## Monthly Archives: March 2017

### I have a sweet tooth !!!

1. Tracey will not get the chocolate-covered mints unless Neil has the plain mints.
2. Alan will not get the toffee unless Robert has the mint-flavoured toffee.
3. Neil will not get the plain mints unless Alan has the mint-flavoured toffee.
4. Robert will not get the toffee unless Tracey gets the plain mints.
5. Robert will not get the chocolate-covered mints unless Tracey gets the toffee.
6. Tracey will not get the toffee unless Neil gets the chocolate.
7. Robert will not get the mint-flavoured toffee unless James gets the plain mints.
8. Alan will not get the plain mints unless Tracey gets the toffee.
9. Tracey will not get the plain mints unless Alan gets the chocolate-covered mints.
10. Alan will not get the mint-flavoured toffee unless Tracey gets the chocolate-covered mints.
11. James will not get the plain mints unless Tracey gets the toffee.
12. Neil will not get the chocolate unless Alan gets the chocoate-covered mints.
13. Roberts will not get the plain mints unless James gets the toffee.
14. James will not get the toffee unless Tracey gets the plain mints.
15. Neil will not get the toffee unless Tracey gets the toffee.
16. Tracey will not get the plain mints unless Roberts gets the toffee.
17. Alan will not get the chocolate-covered mints unless James gets the plain mints.

Who will get what ?

Oh, to be precise, I do not have a sweet tooth. I have sweet-teeth! Ich habe eine stucke schokolade sehr gehn!

Auf wiedersehen.

Nalin Pithwa.

### People’s Pets

Consider the following:

1. Five men each have different first names and different surnames, have five different pets and live at five different addresses. All five pets have a different name.
2. Tom’s surname is Williams and the fish is not  called Spike, Benson or Rodney.
3. Harry has a pet cat and the budgie is called Percy.
4. George’s surname is not Hudson or Smith.
5. Mr. Thompson owns the dog and the owner of the rabbit lives in Pine Avenue.
6. The cat is not called Benson and one of the five men has a pet called Fred.
7. Mr. Anderson does not live in Cedar Road.
8. Mr. Hudson lives in Willow Street.
9. Mr. Anderson owns a pet called Percy and John lives in Cedar Road.
10. Bill’s pet is called Rodney and is not  the dog; the owner of the fish lives in Maple Grove.

So, who lives in Chestnut Crescent and what is the name of their pet?

-Nalin Pithwa.

### All The Twos

Alex, Brad, Colin, Doug, Eric and Frank had a race to school from the bus-stop and then another race from school to the bus-stop. In the first race, Brad wasn’t last and Eric finished before Colin, Frank wasn’t first but finished before Doug, Brad finished before Frank but after Alex and Colin. Alex finished four places ahead of Doug. In the second race, Alex finished two places ahead of Doug and before Eric who in turn was two places ahead of Doug and before Eric who in turn was two places behind Colin. Alex wasn’t first and nor was Brad who finished before Eric. Frank finished one place ahead of Doug.

From the information given:

1. Which two boys finished in a better position in the first race than in the second race?
2. Which two boys finished in the same position in the second race as they did in the first race?

Let me see which of my student(s) is first in this !

Nalin Pithwa.

### Three slices of toast

Mother makes tasty toast in a small pan. After toasting one side of a slice, she turns it over. Each side takes 30 seconds.

The pan can only hold two slices. How can she toast both sides of three slices in one and half instead of 2 minutes?

🙂 🙂 🙂

Nalin Pithwa.

### Fun with English !!!

Done what ?

Arthur said Dave did it, Dave said Bill did it, Bill said Charlie did it, Charlie said that he didn’t do it and Eddie confessed that he did it. If Arthur didn’t do it, one of the five of them did do it and only one of them is telling the truth, who did do it?

🙂 🙂 🙂

Nalin Pithwa

### Want to be a safe-cracker like Richard Feynman ? :-)

You must have heard of “the” American, darling, maverick, physicist, Nobel-Laureate, Mr. Richard Feynman. He was one of the young, brilliant team of physicists assembled at Los Alamos for the secret atomic bomb project. (There were many others also!) In his autobiography, “Surely, you are joking, Mr. Feynman!”, he says that he used to pick locks, and cracks safes containing top-secret stuffs at Los Alamos. He was a prankster!! You too can try to see if you can be a “good” safecracker 🙂

The combination of a safe consists of the numbers 1, 2, 3 and 4. There are only the four numbers in the combination but the safe is computerized and changes the combination each time it is closed. Each number appears only once in each combination. The person opening the safe is only allowed four attempts, after which the safe cannot be opened for at least 48 hours. When you tap into the combination on the push-button panel, there is a display next to the panel which indicates the number of numbers you have “tapped in” in the correct position. The last time that I opened the safe I tapped in the following combinations; on my first attempt I tapped in 3 1 2 4, none of which were in the correct place. I then tapped in 2 3 4 1; again none of the numbers were in the correct place. On my third attempt I only tapped in the number 1 in first place; the safe indicated that it was not  in the correct place, after which I then knew the combination. What was the correct combination?

There is a chapter in Richard Feynman’s autobiography: He fixes radios by thinking!!! So, now u can pick locks and cracks safes by thinking !! By the way, this is also a nice exercise in permutations and combinations.

Cheers,

Nalin Pithwa.

### Movie Magic: The Mathematics Behind Hollywood’s Visual Effects

Distinguished Lecture Series: The Mathematical Association of America

### Some more examples of pair of straight lines questions for IITJEE math

Problem 1:

If one of the straight lines given by the equation $ax^{2}+2hxy+by^{2}=0$ coincides with one of those given by $a^{'}x^{2}+2h^{'}xy+b^{'}y^{2}=0$, and the other lines represented by them be perpendicular, prove that $\frac{ha^{'}b^{'}}{b^{'}-a^{'}}=\frac{h^{'}ab}{b-a}=\frac{1}{2}\sqrt{aa^{'}bb^{'}}$

Solution 1:

Let the lines represented by $ax^{2}+2hxy+by^{2}=0$ be $y=m_{1}x$ and $y=m_{2}x$, so that

$m_{1}+m_{2}=\frac{2h}{b}$ and $m_{1}m_{2}=\frac{a}{b}$…call this equation (A)

The lines represented by $a^{'}x^{2}+2h^{'}xy+b^{'}y^{2}=0$ are $y=m_{1}x$ and $y=-\frac{1}{m_{2}}x_{1}$ so that

$m_{1}-\frac{1}{m_{2}}=-\frac{2h^{'}}{b^{'}}$ and $m_{1}(-\frac{1}{m_{2}})=\frac{a^{'}}{b^{'}}$…call this equation (B)

From (A) and (B), we get $m_{1}m_{2}(-\frac{m_{1}}{m_{2}})=\frac{aa^{'}}{bb^{'}}$

which in turn $\Longrightarrow m_{1}^{2}=\frac{aa^{'}}{bb^{'}}$,

which in turn $\Longrightarrow m_{1}=\sqrt{(-\frac{aa^{'}}{bb^{'}})}$

and from (A), again, we get $\sqrt{-\frac{aa^{'}}{bb^{'}}}-\sqrt{-\frac{aa^{'}}{ba^{'}}}=-\frac{2h}{b}$

and this in turn $\Longrightarrow \sqrt{(-aa^{'}bb^{'})}(\frac{1}{bb^{'}}-\frac{1}{a^{'}b})=-\frac{2h}{b}$, that is, it $\Longrightarrow \frac{1}{2}\sqrt{(-aa^{'}bb^{'})}=\frac{ha^{'}b^{'}}{b^{'}-a^{'}}$.

Similarly, from (B), we get $\frac{1}{2}\sqrt{(-aa^{'}bb^{'})}=\frac{h^{'}ab}{b-a}$. Hence, the required result follows.

Problem 2:

If the equation $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ represents two straight lines, prove that the equation of the third pair of straight lines passing through the points where these meet the axes $xy=0$ is $c(ax^{2}+2hxy+by^{2}+2gx+2fy+c)+4(fg+ch)xy=0$.

Solution 2:

Equation of the lines passing through the intersection of  the given lines and axes is

$ax^{2}+2hxy+by^{2}+2gx+2fy+c+2xy \lambda =0$, or

$ax^{2}+2(h+\lambda)xy+by^{2}+2gx+2fy+c=0$

Since it represents a pair of straight lines

$abc+2fg(h+\lambda)-af^{2}-bg^{2}-c(h+\lambda)^{2}=0$,

$\Longrightarrow abc+2fgh-af^{2}-bg^{2}-ch^{2}+2fg\lambda -2ch\lambda -c\lambda^{2}=0$

$\Longrightarrow \lambda = \frac{2(fg-ch)}{c}$ (since $abc+2fgh-af^{2}-bg^{2}-ch^{2}=0$).

Hence, the required equation of the lines is $c(ax^{2}+2hxy+by^{2}+2gx+2fy+c)+4(fg+ch)xy=0$

Problem 3:

Show that the equation $\sqrt{3}x^{2}-(4+\sqrt{3})x^{2}y+(4+\sqrt{3})xy^{2}-\sqrt{3}y^{2}=0$ represents three straight lines through the origin such that one of them bisects the angle between the other two. Also find the equation of  the lines perpendicular to the given lines through the origin.

Solution 3:

The given equation can be written as $\sqrt{3}(x^{3}-y^{3})-(4+\sqrt{3})xy(x-y)=0$, or

$(x-y)(\sqrt{3}(x^{2}+xy+y^{2}))-(4+\sqrt{3})xy=0$, or

$(x-y)(\sqrt{3}x^{2}-4xy+\sqrt{3}y^{2})=0$, or

$(x-y)(\sqrt{3}x-y)(x-\sqrt{3}y)=0$,or

$x-y=0$, $\sqrt{3}x-y=0$, and $x-\sqrt{3}y=0$

which gives three straight lines passing through the origin with slopes 45 degrees, 60 degrees, and 30 degrees respectively showing that $x-y=0$ bisects the angles between the other two.

Now, equations of the lines through the origin perpendicular to these lines are

$x+y=0$, $x+\sqrt{3}y=0$, $\sqrt{3}x+y=0$ so their joint equation is given by

$(x+y)(x+\sqrt{3}y)(\sqrt{3}x+y)=0$

or, $\sqrt{3}x^{3}+(4+\sqrt{3})x^{2}y+(4+\sqrt{3})xy^{2}+\sqrt{3}y^{3}=0$

That’s all, folks !

Nalin Pithwa.

My friend Jack asked me if I would help him prepare a new C.V. for a job he had seen advertised in a newspaper. I asked him what jobs he had had since leaving school. He said his first job was as a journalist, and his present job was a scene-shifter, but, although he could remember the various  jobs, he had had, he couldn’t remember the order in which he had had them. I told him to tell me what he could remember and then, I would try to prepare a list of the jobs he had had in the correct order.

This is what he said: “After being a journalist, I was a signwriter, an illustrator, a compositor and a pawnbroker, but I was a pawnbroker before I was a ringmaster, a piano-tuner and an interpreter. Before I was an interpreter I was a chargehand, and after I was an interpreter I was a compositor. I can remember being a compositor before I was a gamekeeper, a taxi-driver, an electrician and a roadsweeper. I was a gamekeeper and an electrician after I was a roadsweeper which in turn was after I was a taxi-driver and a woodcarver. I liked my job as a woodcarver which I got after being a blacksmith, a chargehand, a lumberjack and an illustrator. I was an illustrator after I had a job as a cartoonist, and an undertaker, and i was an undertaker before I was a pawnbroker and a chargehand but after I was a signwriter and a cartoonist.

“I can also remember being a fishmonger, a lumberjack, a chargehand and a pawnbroker after I was a cartoonist which in turn was after I was a journalist and a signwriter. I was a signwriter before I was a pawnbroker, a pawnbroker before I was an upholsterer, an upholsterer before I was a programmer, a programmer before I was a fishmonger, a fishmonger before I was a blacksmith, and a blacksmith before I was a gamekeeper, an escapologist, a compositor and a piano-tuner. I was a piano-tuner before I was a lumberjack but that was after I had been a chargehand and an escapologist. I was also a greengrocer after being an upholsterer and an illustrator. Then again, I can also  remember being a greengrocer after I was an escapologist but before I was a piano-tuner.

“At one time, I was a postmaster but that was before I was a pawnbroker, a chargehand and a newscaster which in turn was after I was an illustrator. I had quite a few jobs after I was an illustrator, including being a blacksmith, a gamekeeper, a newscaster, an escapologist and a programmer. On thing that I do know is that I haven’t done the same type of job twice. As I said before, I liked my job as a woodcarver but that was before I was a taxi-driver, an interpreter, a compositor, a gamekeeper and an electrician. I was actually an electrician before I was a gamekeeper.

“One job that I wasn’t too keen on was that time I was an upholsterer, but that was before I was a ringmaster, a newscaster, a blacksmith and a compositor. I was a compositor after I was chargehand and a lumberjack. One quite boring job was that of a programmer, which was after I was a newscaster and a ringmaster, and I was a ringmaster before I was a newscaster. The only other things I can remember is that I was a chargehand after I was a fishmonger and a pawnbroker, and that I was a pawnbroker before I was a fishmonger.”

Given what Jack said, see if you can make a list of the jobs he has had since being a journalist and the order in which he has had them.

🙂 🙂 🙂

Nalin Pithwa.

### Some solved examples based on pairs of straight lines

The following questions use the previous (previous blog)theory, concepts and formulae related to pair of straight lines:

Problem 1:

Prove that the straight lines joining the origin to the points of intersection of the straight line $hx + ky =2hk$ and the curve $(x-k)^{2}+(y-h)^{2}=c^{2}$ are at right angles if $h^{2}+k^{2}=c^{2}$.

Solution 1:

Making the equation of the curve homogeneous with the help of the equation of the line, we get

$x^{2}+y^{2} - 2(kx+hy)(\frac{hx+ky}{2hk})+(h^{2}+k^{2}-c^{2})(\frac{hx+ky}{2hk})^{2}=0$, or

$4h^{2}k^{2}x^{2} + 4h^{2}k^{2}y^{2}-4hk^{2}x(hx+ky) - 4h^{2}ky(hx+ky)+(h^{2}+k^{2}-c^{2})(h^{2}x^{2}+k^{2}y^{2}+2hxy)=0$

This is the equation of the pair of lines joining the origin to the points of intersection of the given line and the curve. They will be at right angles if

coefficient of $x^{2}$ + coefficient of $y^{2}$ is zero, that is,

$\Longrightarrow (h^{2}+k^{2})(h^{2}+k^{2}-c^{2})=0$

$\Longrightarrow h^{2}+k^{2}=c^{2}$ (since $h^{2}+k^{2} \neq 0$).

Problem 2:

Prove that the two straight lines $(x^{2}+y^{2})(\cos^{2}{\theta}\sin^{2}{\alpha}+\sin^{2}{\theta})=(x\tan{\alpha}+y\sin{\theta})^{2}$ include an angle $2\alpha$.

Solution 2:

The given equation can be written as

$(\cos^{2}{\theta}\sin^{2}{\alpha} + \sin^{2}{\theta} - \tan^{2}{\alpha})x^{2} + 2\tan{\alpha}\sin{\theta}xy + \cos^{2}{\theta}\sin^{2}{\alpha}y^{2}=0$

Here, $a=\cos^{2}{\theta}\sin^{2}{\alpha} + \sin^{2}{\theta}-\tan^{2}{\alpha}$,

$b=\cos^{2}{\theta}\sin^{2}{\alpha}$ and $h=\tan{\alpha}\sin{\theta}$.

We need to show  that

$2\alpha = \arctan {2\frac{\sqrt{h^{2}-ab}}{a+b}}$

or $\tan{2\alpha} = \frac{2\sqrt{h^{2}-ab}}{a+b}$

Now, $\frac{2\sqrt{h^{2}-ab}}{a+b}$

$= \frac{2\sqrt{\tan^{2}{\alpha}\sin^{2}{\theta}-(\cos^{2}{\theta}\sin^{2}{\alpha} + \sin^{2}{\theta}-\tan^{2}{\alpha})\cos^{2}{\theta}\sin^{2}{\alpha}}}{2\cos^{2}{\theta}\sin^{2}{\alpha}+\sin^{2}{\theta}-\tan^{2}{\alpha}}$

$= \frac{2\tan{\alpha}\sqrt{\sin^{2}{\theta}-\cos^{4}{\theta}\sin^{2}{\alpha}\cos^{2}{\alpha}-\sin^{2}{\theta}\cos^{2}{\theta}\cos^{2}{\alpha}+\sin^{2}{\alpha}\cos^{2}{\theta}}}{2\cos^{2}{\theta}\sin^{2}{\alpha}+1-\cos^{2}{\theta}-\tan^{2}{\alpha}}$

$= \frac{2\tan{\alpha}\sqrt{1-\cos^{2}{\theta}-\cos^{4}{\theta}{\cos^{2}{\alpha(1-\cos^{2}{\alpha})-(1-\cos^{2}{\theta})\cos^{2}{\theta}\cos^{2}{\alpha}+(1-\cos^{2}{\alpha})\cos^{2}{\theta}}}}}{1-\tan^{2}{\alpha}-\cos^{2}{\theta}(1-2\sin^{2}{\alpha})}$

$=\frac{2\tan{\alpha}\sqrt{1+\cos^{4}{\theta}\cos^{4}{\alpha}-2\cos^{2}{\theta}\cos^{2}{\alpha}}}{1-\tan^{2}{\alpha}-\cos^{2}{\theta}(\cos^{2}{\theta}-\sin^{2}{\alpha})}$

$= \frac{2\tan{\alpha}(1-\cos^{2}{\theta}\cos^{2}{\alpha})}{(1-\tan^{2}{\alpha})(1-\cos^{2}{\theta}\cos^{2}{\alpha})}=\tan{2\alpha}$

Hence, the given lines include an angle $2\alpha$.

More later,

Nalin Pithwa.