“We had to answer all the questions, just “Yes” or “No”.” Fran explained. “The first paper I got only five wrong.”

Ben smiled. “That sounds good. What about the other paper?”

“Tough, but the same number of questions,” the girl replied. “I got just a third of them wrong, so over all I got 75 percent right for the two papers.”

How many questions in each paper?

Cheers,

Nalin Pithwa.

“It’s only teenage infatuation,” Tom declared. “I do like Bob, but he is double your age.”

“Come on, Dad, that’s nothing these days,” Cathy replied. “He doesn’t look it, and anyway three years ago the digits of my age added up to the same as the digits of his age did.”

What they call a non-sequitur? How old was she?

Reference: Entertaining Mathematical Teasers, and how to solve them by J.A.H. Hunter, Dover Publications Inc., New York, available at Amazon India.

Cheers,

Nalin Pithwa.

## Fun with Math websites: MAA

https://www.maa.org/programs/students/fun-math

cheers to MAA! 🙂

Nalin Pithwa.

## Cartesian System, Straight Lines: IITJEE Mains: Problem Solving Skills II

I have a collection of some “random”, yet what I call ‘beautiful” questions in Co-ordinate Geometry. I hope kids preparing for IITJEE Mains or KVPY or ISI Entrance Examination will also like them.

Problem 1:

Given n straight lines and a fixed point O, a straight line is drawn through O meeting lines in the points $R_{1}$, $R_{2}$, $R_{3}$, $\ldots$, $R_{n}$ and on it a point R is taken such that $\frac{n}{OR} = \frac{1}{OR_{1}} + \frac{1}{OR_{2}} + \frac{1}{OR_{3}} + \ldots + \frac{1}{OR_{n}}$

Show that the locus of R is a straight line.

Solution 1:

Let equations of the given lines be $a_{i}x+b_{i}y+c_{i}=0$, $i=1,2,\ldots, n$, and the point O be the origin $(0,0)$.

Then, the equation of the line through O can be written as $\frac{x}{\cos{\theta}} = \frac{y}{\sin{\theta}} = r$ where $\theta$ is the angle made by the line with the positive direction of x-axis and r is the distance of any point on the line from the origin O.

Let $r, r_{1}, r_{2}, \ldots, r_{n}$ be the distances of the points $R, R_{1}, R_{2}, \ldots, R_{n}$ from O which in turn $\Longrightarrow OR=r$ and $OR_{i}=r_{i}$, where $i=1,2,3 \ldots n$.

Then, coordinates of R are $(r\cos{\theta}, r\sin{\theta})$ and of $R_{i}$ are $(r_{i}\cos{\theta},r_{i}\sin{\theta})$ where $i=1,2,3, \ldots, n$.

Since $R_{i}$ lies on $a_{i}x+b_{i}y+c_{i}=0$, we can say $a_{i}r_{i}\cos{\theta}+b_{i}r_{i}\sin{\theta}+c_{i}=0$ for $i=1,2,3, \ldots, n$

$\Longrightarrow -\frac{a_{i}}{c_{i}}\cos{\theta} - \frac{b_{i}}{c_{i}}\sin{\theta} = \frac{1}{r_{i}}$, for $i=1,2,3, \ldots, n$

$\Longrightarrow \sum_{i=1}^{n}\frac{1}{r_{i}}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}$

$\frac{n}{r}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}$ …as given…

$\Longrightarrow (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})r\cos{\theta}+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})r\sin{\theta} + n=0$

Hence, the locus of R is $(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})x+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})y+n=0$ which is a straight line.

Problem 2:

Determine all values of $\alpha$ for which the point $(\alpha,\alpha^{2})$ lies inside the triangle formed by the lines $2x+3y-1=0$, $x+2y-3=0$, $5x-6y-1=0$.

Solution 2:

Solving equations of the lines two at a time, we get the vertices of the given triangle as: $A(-7,5)$, $B(1/3,1/9)$ and $C(5/4, 7/8)$.

So, AB is the line $2x+3y-1=0$, AC is the line $x+2y-3=0$ and BC is the line $5x-6y-1=0$

Let $P(\alpha,\alpha^{2})$ be a point inside the triangle ABC. (please do draw it on a sheet of paper, if u want to understand this solution further.) Since A and P lie on the same side of the line $5x-6y-1=0$, both $5(-7)-6(5)-1$ and $5\alpha-6\alpha^{2}-1$ must have the same sign.

$\Longrightarrow 5\alpha-6\alpha^{2}-1<0$ or $6\alpha^{2}-5\alpha+1>0$ which in turn $\Longrightarrow (3\alpha-1)(2\alpha-1)>0$ which in turn $\Longrightarrow$ either $\alpha<1/3$ or $\alpha>1/2$….call this relation I.

Again, since B and P lie on the same side of the line $x+2y-3=0$, $(1/3)+(2/9)-3$ and $\alpha+2\alpha^{2}-3$ have the same sign.

$\Longrightarrow 2\alpha^{2}+\alpha-3<0$ and $\Longrightarrow (2\alpha+3)(\alpha-1)<0$, that is, $-3/2 <\alpha <1$…call this relation II.

Lastly, since C and P lie on the same side of the line $2x+3y-1=0$, we have $2 \times (5/4) + 3 \times (7/8) -1$ and $2\alpha+3\alpha^{2}-1$ have the same sign.

$\Longrightarrow 3\alpha^{2}+2\alpha-1>0$ that is $(3\alpha-1)(\alpha+1)>0$

$\alpha<-1$ or $\alpha>1/3$….call this relation III.

Now, relations I, II and III hold simultaneously if $-3/2 < \alpha <-1$ or $1/2<\alpha<1$.

Problem 3:

A variable straight line of slope 4 intersects the hyperbola $xy=1$ at two points. Find the locus of the point which divides the line segment between these two points in the ratio $1:2$.

Solution 3:

Let equation of the line be $y=4x+c$ where c is a parameter. It intersects the hyperbola $xy=1$ at two points, for which $x(4x+c)=1$, that is, $\Longrightarrow 4x^{2}+cx-1=0$.

Let $x_{1}$ and $x_{2}$ be the roots of the equation. Then, $x_{1}+x_{2}=-c/4$ and $x_{1}x_{2}=-1/4$. If A and B are the points of intersection of the line and the hyperbola, then the coordinates of A are $(x_{1}, \frac{1}{x_{1}})$ and that of B are $(x_{2}, \frac{1}{x_{2}})$.

Let $R(h,k)$ be the point which divides AB in the ratio $1:2$, then $h=\frac{2x_{1}+x_{2}}{3}$ and $k=\frac{\frac{2}{x_{1}}+\frac{1}{x_{2}}}{3}=\frac{2x_{2}+x_{1}}{3x_{1}x_{2}}$, that is, $\Longrightarrow 2x_{1}+x_{2}=3h$…call this equation I.

and $x_{1}+2x_{2}=3(-\frac{1}{4})k=(-\frac{3}{4})k$….call this equation II.

Adding I and II, we get $3(x_{1}+x_{2})=3(h-\frac{k}{4})$, that is,

$3(-\frac{c}{4})=3(h-\frac{k}{4}) \Longrightarrow (h-\frac{k}{4})=-\frac{c}{4}$….call this equation III.

Subtracting II from I, we get $x_{1}-x_{2}=3(h+\frac{k}{4})$

$\Longrightarrow (x_{1}-x_{2})^{2}=9(h+\frac{k}{4})^{2}$

$\Longrightarrow \frac{c^{2}}{16} + 1= 9(h+\frac{k}{4})^{2}$

$\Longrightarrow (h-\frac{k}{4})^{2}+1=9(h+\frac{k}{4})^{2}$

$\Longrightarrow h^{2}-\frac{1}{2}hk+\frac{k^{2}}{16}+1=9(h^{2}+\frac{1}{2}hk+\frac{k^{2}}{16})$

$\Longrightarrow 16h^{2}+10hk+k^{2}-2=0$

so that the locus of $R(h,k)$ is $16x^{2}+10xy+y^{2}-2=0$

More later,

Nalin Pithwa.

## Prioritize your passions and commitments, says Dr. Shawna Pandya

Kids now-a-days need counselling for a choice of career. In my humble opinion, excellence in any field of knowledge/human endeavour gives deep satisfaction as well as a means of livelihood. But, I am a mere mortal, most of my students are bright, ambitious, multi-talented and hard-working. I would like to present to them the views of one of my “idols”, though not a mathematician…

Dr. Shawna Pandya:

http://www.indiatimes.com/news/india/introducing-dr-shawna-pandya-the-third-indian-origin-astronaut-to-go-into-space-271207.html

Hats off to Dr. Shawna Pandya, belated though from me…:-)

— Nalin Pithwa.