Stephen Hawking in his own words: excerpts: My Brief History

I. Another early memory was getting my first train set. Toys were not manufactured during the war, at least not for the home market. But, I had a passionate interest in model trains. My father tried making me a wooden train, but that didn’t satisfy me, as I wanted something that moved on its own. So, he got a secondhand clockwork train, repaired it with a soldering iron, and gave it to me for Christmas when I was nearly three. That train didn’t work very well. But my father went to America just after the war, and when he came back on the Queen Mary he bought my mother some nylons, which were not obtainable in Britain at that time. He bought my sister Mary a doll that closed its eyes when you laid it down. And he brought me an American train, complete with a cowcatcher and a figure-eight track. I can still remember my excitement as I opened the box.

Clockwork trains, which you had to wind up, were all very well, but I really wanted were electric trains. I used to spend hours watching a model railway club layout in Crouch End, near Highgate. I dreamed about electric trains. Finally, when both my parents were away somewhere, I took the opportunity to draw out of the Post Office bank all of the very modest amount of money that people had given me on special vacations, such as my christening. I used the money to buy an electric train set, but frustratingly enough, it didn’t work very well either. I should have taken the set back and demanded that the shop or the manufacturer replace it, but in those days the attitude was that it was a privilege to buy something and it was just your bad luck, if it turned out to be faulty. So, I paid for the electric motor of the engine to be serviced, but it never worked very well, even then.

Later on, in my teens, I built model aeroplanes and boats. I was never very good with my hands, but I did this with my school friend John McClenahan, who was much better and whose father had a workshop in their house. My aim was always to build working models that I could control. I didn’t care what they looked like. I think it was the same drive that led me to invent a series of very complicated games with another school friend, Roger Ferneyhough. There was a manufacturing game, complete with factories in which units of different colours were made, roads and railways on which they were carried, and a stock market. There was a war game, played on a board of 4000 squares, and even a feudal game, in which each player was a whole dynasty, with a family tree. I think these games, as well as the trains, boats, and aeroplanes, came from an urge to know how systems worked and how to control them. Since I began my PhD, this need has been met by research into cosmology. If you understand how the universe operates, you control it, in a way.

II. When I was thirteen, my father wanted me to try for Westminister School, one of Britain’s main public schools (what in the United States are called private schools). At that time, there was a sharp division in education along class lines, and my father felt that the social graces such a school would give me would be an advantage in life. My father believed that his own lack of poise and connections had led to him being passed over in his career in favour of people of less ability. He had a bit of a chip on his shoulder because he felt that other people who were not as good but who had the right background and connections had got ahead of him. He used to warn me against such people.

Because my parents were not well-off, I would have to win a scholarship in order to attend Westminister. I was ill at the time of the scholarship examination, however, and did not take it. Instead, I remained at St. Albans School, where I got an education that was as good as, if not better than, the one I would have had at Westminister. I have never found that my lack of social graces has been a hindrance. But I think physics is a bit different from medicine. In physics it doesn’t matter what school you went to or whom you are related. It matters what you do.

I was never more than about halfway up the class (it was a very bright class.) My classwork was untidy, and my handwriting was the despair of my teachers. But, my classmates gave me the nickname Einstein, so presumably they saw signs of something better. When I was twelve, one of my friends bet another friend a bag of sweets that I would never amount to anything. I don’t know if this bet was ever settled, and if so, which way it was decided.

I had six or seven close friends, most of whom I’m still in touch with. We used to have long discussions and arguments about everything from radio-controlled models to religion and from parapsychology to physics. One of the things we talked about was the origin of the universe and whether it had required a God to create it and set it going. I had heard that light from distant galaxies was shifted towards the red end of the spectrum and that this was supposed to indicate that the universe was expanding. (A shift towards the blue would have meant it was contracting.) But, I was sure there must be some other reason for the red shift. An essentially unchanging and everlasting universe seemed so much more natural. Maybe light just got tired, and more red, on its way to us, I speculated. It was only after about two years of PhD research that I realized that I had been wrong.

I was always very interested in how things operated, and I used to take them apart to see how they worked, but I was not so good at putting them back together again. My practical abilities never matched up to my theoretical inquiries. My father encouraged my interest in science, and he even coached me in mathematics until I got to a stage beyond his knowledge. With this background and my father’s job, I took it as natural that I would go into scientific research.

When I came to the last two years of school, I wanted to specialize in mathematics and physics. There was an inspiring maths teacher, Mr. Tanta, and the school had also just built a new maths room, which the maths set had as their classroom. But my father was very much against it because he thought there wouldn’t be any jobs for mathematicians except as teachers. He would really have liked me to do medicine, but I showed no interest in biology, which seemed to me to be too descriptive and not sufficiently fundamental. It also had a rather low status at school. The brightest boys did mathematics and physics; the less bright did biology.

My father knew I wouldn’t do biology, but he made me do chemistry and only a small amount of mathematics. He felt this would keep my scientific options open. I am now professor of mathematics, but I have not had any formal instruction in mathematics since I left St. Albans School at the age of seventeen. I have had to pick up what I know as I went along. I used to supervise undergraduates at Cambridge and kept one week ahead of them in the course.

Physics was always the most boring subject at school because it was so easy and obvious. Chemistry was much more fun because unexpected things, such as explosions kept happening. But physics and astronomy offered the hope of understanding where we came from and why we are here. I wanted to fathom the depths of the universe. Maybe I have succeeded to a small extent, but there’s still plenty I want to know.

III. My early work showed that classical general relativity broke down at singularities in the Big Bang and black holes. My later work has shown how quantum theory can predict what happens at the beginning and end of time. It has been a glorious time to be alive and doing research in theoretical physics. I’m happy if I have added something to our understanding of the universe.

A humble tribute to Professor Hawking …to understand him from a layman’s viewpoint…by Nalin Pithwa.

Reference: My Brief History by Stephen Hawking, Bantam Press.

Amazon India link:

An immortal genius Professor Stephen Hawking is no more in this world

Some tributes that I can only humbly share …

 

 

 

 

Happy Pie with Pi…

via Everything you wanted to know about pi but were afraid to ask

Co-ordinate Geometry : IITJEE Mains practice: some random problems again

Problem 1:

The line Ax+By+C=0 cuts the circle x^{2}+y^{2}+ax+by+c=0 at P and Q. The line A^{'}x+B^{'}y+C^{'}=0 cuts the circle x^{2}+y^{2}+a^{'}x+b^{'}y+c^{'}=0 at R and S. If P, Q, R and S are concyclic, prove that

\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0.

Solution I;

An equation of a circle through P and Q is x^{2}+y^{2}+ax+by+c +\lambda (Ax+By+C)=0…call this equation I.

And, an equation of a circle through R and S is x^{2}+y^{2}+a^{'}x+b^{'}y + c^{'}+\mu (A^{'}x+B^{'}y+C^{'})=0…call this equation II.

If P, Q, R and S are concyclic, then I and II represent the same circle for same values of \lambda and \mu.

\Longrightarrow a+ \lambda A=a^{'}+\mu A^{'} or a-a^{'} + \lambda A - \mu A^{'}=0

so also,

b + \lambda B = b^{'} + \mu B^{'} or b-b^{'}+\lambda B - \mu B^{'}=0

c + \lambda C = c^{'} + \mu C^{'} or c-c^{'} + \lambda C - \mu C^{'}=0.

Eliminating \lambda and \mu, we get the following:

\left | \begin{array}{ccc} a-a^{'} & A & -A^{'}\\ b-b^{'} & B & -B^{'}\\ c-c^{'} & C & -C^{'} \end{array} \right |=0, that is,

\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0

Problem II:

A straight line is such that the algebraic sum of the perpendiculars falling upon it from any number of fixed points is zero. Show that it always passes through a fixed point.

Solution II:

Let (x_{i},y_{i}) where i=1,2,3,\ldots be n fixed points. Let ax+by+c=0 be the given line. Thus, as per given hypthesis, we have

\sum_{i=1}^{n}\frac{ax{i}+by_{i}+c}{\sqrt{(a^{2}+b^{2})}}=0 \Longrightarrow a\sum_{i=1}^{n}x_{i}+b\sum_{i=1}^{n}y_{i}+nc=0 \Longrightarrow a\overline{x}+b\overline{y}+c=0 where \overline{x}=\frac{1}{n}\sum_{i=1}^{n}x_{i} and \overline{y}=\frac{}{}\sum_{i=1}^{n}y_{i}

which shows that the given line passes through the fixed point (\frac{1}{n}\sum_{i=1}^{n}x_{i}, \frac{1}{n}\sum_{i=1}^{n}y_{i}).

Problem III:

The straight lines L \equiv ax+by+c=0 and L_{1} \equiv a_{1}x+b_{1}y+c_{1}=0 are intersecting. Find the straight line L_{2} such that L is the bisector of the angle between L_{1} and L_{2}.

Solution III:

Let the equation of the line L_{2} be L_{1}+ \lambda L=0 \Longrightarrow (a_{1}+\lambda a)x+(b_{1}+\lambda b)y+\lambda c=0 where the slopes of L_{2}, L, L_{1} are respectively

-\frac{a_{1}+\lambda a}{b_{1}+\lambda b}, \frac{-a}{b}, -\frac{a_{1}}{b_{1}}.

Since L is the bisector of the angle between L_{2} and L_{1} we have

\frac{-(\frac{a_{1}+\lambda a}{b_{1}+\lambda b})+\frac{a}{b}}{1+\frac{a(a_{1}+\lambda a)}{b(b_{1}+\lambda b)}}=\frac{-\frac{a}{b}+\frac{a_{1}}{b_{1}}}{1+\frac{aa_{1}}{bb_{1}}}

\Longrightarrow \frac{-b(a_{1}+\lambda a)+a(a_{1}+\lambda b)}{b(b_{1}+\lambda b)+a(a_{1}+\lambda a)}=-\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}

\Longrightarrow \frac{ab_{1}-a_{1}b}{\lambda (a^{2}+b^{2})+aa_{1}+bb_{1}} = -\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}

\Longrightarrow \lambda = - \frac{2(aa_{1}+bb_{1})}{a^{2}+b^{2}}

Hence, the equation of the required line L_{1} is (a^{2}+b^{2})(a_{1}x+b_{1}y+c_{1})=2(aa_{1}+bb_{1})(ax+by+c).

Problem IV:

If a, b are real numbers and c>0, find the locus represented by |ay-bx|=c\sqrt{(x-a)^{2}+(y-b)^{2}}.

PS: Please draw a right angled triangle PMA, with right angle at M, and P being (x,y) and A being (a,b).

Solution IV:

Let x=a+r\cos {\theta} and y=b+r\sin {\theta}, then the given equation becomes a\sin {\theta}-b\cos {\theta}=c.

\Longrightarrow r\sin{(\theta-\alpha)}=c where r=\sqrt{a^{2}+b^{2}} and \tan {(\alpha)}=\frac{b}{a} which is the slope of ay-bx, which in turn implies \frac{c}{r}=\sin (\theta -\alpha) \leq 1

\Longrightarrow c \leq r, or c \leq \sqrt{a^{2}+b^{2}}. The given equation now becomes

\frac{|ay-bx|}{\sqrt{a^{2}+b^{2}}}=\frac{c}{\sqrt{a^{2}+b^{2}}}\sqrt{(x-a)^{2}+(y-b)^{2}}….call this as relation I.

If M is the foot of the perpendicular from a point P(x,y) on the line ay-bx=0 and A is the point (a,b) which clearly lies on this line, then from relation I, we have

\frac{PM}{PA}=\frac{c}{\sqrt{a^{2}+b^{2}}}=\sin {(\theta - \alpha)}. Hence, the locus of P is a straight line through the point (a,b) inclined at an angle \arcsin {\frac{c}{\sqrt{a^{2}+b^{2}}}} with the line ay-bx=0.

Problem V:

Find the co-ordinates of the orthocentre of the triangle formed by the lines y=0 and (1+t)x-ty+t(1+t)=0 and (1+u)x-uy+u(1+u)=0, where t \neq u, and show that for all values of t and u, the orthocentre lies on the line x+y=0.

Solution V:

Let the equation of the side BC be y=0. Then, the coordinates of B and C are (-t,0) and (-u,0), respectively, where (1+t)x-ty+t(1+t)=0 and (1+u)x-uy+u(1+u)=0 are equations of AB and AC, respectively.

PS: Please draw the diagram on your own for a better understanding of the solution presented.

Now, equation of BE is y={\frac{-u}{1+u}}(x+t)…let us call this equaiton I.

And, equation of CF is y=\frac{-t}{1+t}(x+u)…let us call this equation II.

Solving I and II, we get the following:

x(\frac{u}{1+u}-\frac{t}{1+t})=\frac{tu}{1+t}-\frac{tu}{1+u}, which in turn implies that

x=tu and y=-tu, so that the orthocentre is the point (tu,-tu) which lies on the line x+y=0.

Cheers,

Nalin Pithwa

Higher paying job than doctor/lawyer: some Singapore data

Higher paying job than Doctor / Lawyer

Some random problems/solutions in Coordinate Geometry II: IITJEE mathematics training

Question I:

Find the equation of the tangent to the circle x^{2}-y^{2}-4x-8y+16=0 at the point (2+\sqrt{3},3). If the circle rolls up along this tangent by 2 units, find its equation in the new position.

Solution I:

The centre C_{1} of the given circle is (2,4) and its radius is 2. Equation of the tangent at A(2+\sqrt{3},3) to the circle is

x(2+\sqrt{3})+3y-2(x+2+\sqrt{3})-4(y+3)+16=0

or \sqrt{3}x-y-2\sqrt{3}=0.

The slope of this line is \sqrt{3} showing that it makes an angle of 60 degrees with the x-axis. After the circle rolls up along the tangent at A through a distance 2 units, its centre moves from C_{1} to C_{2}. We now find the co-ordinates of C_{2}. Since C_{1}C_{2} is parallel to the tangent at A and it passes through C_{1}(2,4), its equation is \frac{x-2}{\cos {\theta}} = \frac{y-4}{\sin {\theta}}, where \theta=60 \deg; C_{2} being at a distance 2 units on this line from C_{1}; its co-ordinates are

(2\cos {\theta}+2, 2\sin{\theta}+4), that is, (3, 4+\sqrt{3}).

Hence, the equation of the circle in the new position is

(x-3)^{2}+(y-(4+\sqrt{3})^{2})=2^{2}, which in turn implies that

x^{2}+y^{2}-6x-2(4+\sqrt{3})y+8(3+\sqrt{3})=0.

Question 2:

A triangle has two of its sides along the axes, its third side touches the circle x^{2}+y^{2}-2ax-2ay+a^{2}=0. Prove that the locus of the circumcentre of the triangle is

a^{2}-2a(x+y)+2xy=0.

Solution 2:

The given circle has its centre at C(a,a) and its radius is a so that it touches both the axes along which lie the two sides of the triangle. Let the third side be \frac{x}{p} + \frac{y}{p}=1.

So that A is (p,a) and B is (a,q) and the line AB touches the given circle. Since \angle AOB is a right angle, AB is diameter of the circumcentre of the triangle AOB. So, the circumcentre P(h,k) of the triangle AOB is the mid-point of AB,

that is, 2h=p, 2k=q.

Now, the equation of AB is \frac{x}{p} + \frac{y}{q}=1, which touches the given circle,

\frac{a(p+q)-pq}{\sqrt{p^{2}+q^{2}}}=a

\Longrightarrow a^{2}(p+q)^{2}+p^{2}-2apq(p+q)=a^{2}(p^{2}+q^{2})

\Longrightarrow 2a^{2}-2a(p+q)+pq=0

2a^{2}-2a(2h+2k)+2h-2k=0.

Hence, the locus of P(h,k) is a^{2}-2a(x+y)+2xy=0.

Question 3:

A circle of radius 2 units rolls on the outerside of the circle x^{2}+y^{2}+4x=0, touching it externally. Find the locus of the centre of this outside circle. Also, find the equations of the common tangents of these two circles when the line joining the centres of the two circles make an angle of 60 degrees with x-axis.

Solution 3:

The centre C of the given circle is (-2,0) and its radius is 2. Let P(h,k) be the centre of the outer circle touching the given circle externally then CP=2+2=4, which in turn implies,

(h+2)^{2}+k^{2}=4^{2}

So, the locus of P is (x+2)^{2}+y^{2}=16, or x^{2}+y^{2}+4x-12=0.

Since the two circles touch each other externally,, there are 3 common tangents to these circles.

One will be perpendicular to the line joining the centres and the other two will be parallel to the line joining the centres as the radii of the two circles are equal, co-ordinates of P are given by

\frac{h+2}{\cos{60 \deg}}  = \frac{k-0}{\sin{60 \deg}}=4 \Longrightarrow h=0, k=2\sqrt{3},

co-ordinates of M, the mid-point of CP is (-1,\sqrt{3}).

Hence, the equation of the common tangent perpendicular to CP is

y-\sqrt{3}=-\frac{1}{\sqrt{3}}(x+1) \Longrightarrow x+\sqrt{3}y-2=0.

Let the equation of the common tangent parallel to CP be \sqrt{3}x-y+\lambda=0.

Since it touches the given circle \frac{}{}= \pm 2 \Longrightarrow \lambda = 2\sqrt{3} \pm 4.

Hence, the other common tangents are \sqrt{3}x -y \pm 4 + 2\sqrt{3}=0.

Question 4:

If S=0 and S^{'}=0 are the equations of two circles with radii r and r^{'} respectively, then show that the circles \frac{S}{r} \pm \frac{S^{'}}{r}=0 cut orthogonally.

Solution 4:

Let the line of centres of the given circle be taken as the x-axis and its mid-point as the origin…Note this is the key simplifying assumption. 

If the distance between the centres is 2a, the co-ordinates of the centre are (a,0) and (-a,0). Hence, we get the following:

S \equiv (x-a)^{2}+y^{2}-r^{2}=0, that is,

S^{'} \equiv x^{2}+y^{2}-2ax +a^{2}-r^{2}=0

and S^{'} \equiv x^{2}+y^{2}+2ax+a^{2}-(r^{'})^{2}=0 so that \frac{}{} + \frac{}{}=0 \Longrightarrow Sr^{'}+S^{'}r^{'}=0, that is,

\Longrightarrow (r+r^{'})(x^{2}+y^{2}+a^{2})-2ax(r^{'}-r)-rr^{'}(r+r^{'})=0

\Longrightarrow x^{2}+y^{2}-2a\frac{r^{'}-r}{r^{'}+r}x + a^{2}-rr^{'}=0…call this I.

and \frac{S}{r} - \frac{S^{'}}{r^{'}}=0 and in turn \Longrightarrow x^{2}+y^{2}-2a\frac{r^{'}+r}{r^{'}-r}+a^{2}+rr^{'}=0…call this II.

Now, since 2(\{ -a\frac{r^{'}-r}{r^{'}+r}\})(\{-a\frac{r^{'}+r}{r^{'}-r} \})+ 2\times 0 \times 0 =2a^{2}=(a^{2}-rr^{'}) +(a^{2}+rr^{'}).

The circles I and II intersect orthogonally.

Question 5:

Let P, Q, R, S be the centres of the four circles each of which is cut by a fixed circle orthogonally. If t_{1}^{2}, t_{2}^{2}, t_{3}^{2}, t_{4}^{2} be the squares of the lengths of the tangents to the four circles from a point in their plane, then prove that

t_{1}^{2}\Delta QRS +t_{2}^{2}\Delta RSP + t_{3}^{2}\Delta SPQ + t_{4}^{2}\Delta PQR=0

Solution 5:

Let the equations of the four circles be

x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0, i=1,2,3,4, then centres of these circles are as follows:

P(-g_{1},-f_{1}), Q(-g_{2},-f_{2}), R(-g_{3},-f_{3}), and S(-g_{4},-f_{4})

Let the fixed point in the plane be taken as the origin, then t_{1}^{2}=c_{1}, t_{2}^{2}=c_{2}, t_{3}^{2}=c_{3} and t_{4}^{2}=c_{4}. Let the equation of the fixed circle cutting the four circles orthogonally be

x^{2}+y^{2}+2gx + 2fy +c=0, then 2gg_{1}+2ff_{1}=c+c_{1}=c+t_{1}^{2}, or

we get the following:

2gg_{i}+2ff_{i}-c-t_{i}^{2}=0, for i=1,2,3,4.

Eliminating the unknowns g, f, c we get

\left| \begin{array}{cccc}    2g_{1} &  2f_{1}  &  -1  & -t_{1}^{2} \\    2g_{2} & 2f_{2}  &  -1  & -t_{2}^{2}\\    2g_{3} & 2f_{3}  &  -1  & -t_{3}^{2}\\    2g_{4} & 2f_{4} & -1    & -t_{4}^{2}    \end{array} \right|

or, t_{1}^{2}|D_{1}|-t_{2}^{2}|D_{2}|+t_{3}^{2}|D_{3}|-t_{4}^{2}|D_{4}|=0

where |D_{1}|= \left| \begin{array}{ccc} g_{2} & f_{2} & 1\\ g_{1} & f_{1} & 1 \\ g_{4} & f_{4} & 1 \end{array} \right |=2\Delta QRS,

and |D_{2}|=2\Delta PRS, |D_{3}|=2\Delta PQS and |D_{4}|=2\Delta PQR

Hence, we get the following:

t_{1}^{2}\Delta QRS + t_{2}^{2}\Delta RSP + t_{3}^{2}SPQ + t_{4}^{2}PQR=0.

Homework Quiz Coordinate Geometry:

  1. OAB is any chord of a circle which passes through O, a point in the plane of the circle and meets it in points A and B. A point P is taken on this chord such that OP is (i) arithmetic mean (ii) geometric mean of OA and OB. Prove that the locus of P in either case is a circle. Determine the circle.
  2. Let 2x^{2}+y^{2}-3xy=0 be the equation of a pair of tangents drawn from the origin O to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, find the length OA.
  3. Let P, Q and R be the centres and r_{1}, r_{2}, r_{3} are the radii respectively of three coaxial circles. Show that r_{1}^{2}QR + r_{2}^{2}RP + r_{3}^{2}PQ=-PQ. QR. RP
  4. If ABC be any triangle and A^{'}B^{'}C^{'} be the triangle formed by the polars of the points A, B, C with respect to a circle, so that B^{'}C^{'} is the polar of A; C^{'}A^{'} is the polar of B and A^{'}B^{'} is the polar of C. Prove that the lines AA^{'}, BB^{'} and CC^{'} meet in a point.

That’s all, folks !

Nalin Pithwa.

Listening math : Gaurish Korpal way :-) :-) :-)

https://gaurish4math.wordpress.com/2018/02/17/listening-maths/

Thanks Gaurish —

From — Nalin Pithwa.

PS: Hope my readers also read your blog regularly !! 🙂

Some random sample problems-solutions in Coordinate Geometry: I: IITJEE Mains Maths tutorials

Question I:

The point (4,1) undergoes the following transformations, successively:

a) reflection about the line y=x.

b) translation through a distance 2 units along the positive directions of the x-axis.

c) rotation through an angle of \pi/4 about the origin in the anticlockwise direction.

d) reflection about x=0.

Hint: draw the diagrams at very step!

Ans: (1/\sqrt{2}, 7/\sqrt{2})

Question 2:

A_{1}, A_{2}, A_{3}, \ldots, A_{n} are n points in a plane whose co-ordinates are (x_{1}, y_{1}), (x_{2},y_{2}), \ldots, (x_{n},y_{n}) respectively. A_{1}, A_{2} is bisected at the point G_{1}, G_{1}A_{3} is divided in the ratio 1:2 at G_{2}, G_{2}A_{4} is divided in the ratio 1:3 at G_{3}, G_{3}A_{3} is divided in the ratio 1:4 at G_{4} and so on until all n points are exhausted. Show that the co-ordinates of the final point so obtained are

(\frac{1}{n}(x_{1}+x_{2}+ \ldots + x_{n}) , \frac{1}{n}(y_{1}+y_{2}+ \ldots + y_{n}) ).

Solution 2:

The co-ordinates of G_{1} are (\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}).

Now, G_{2} divides G_{1}A_{3} in the ratio 1:2. Hence, the co-ordinates of G_{2} are

( \frac{1}{3}(\frac{2(x_{1}+x_{2})}{2}+x_{3}), \frac{1}{3}(\frac{3(y_{1}+y_{2})}{2}+y_{3})), or (\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3}).

Again, G_{3} divides G_{2}A_{4} in the ratio 1:4. Therefore, the co-ordinates of G_{3} are (\frac{1}{4}(\frac{3(x_{1}+x_{2}+x_{3})}{3}+x_{4}) ,\frac{1}{4}(\frac{3(y_{1}+y_{2}+y_{3})}{3}+y_{4}) ), or

( \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\frac{y_{1}+y_{2}+y_{3}+y_{4}}{4} ).

Proceeding in this manner,we can show that the coordinates of the final point obtained will be

(\frac{1}{n}(x_{1}+x_{2}+x_{3}+\ldots + x_{n}),\frac{1}{n}(y_{1}+y_{2}+y_{3}+\ldots + y_{n})).

Remark: For a rigorous proof, prove the above by mathematical induction.

Question 3:

A line L intersects the three sides BC, CA, and AB of a triangle ABC at P, Q and R, respectively. Show that \frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=-1

Solution 3:

Let A(x_{1},y_{1}), B(x_{2},y_{2}), and C(x_{3},y_{3}) be the vertices of \triangle ABC, and let lx+my+n=0 be equation of the line L. If P divides BC in the ratio \lambda:1, then the coordinates of P are (\frac{\lambda x_{3}+x_{2}}{\lambda + 1} ,\frac{\lambda y_{3}+y_{2}}{\lambda + 1}).

Also, as P lies on L, we have l(\frac{\lambda x_{3}+x_{2}}{\lambda + 1})+m(\frac{\lambda y_{3}+y_{2}}{\lambda + 1})+n=0

\Longrightarrow \frac{lx_{2}+my_{2}+n}{lx_{3}+my_{3}+n}=\lambda=\frac{BP}{PC}…..call this relation I.

Similarly, we can obtain \frac{CQ}{QA}=-\frac{lx_{3}+my_{3}+n}{lx_{1}+my_{1}+n}….call this relation II.

and so, also, we can prove that \frac{AR}{RB}=-\frac{lx_{1}+my_{1}+n}{lx_{2}+my_{2}+n}…call this III.

Multiplying, I, II and III, we get the desired result.

The above is the famous Menelaus’s theorem of plane geometry proved with elementary tools of co-ordinate geometry. As a homework quiz, try proving the equally famous Ceva’s theorem of plane geometry with elementary tools of co-ordinate geometry.

Question 4:

A triangle has the lines y=m_{1}x and y=m_{2}x as two of its sides, with m_{1} and m_{2} being roots of the equation bx^{2}+2hx+a=0. If H(a,b) is the orthocentre of the triangle, show that the equation of the third side is (a+b)(ax+by)=ab(a+b-2h).

Solution 4:

Since the given lines intersect at the origin, one of the triangle lies at the origin O(0,0). Let OA and OB be the given lines y=m_{1}x and y=m_{2}x, respectively. Let the equation of AB be lx+my=1. Now, as OH is perpendicular to AB, we have

\frac{b}{a}=\frac{m}{l}, \Longrightarrow \frac{l}{a}=\frac{m}{b}=k, say…call this equation I

Also, the coordinates of A and B are respectively,

(\frac{1}{l+mm_{1}}, \frac{m_{1}}{l+mm_{1}}) and (\frac{1}{l+mm_{2}} , \frac{m_{2}}{l+mm_{2}})

Therefore, the equation of AB is

(y-\frac{m_{1}}{l+mm_{1}})=-\frac{1}{m_{2}}(x-\frac{1}{l+mm_{1}})

or x+m_{2}y=\frac{1+m_{1}m_{2}}{1+mm_{1}}…call this II.

Similarly, the equation of BH is x+m_{1}y=\frac{1+m_{1}m_{2}}{1+mm_{2}}….call this III.

Solving II and III, we get the coordinates of H. Subtracting III from II, we get

y=\frac{(1+m_{1}m_{2})m}{l^{2}+lm(m_{1}+m_{2})+m^{2}m_{1}m_{2}}

Since m_{1} and m_{2} are the roots of the equation bx^{2}+2hx+a=0, we have m_{1}+m_{2}=-\frac{2h}{b} and m_{1}m_{2}=a/b.

\Longrightarrow y=\frac{(a+b)m}{bl^{2}-2hlm+am^{2}} \Longrightarrow \frac{m}{b}=\frac{bl^{2}-2hlm+am^{2}}{a+b} because y=b for H.

\Longrightarrow k=\frac{k^{2}(ba^{2}-2hab+ab^{2})}{a+b} \Longrightarrow k=\frac{a+b}{ab(a-2h+b)}.

Hence, the equation of AB is

ax+by=\frac{1}{k}=\frac{ab(a+b-2h)}{a+b}

\Longrightarrow (a+b)(ax+by)=ab(a+b-2h)

More later,

Nalin Pithwa.

World Maths day : Mar 7 competitions

Thanks to Ms. Colleen Young:

https://colleenyoung.wordpress.com/2018/02/15/world-maths-day-2018/

Happy “e” day via Math: thanks to PlusMaths :-)

https://plus.maths.org/content/happy-e-day?nl=0