Monthly Archives: February 2018

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

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}… 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


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


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


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

More later,

Nalin Pithwa.

World Maths day : Mar 7 competitions

Thanks to Ms. Colleen Young:

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

We owe a lot to the Indians who taught us counting — Albert Einstein

Jim Simons: another mathematical master and brightest billionaire !!

Just sharing this AMS blog, one of the best motivational, living legends of Mathematics and its applications in minting money..

With lots of regards to Prof. Jim Simons, and AMS blogger, Jacob Gross — from Nalin Pithwa.

PS: I hope/wish such motivational examples exist in India also…It would be a great service to Indian mathematics..

Paul Erdos, Mathematics, Russia and USA:

It is true …universally, including India…

Hats off to the “Intellect of the Wise Mathematicians”, late, adorable professor of mathematics, Paul Erdos.

— humble tribute …from Nalin Pithwa.

Paul Erdos, most profound quote ever made by any mathematician