## Monthly Archives: April 2017

### Pick’s theorem to pick your brains!!

Pick’s theorem:

Consider a square lattice of unit side. A simple polygon (with non-intersecting sides) of any shape is drawn with its vertices at the lattice points. The area of the polygon can be simply obtained as $(B/2)+I-1$ square units, where B is number of lattice points on the boundary, I is number of lattice points in the interior of the polygon. Prove this theorem!

Do you like this challenge?

Nalin Pithwa.

### Safeguarding children from digital addiction — Kiran Bajaj, Principal, Greenlawns School.

Reproduced from The DNA newspaper, print edition, Mumbai, date April 17 2017, Monday; authored by Kiran Bajaj:

(From the Principal’s Desk) (The writer is Principal of Greenlawns High School, Mumbai.)

Generally, the word addiction brings to mind habits such as smoking, drinking or gambling, but in today’s digital age, addiction is more connected to technology. A child  could get addicted to iPads, television, or any kind of “screen”. Children develop an uncontrolled habit of indulging in certain activities even when they are warned that doing those things are not good for them.

Children are slowly becoming digital addicts. These days, some kids don’t just play with electronic toys but make it a part of their lives. Children carry their smartphones every where including the washroom as they feel disconnected from their social network without their phones. Today, young addicts include three year olds who scream when they can’t have their tablets to play on, or secondary school children who can’t quit “WhatsApping” or posting messages on Facebook, and others who compulsively play online games.

Toddlers are becoming couch potatoes almost as soon as they have the pram. No two children are alike and different children perceive the television medium differently. Studies have shown that watching TV at an early age does form a habit and that it has potentially damaging effects on their health. Some children watch TV while eating dinner, while doing homework, doing chores, etc. To prevent children from becoming couch potatoes, parents can encourage children to play board games, outdoor activities and socialize with friends.  Try to make the alternatives really fun at first, to help your child transition into watching less TV.

Families can help prevent addiction when there is a strong bond between children and parents, and a lot of parental involvement in the child’s life and discipline. Bringing in new games, books, and other activities will give children something better to do with their time while at home. For younger ones, simple sticker books and colouring books will keep them entertained for hours. Art and craft activities encourage children to use their imagination, as well as learning toys, puppet theaters, anything that gets your child thinking rather than just watching.

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Regards,

Nalin Pithwa.

PS: Thanks to DNA and Kiran Bajaj !

### Genius of Srinivasa Ramanujan

1. In December 1914, Ramanujan was asked by his friend P.C. Mahalanobis to solve a puzzle that appeared in Strand magazine as “Puzzles at a Village Inn”. The puzzle stated that n houses on one side of the street are numbered sequentially starting from 1. The sum of the house numbers on the left of a particular house having the number m, equals that of the houses on the right of this particular house. It is given that n lies between 50 and 500 and one has to determine the values of m and n. Ramanujan immediately rattled out a continued fraction generating all possible values of m without having any restriction on the values of n. List the first five values of m and n.
2. Ramanujan had posed the following problem in a journal: $\sqrt{1+2\sqrt{1+3\sqrt{\ldots}}}=x$, find x. Without receiving an answer from the readers, after three months he gave answer as 3. This he could say because he had an earlier general result stating $1+x=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+\ldots}}}}$ is true for all x. Prove this result, then $x=2$ will give the answer to Ramanujan’s problem.

Try try until you succeed!!

Nalin Pithwa.

### Men, monkey and coconuts problem

Five men and a monkey collected a heap of coconuts in a desert island. The men went to sleep in the night. One of them got up in the middle of the night and divided the heap into five equal parts and found an extra coconut, which he gave to the monkey. He hid his share and made one heap combining the rest of the coconuts. Then, the second person got up and divided this new heap into five equal parts and again found an extra coconut, which he gave to the monkey. He also hid his share and made a new heap of the remaining coconuts. This process continued until the last person did exactly the same thing as the others. What must be the minimum number of coconuts in the original heap if

(a) the next morning when they divided the last remaining heap into five equal parts, there was no coconut left for the monkey.

(b) the next morning they divided the last remaining heap into five equal parts, they again found an extra one, which they gave to the monkey.

Cheers,

Nalin Pithwa.

PS: One good way to increase concentration, motivation, intellectual stamina for solving such demanding puzzles or math problems for the IITJEE or RMO/INMO or even Mensa challenges is to give yourself a “reward” after you actually solve that problem. So, you may go ahead and have chilled coconut juice after you solve this puzzle. Anyway, these are the hot summer days in India.

### Some problems on conics (parabola, ellipse, hyperbola) : IITJEE Mains — Basics 2

I am solving some “nice” problems below:

Problem 1:

Show that the locus of a point that divides a chord of slope 2 of the parabola $y^{2}=4x$ internally in the ratio $1:2$ is a parabola. Find the vertex of this parabola.

Solution 1:

Let $P(t_{1}^{2}, 2t_{1})$ and $Q(t_{2}^{2}, 2t_{2})$ be the extremities of the chord with slope 2.

Hence, $\frac{2t_{1}-2t_{2}}{t_{1}^{2}-t_{2}^{2}}=2 \Longrightarrow t_{1}+t_{2}=1$

Let $(t_{1},t_{2})$ be co-ordinates of the point which divides PQ in the ratio $1:2$. Then,

$h=\frac{2t_{1}^{2}+t_{2}^{2}}{3}$ and $k=\frac{4t_{1}+2t_{2}}{3}$

$\Longrightarrow 3h = 2t_{1}^{2}+(1-t_{1})^{2}$ and $3k = 4t_{1}+2(1-t_{1})$

$\Longrightarrow 3h = 3t_{1}^{2} - 2t_{1} + 1$ and $3k = 2t_{1}+2$

$\Longrightarrow 3h = 3(\frac{3k-2}{2})^{2} -2(\frac{4k-2}{2})+1$

$\Longrightarrow 12h = 3(9k^{2}-12k+4)-12k+8+4$

$4h = 9k^{2}-16k+8$

Hence, the locus of $(h,k)$ is $9y^{2}-16y-4x+8=0$

$\Longrightarrow (3y-\frac{8}{3})^{2}=4x-8+\frac{64}{9}$

$\Longrightarrow (y-\frac{8}{9})^{2}=\frac{4}{9}(x-\frac{2}{9})$, which is a parabola with vertex $(\frac{2}{9}, \frac{8}{9})$.

Problem 2:

If $P_{1}P_{2}$ and $P_{3}P_{4}$ are two focal chords of the parabola $y^{2}=4ax$,  then show that the chords $P_{1}P_{3}$ and $P_{2}P_{4}$ intersect on the directrix of the parabola.

Solution 2:

Let the co-ordinates of $P_{i}$ be $(at_{i}^{2},2at_{i})$ for $i=1,2,3,4$.

Since $P_{1}P_{2}$ is a focal chord, $t_{1}t_{2}=-1$ —– call this Equation I.

Similarly, $t_{3}t_{4}=-1$ —- call this Equation II.

Equation of $P_{1}P_{3}$ is $y(t_{1}+t_{3})=2(x+at_{1}t_{3})$ —- call this Equation III.

and that of $P_{2}P_{4}$ is $y(t_{2}+t_{4})=2(x+\alpha t_{2}t_{4})$ —- call this Equation IV.

Using I and II, IV reduces to $y(-\frac{1}{t_{1}}-\frac{1}{t_{2}})=2(x+\frac{a}{t_{1}}t_{3})$

that is, $-y(t_{1}+t_{3})=2(xt_{1}t_{3}+a)$ — call this Equation V.

Adding III and V we get:

$0=2[x+at_{1}t_{3}+xt_{1}t_{3}+a]$, which in turn implies, $(x+a)(1+t_{1}t_{3})=0$, which in turn implies, that $x=-a$. Hence, III and V intersect on the directrix $x+a=0$.

More later,

Nalin Pithwa.

### Some words of advice for students and parents and teachers – from Sudha Murthy

Life isn’t just about getting a degree: Sudha Murthy

(Reproduced from today’s, April 3, 2017 newspaper, The DNA, Mumbai (print)edition just to share with my readers/students. There is no other purpose in reproducing this interview here.)

(Comment: Though not directly related to the express purpose of this blog, it does matter a lot. Read and think about it!! )

Educationist, celebrated authorand Infosys chairperson Sudha Murthy speaks to Laveena Francis about India’s education system:

Question:

What are the lessons you learnt as a teacher?

The first thing is that you should make children love you first, and then they will love your subject. If you are not smiling, if you are extremely punishing, or if you never bring confidence in them, then they won’t like your subject as well. Secondly, you need to have a lot of patience in class. Not all students have the same pace while studying, somebody may be a slow learner. You should always remember that you are paid for the last candidate and not the first candidate. These are the two important learnings that I found as a teacher.

Question:

If you were to be part of a committee to design a school syllabus, what are the elements you would include, and exclude?

I would like to include History in a proper way. It’s not just about the wars. It’s also about how people lived. We have to tell the facts without hiding anything. We have to accept whatever has happened in the past. The quality of acceptance that happened in the past should be known to children. I would also want to include yoga, because yoga is a good exercise for both physical and mental health. And, of course, a sport. Any sport which the child likes. Playing at least one sport should be made compulsory.

Stitching, drawing and yoga are a must, as they help in the later part of life.

Usually, stitching is associated with females. But, it’s not like that. Boys also stitch very well. Manish Malhotra and Rohit Bahl are one of the best fashion designers. Stitching is nothing to do with men or women. It’s a kind of creativity according to me. And, excludes the excessive use of computers.

Question:

Do you think community service should be made compulsory in the education system so that children learn it from the very beginning?

Community service shouldn’t be a lip service. Children should go to a village and see the lifestyle of villages. It should not  happen that they are sitting at home and talking about helping. They should go out and see how others suffer, and then they will learn sensitivity from the have-nots.

Question:

But, how do we instill sensitivity towards humanity and the environment among students?

One needs to explain it. It’s the duty of both teachers and parents. I used to exercise with my students. Apart from teaching Computer Science, once in a while I used to take them to orphanage and tell them to see how people live and then count their blessings.

Question:

The race for marks has stripped the joy of learning. Students are result-oriented and barely focused on the process.

It would be very wrong for me to comment because I really don’t have any responsibility of children now. So, I can afford to sit and talk.

As a mother, I did not bother about marks. I would tell them, “Children, look, do not worry much about the marks, learn to love the subjects and it will remain with you. Of course, it’s a mad rat race now. It takes a toll on the students, parents and teachers. Life is not about getting a degree. There are so many ways one can live, but we never bother about that. We generally want our children to be either engineers or doctors. And, now, computer engineers. Not all are meant for that. I feel good education and the art of learning new things is more important. I think children should have the love for knowledge more than the love for marks. They should have a love and thirst for knowledge.

Question:

What kind of ambiance would you suggest parents create in their homes for their children to learn and grow?

Parents should sit with their children. Studies should be made interesting. Explain everything to them in a simpler way. Buy less for children. Don’t buy them things before they even know they need it. Children should look forward to buying new things. Parents have to under-go 10-12 years with them. It’s like going to school. And, don’t compare. Don’t say, look at the neighbour’s son, he does so well, or look at your own cousins. Don’t do that. A child is like a bud, every flower has its own beauty, so please allow them to bloom.

Question:

The importance of pursuing offbeat courses such as art, music, animal training etc.

They should pursue something only if they like it. I see so many RJ’s, they are all engineers, they are passionate about music, but parents say, “first you become an engineer, start earning, and then we’ll see. So, if you have a passion, follow the passion.

Question:

Normally, I don’t give any advice because it is not my nature. I always say make life meaningful. Make life colourful. Make it enjoyable. Don’t make it like a curse. Make your work enjoyable. You should look forward to going to school or college.

Teachers should have patience and should be smiling. Don’t punish or use harsh words, after all, they are children. And, don’t be judgmental.

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By the way, for teachers, I would recommend the book “Secrets of Good Teaching” by Viney Kirpal. I love of one of the gems that Dr. Vikram Gadre, Prof. Dept. of Electrical Engineering, IIT-Bombay explains in an essay in that book —- “teaching is like gardening”.

More later,

Nalin Pithwa.

### Conic Sections (Parabola, Ellipse, Hyperbola): some basic theorems/questions: IITJEE Mains Maths

Problem 1:

(A) Proposition 1: Show  that the tangents at the extremities of a focal chord of a parabola intersect at right angles on the directrix.

Proof 1:

Let $t_{1}$ and $t_{2}$ be the extremities of a focal chord of the parabola $y^{2}=4ax$. Then, it can be shown that $t_{1}t_{2}=-1$. (Try this part on your own and let me know; if you can’t produce the proof, I will send it to you.) The equations of the tangents at $t_{1}$ and $t_{2}$ are $t_{1}y=x+at_{1}^{2}$ and $t_{2}y=x+at_{2}^{2}$. The product of the slopes is $\frac{1}{t_{1}}-\frac{1}{t_{2}}=-1$

Therefore, the tangents are at right angles. Also, the point of intersection of  these tangents is $x=at_{1}t_{2}$, $y=a(t_{1}+t_{2})$, that is, $x=-a$, $y=a(t_{1}+t_{2})$, which clearly lies on the directrix $x=-a$.

(B) Proposition 2: The tangent at any point of a parabola bisects the angle between the focal distance of the point and the perpendicular on the directrix from that point. Homework !

(C) Proposition 3: The portion of a tangent to a parabola cut off between the directrix and the curve, subtends a right angle at the focus. Homework!

Problem 2:

Theorem:

Show that in general, three normals can be drawn to a given parabola from a given point, one of which is always real. Also, show that the sum of the ordinates of the feet of these co-normal points is zero.

Proof:

Let $y^{2}=4ax$ be the given parabola. The equation of a normal to this parabola at $(am^{2},-2am)$ is $y=mx-2am-am^{2}$. If it passes through a given point $(\alpha, \beta)$ then

$\beta = m\alpha -2am-am^{3}$, or $am^{3}+(2a-\alpha)m + \beta = 0$…call this Equation (I)

which, being cubic in m, gives three values of m, say $m_{1}$. $m_{2}$, and $m_{3}$, and hence, three points on the parabola, the normals at which pass through $(\alpha, \beta)$. Since the complex roots of the equation with real coefficients occur in pairs and the degree of the above equation is odd, at least one of the roots is real so there is at least one real normal to the parabola passing through the given point $(\alpha, \beta)$.

From (I), we have

$m_{1}+m_{2}+m_{3}=0 \Longrightarrow -2am_{1}-2am_{2}-2am_{3}=0 \Longrightarrow y_{1}+y_{2}+y_{3}=0$, where $y_{i}=-2am_{i}$ for $i=1, 2, 3$. Hence, the result.

Problem 3:

If the tangents and normals at the extremities of a focal chord of the parabola $y^{2}=4ax$ intersect at $(x_{1},y_{1})$ and $(x_{2},y_{2})$ respectively, then show that $y_{1}=y_{2}$.

Let $P(at_{1}^{2}, 2at_{1})$ and $Q(at_{2}^{2},2at_{2})$ be the extremities of a focal chord of the parabola $y^{2}=4ax$, then

$t_{1}t_{2}=-1$….call this Relation (I).

Next, equations of the tangents at P and Q are respectively, $t_{1}y=x+at_{1}^{2}$ and $t_{2}y=x+at_{2}^{2}$

Solving these equations, we get

$(t_{1}-t_{2})y=a(t_{1}^{2}-t_{2}^{2}) \Longrightarrow y=a(t_{1}+t_{2})$ and

$x=t_{1}a(t_{1}+t_{2})-at_{1}^{2}=at_{1}t_{2}=-a$

So that $x_{1}=-a$ and $y_{1}=a(t_{1}+t_{2})$….call this Relation (II).

Now, equations of the normals at P and Q are respectively

$y = -t_{1}x+2at_{1}+at_{1}^{3}$ and $y=-t_{2}x+2at_{2}+at_{2}^{3}$

Solving these equations we get

$-(t_{1}-t_{2})x+2a(t_{1}-t_{2})+a(t_{1}^{3}-t_{2}^{3})=0$

$\Longrightarrow x =2a +a(t_{1}^{2}+t_{2}^{2}+t_{1}t_{2}) = a(t_{1}^{2}+t_{2}^{2}+1)$ (using relation I)

and $y = -t_{1}a(t_{1}^{2}+t_{2}^{2}+1) + 2at_{1}+at_{1}^{3} = -at_{1}t_{2}^{2}+at_{1}=a(t_{1}+t_{2})$ (again by using relationship I),

So that $x_{2}=a(t_{1}^{2}+t_{2}^{2}+1)$ and $y_{2}=a(t_{1}+t_{2})$ so that we get $y_{1}=y_{2}$.

Note:

Results or relations I and II should be memorized as they are frequently used in the theory and applications.

More later,

Nalin Pithwa.

### Alice in Wonderland and probability stuff !!

Lewis Carroll of Alice’s Adventures in the Wonderland was a mathematician in Cambridge. He had posed the following problem in one of his books:

A box contains a handkerchief, which is rather white or black. You put a white handkerchief in this box and mix up the contents. Then you draw a handkerchief, which turns out to  be white. Now, if you draw the remaining handkerchief, what is the probability of this one being white?

How likely is that you would enjoy such questions ?! 🙂

Nalin Pithwa.

### Practice Quiz on Conic Sections (Parabola, Ellipse, Hyperbola): IITJEE Mains — basics 1

Multiple Choice Questions:

Problem 1:

A line bisecting the ordinate PN of a point P $(at^{2}, 2at)$, $t>0$, on the parabola $y^{2}=4ax$ is drawn parallel to the axis to meet the curve at Q. If NQ meets the tangent at the vertex at the point T, then the coordinates of T are

(a) $(0, \frac{4}{3}at)$ (b) Slatex (0,2at)\$ (c) $(\frac{1}{4}at^{2},at)$ (d) $(0,at)$

Problem 2:

If P, Q, R are three points on a parabola $y^{2}=4ax$ whose ordinates are in geometrical progression, then the tangents at P and R meet on

(a) the line through Q parallel to x-axis

(b) the line through Q parallel to y-axis

(c) the line through Q to the vertex

(d) the line through Q to the focus.

Problem 3:

The locus of the midpoint of the line segment joining the focus to a moving point on the parabola $y^{2}=4ax$ is another parabola with directrix:

(a) $x=-a$ (b) $x=-a/2$ (c) $x=0$ (d) $x=a/2$

Problem 4:

Equation of the locus of the pole with respect to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$, of any tangent line to the auxiliary circle is the curve $\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}}=\lambda^{2}$ where

(a) $\lambda^{2}=a^{2}$ (b) $\lambda^{2}=\frac{1}{a^{2}}$ (c) $\lambda^{2}=b^{2}$ (d) $\lambda^{2}=\frac{1}{b^{2}}$

Problem 5:

The locus of the points of the intersection of the tangents at the extremities of the chords of the ellipse $x^{2}+2y^{2}=6$, which touch the ellipse $x^{2}+4y^{2}=4$ is

(a) $x^{2}+y^{2}=4$ (b) $x^{2}+y^{2}=6$ (c) $x^{2}+y^{2}=9$ (d) none of these.

Problem 6:

If an ellipse slides between two perpendicular straight lines, then the locus of its centre is

(a) a parabola (b) an ellipse (c) a hyperbola (d) a circle

Problem 7:

Let $P(a\sec{\theta}, b\tan{\theta})$ and $Q(a\sec{\phi}, b\tan{\phi})$ where $\theta + \phi=\pi/2$, be two points on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}}=1$. If $(h,k)$ is the point of intersection of normals at P and Q, then k is equal to

(a) $\frac{a^{2}+b^{2}}{a}$ (b) $-(\frac{a^{2}+b^{2}}{a})$ (c) $\frac{a^{2}+b^{2}}{b}$ (d) $-(\frac{a^{2}+b^{2}}{b})$

Problem 8:

A straight line touches the rectangular hyperbola $9x^{2}-9y^{2}=8$, and the parabola $y^{2}=32x$. An equation of the line is

(a) $9x+3y-8=0$ (b) $9x-3y+8=0$ (c) $9x+3y+8=0$ (d) $9x-3y-8=0$

There could be multiple answers to this question.

Problem 9:

Two parabolas C and D intersect at the two different points, where C is $y=x^{2}-3$ and D is $y=kx^{2}$. The intersection at which the  x-value is positive is designated point A, and $x=a$ at this intersection. The tangent line l at A to the curve D intersects curve C at point B, other than A. If x-value of point B is 1(one), then what is a equal to?

Problem 10:

The triangle formed by the tangent to the parabola $y=x^{2}$ at the point whose abscissa is $x_{0}(x_{0} \in [1,2])$, the y-axis and the straight line $y=a_{0}^{2}$ has the greatest area if $x_{0}= ?$. Fill in the question mark!

More later,

Nalin Pithwa.

PS: I think let’s continue this focus on co-ordinate geometry of IITJEE Mains Maths for some more time.

### Gruffs

The reporter for the local Gazette was at the national dog show yesterday. Unfortunately, he also had to report on quite a few other events on the same day. He made some notes when he was at the dog show which are shown below, but the editor of the Gazette has asked for a list of the 26 finalists and the positions in which they finished overall. Using the reporter’s notes below, see if you can construct the list for the editor, as the reporter didn’t have time to make a note of the overall positions.

The Afghan Hound finished before the Alsatian, the Poodle and the Beagle. The Beagle finished before the Bull Mastiff and the Labrador. The Labrador finished before the Poodle and the Pug. The Pug finished before the Alsatian, the Dobermann Pinscher, the St. Bernard, the Sheepdog and the Griffon. The Griffon finished before the Sheepdog. The Sheepdog finished after the Poodle. The Poodle finished before the St. Bernard, the Collie, the Pug, the Griffon, the Alsatian and the Dobermann Pinscher. The Dobermann Pinscher finished after the Alsatian. The Alsatian finished after the Chow. The Chow finished before the Afghan Hound, the Bulldog, the Chihuahua, the Poodle, the Beagle and the Dachshund. The Dachshund finished before the Spaniel. The Spaniel finished before the Foxhound.

The Foxhound finished before the Labrador. The Labrador finished after the Whippet. The Whippet finished before the Great Dane. The Great Dane finished after the Bull Terrier and before the Chow. The Chow finished before the Bull Mastiff. The Bull Mastiff finished before the Foxhound. The Foxhound finished after the Yorkshire Terrier. The Yorkshire Terrier finished before the Greyhound. The Greyhound finished before the Dachshund. The Dachshund finished after the Kind Charles Spaniel. The King Charles Spaniel finished before the Bull Mastiff, the Greyhound, the Pug, the Chihuahua and the Afghan Hound. The Afghan Hound finished after the Dalmatian. The Dalmatian finished before the Pug, the Labrador, the Poodle and the Collie.

The Collie finished before the Pug. The Pug finished after the Retriever. The Retriever finished before the Bull Terrier, the Chow, the Yorkshire Terrier and the Whippet. The Whippet finished before the Chow, the Spaniel, the Dalmatian and the Yorkshire Terrier. The Yorkshire Terrier finished after the Bulldog. The Bulldog finished before the Chihuahua and the Dalmatian. The Dalmatian finished after the Great Dane. The Great Dane finished before the Yorkshire Terrier. The Yorkshire Terrier finished before the Collie, the King Charles Spaniel, the Spaniel and the Dalmatian. The Dalmatian finished after the Spaniel. The Griffon finished before the Alsatian. The Alsatian finished after the Sheepdog. The Griffon finished after the St. Bernard. The Greyhound finished before the Chihuahua. The Chihuahua finished before the Dachshund. The Bull Terrier finished before the Whippet and the Yorkshire Terrier. The Afghan Hound finished before the Pug and the Foxhound.

So, if u try this, it might ‘hound’ u !!! Good puzzles ain’t easy! But, such puzzles help u develop the habit and power of sustained thinking on a problem for long hours.

Nalin Pithwa.