavior and the terminating number of the sequence if the rule is the following:

Consider the following operation on an arbitrary positive integer:

1. If the number is even, divide it by two.
2. If the number is odd, add 1.

Exercise: Try to understand the behavior and the terminating number of the sequence if the rule is the following:
Consider the following operation on an arbitrary positive integer:

1. If the number is even, divide it by two.
2. If the number is odd, add any 3.

Exercise: Try to understand the behavior and the terminating number of the sequence if the rule is the following:
Consider the following operation on an arbitrary positive integer:

1. If the number is even, divide it by two.
2. If the number is odd, add ANY ODD NUMBER.

Categories

## The Dhaba Problem | ISI and CMI Entrance

Suppose on a highway, there is a Dhaba. Name it by Dhaba A.

You are also planning to set up a new Dhaba. Where will you set up your Dhaba?

Model this as a Mathematical Problem. This is an interesting and creative part of the BusinessoMath-man in you.

You have to assume something for Mathematical simplicity to model a real life phenomenon via math.

Assumptions:

• The length of the highway is 1 unit. Assume the highway is denoted by the interval [0,1].
• The number of Persons in a road is proportional to the road length. P = c. R, where P is the Persons and R is the road length, c is an arbitrary constant.
• The profit is modeled as the Total Number of Persons arriving in the Dhaba.
• A person will visit a dhaba which is nearer over another which is further away.
• When a customer is at the same distance from two shops then the customer choose randomly.

Observe that the assumptions are valid and are actually followed in real life. Just sit and think for some time placing yourself in that position.

The Explicit Mathematical Problem

Profit Calculation of your Dhaba B

Now based on the lengths of the roads above and the assumptions made, we have to calculate the profit of Dhaba B.

So the profit made by B = c.R , where R is the length of the road on which B has monopoly. Here, as shown R = \c.( x + \frac{|d-x|}{2}\).

We have assumed in this case that $0 \leq x \leq d$ by the diagram.

So, the profit for the Dhaba B is c.$\frac{d+x}{2}$ .

Hence as $0 \leq x \leq d$, the profit is maximized if x = d.

Exercise: Show that the profit of Dhaba B as a function of x is c.($1 – \frac{d+x}{2}$) if $d \leq x \leq 1$ .

Also, observe that in this case the profit is maximized at x = d.

Exercise: Calculate the maximum profit in both cases. Do you observe something fishy? What steps and what arguments will you give to understand the fishiness?

Eager to listen to your beautiful ideas.

Categories

## The Organic Math of Origami

Did you know that there exists a whole set of seven axioms of Origami Geometry just like that of the Euclidean Geometry?

Instead of being very mathematically strict, today we will go through a very elegant result that arises organically from Origami.

Before that, let us travel through some basic terminologies. Be patient for a few more minutes and wait for the gem to arrive.

In case you have forgotten what Origami is, the following pictures will remove the dust from your memories.

Origami (from the Japanese oru, “to fold,” and kami, “paper”) is a traditional Japanese art of folding a sheet of paper, usually square, into a representation of an object such as a bird or flower.

Flat origami refers to configurations that can be pressed flat, say between the pages of a book, without adding any new folds or creases.

When an origami object is unfolded, the resulting diagram of folds or creases on the paper square is called a crease pattern.

We denote mountain folds by unbroken lines and valley folds by dashed
lines. A vertex of a crease pattern is a point where two or more folds intersect, and a flat vertex fold is a crease pattern with just one vertex.

In a crease pattern, we see two types of folds, called mountain folds and valley folds.

Now, if you get to play with your hands, you will get to discover a beautiful pattern.

The positive difference between the dotted lines and the full lines is always 2 at a given vertex. The dotted line denotes the mountain fold and the full line denotes the valley fold.

This is encoded in the following theorem.

Maekawa’s Theorem: The difference between the number of mountain
folds and the number of valley folds in a flat vertex fold is two.

Isn’t it strange ?

Those who are familiar graph theory may think it is related to the Euler Number.

We will do the proof step by step but you will weave together the steps to understand it yourself. The proof is very easy.

Step 1:

Let n denote the number of folds that meet at the vertex, m of which
are mountain folds and v that are valley folds, so that n = m + v. (m for mountain folds and v for valley folds.)

Step 2:

Consider the cross section of a flat vertex.

Step 3:

Consider the creases as shown and fold it accordingly depending on the type of fold – mountain or valley.

Step 4:

Now observe that we get the following cross section. Observe that the number of sides of the formed polygon is n = m + v.

Step 5:

We will also count the angle sum of the n sided polygon in the following way. Consider the polygon formed.

Observe that the vertex 2 is the Valley Fold and other vertices are Mountain Fold. Also the angle subtended the vertex due to valley fold is 360 degrees and that of due to the mountain fold is 0 degrees.

Therefore, the sum of the internal angles is v.360 degrees.

Step 6:

Now we also know that the sum of internal angles formed by n vertices is 180.(n-2), which is = v.360.

Hence we get by replacing n by m+v, that m – v = 2.

QED

So simple and yet so beautiful and magical right?

But it is just the beginning!

There are lot more to discover …

Do you observe any pattern or any different symmetry about the creases or even in other geometry while playing with just paper and folding them?

We will love to hear it from you in the comments.

Do you know Cheenta is bringing out their third issue of the magazine “Reason, Debate and Story” this summer?

Do you want to write an article for us?

Email us at babinmukherjee08@gmail.com.

Categories

## Natural Geometry of Natural Numbers

### How does this sound?

#### The numbers 18 and 30 together looks like a chair.

The Natural Geometry of Natural Numbers is something that is never advertised, rarely talked about. Just feel how they feel!

Let’s revise some ideas and concepts to understand the natural numbers more deeply.

We know by Unique Prime Factorization Theorem that every natural number can be uniquely represented by the product of primes.

So, a natural number is entirely known by the primes and their powers dividing it.

Also if you think carefully the entire information of a natural number is also entirely contained in the set of all of its divisors as every natural number has a unique set of divisors apart from itself.

We will discover the geometry of a natural number by adding lines between these divisors to form some shape and we call that the natural geometry corresponding to the number.

#### Let’s start discovering by playing a game.

Take a natural number n and all its divisors including itself.

Consider two divisors a < b of n. Now draw a line segment between a and b based on the following rules:

• a divides b.
• There is no divisor of n, such that a < c < b and a divides c and c divides b.

Also write the number $\frac{b}{a}$ over the line segment joining a and b.

#### Let’s draw for number 6.

Now, whatever shape we get, we call it the natural geometry of that particular number. Here we call that 6 has a natural geometry of a square or a rectangle. I prefer to call it a square because we all love symmetry.

What about all the numbers? Isn’t interesting to know the geometry of all the natural numbers?

#### Let’s draw for some other number say 30.

Observe this carefully, 30 has a complicated structure if seen in two dimensions but its natural geometrical structure is actually like a cube right?

The red numbers denote the divisors and the black numbers denote the numbers to be written on the line segment.

#### Beautiful right!

Have you observed something interesting?

• The numbers on the line segments are always primes.

#### Exercise: Prove from the rules of the game that the numbers on the line segment always correspond to prime numbers.

Did you observe this?

• In the pictures above, the parallel lines have the same prime number on it.

#### Exercise: Prove that the numbers corresponding to the parallel lines always have the same prime number on it.

Actually each prime number corresponds to a different direction. If you draw it perpendicularly we get the natural geometry of the number.

Let’s observe the geometry of other numbers.

Try to draw the geometry of the number 210. It will look like the following:

Obviously, this is not the natural geometry as shown. But neither we can visualize it. The number 210 lies in four dimensions. If you try to discover this structure, you will find that it has four different directions corresponding to four different primes dividing it. Also, you will see that it is actually a four-dimensional cube, which is called a tesseract. What you see above is a two dimensional projection of the tesseract, we call it a graph.

A person acquainted with graph theory can understand that the graph of a number is always k- regular where k is the number of primes dividing the number.

Now it’s time for you to discover more about the geometry of all the numbers.

Exercise: Show that the natural geometry of $p^k$ is a long straight line consisting of k small straight lines, where p is a prime number and k is a natural number.

Exercise: Show that all the numbers of the form $p.q$ where p and q are two distinct prime numbers always have the natural geometry of a square.

Exercise: Show that all the numbers of the form $p.q.r$ where p, q and r are three distinct prime numbers always have the natural geometry of a cube.

Research Exercise: Find the natural geometry of the numbers of the form $p^2.q$ where p and q are two distinct prime numbers. Also, try to generalize and predict the geometry of $p^k.q$ where k is any natural number.

Research Exercise: Find the natural geometry of $p^a.q^b.r^c$ where p,
q, and r are three distinct prime numbers and a,b and c are natural numbers.

Let’s end with the discussion with the geometry of {18, 30}. First let us define what I mean by it.

We define the natural geometry of two natural numbers quite naturally as a natural extension from that of a single number.

Take two natural numbers a and b. Consider the divisors of both a and b and follow the rules of the game on the set of divisors of both a and b. The shape that we get is called the natural geometry of {a, b}.

You can try it yourself and find out that the natural geometry of {18, 30} looks like the following:

Sit on this chair, grab a cup of coffee and set off to discover.