## Monthly Archives: May 2016

### Moduli and conjugates

Problem:

Let $z_{1}$, $z_{2}$, $z_{3}$ be complex numbers such that

$z_{1}+z_{2}+z_{3}=z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=0$

Prove that $|z_{1}|=|z_{2}|=|z_{3}|$.

Proof:

Method I:

Substituting $z_{1}+z_{2}=-z_{3}$ in $z_{1}z_{2}+z_{3}(z_{1}+z_{2})=0$ gives $z_{1}z_{2}=z_{3}^{2}$, so $|z_{1}|.|z_{2}|=|z_{3}|^{2}$.

Likewise, $|z_{2}|.|z_{3}|=|z_{1}|^{2}$ and $|z_{3}|.|z_{1}|=|z_{2}|^{2}$. Then,

$|z_{1}|^{2}+|z_{2}|^{2}+|z_{3}|^{2}=|z_{1}|.|z_{2}|+|z_{2}|.|z_{3}|+|z_{3}|.|z_{1}|$, that is,

$(|z_{1}|-|z_{2}|)^{2}+(|z_{2}|-|z_{3}|)^{2}+(|z_{3}|-|z_{1}|)^{2}=0$ yielding $|z_{1}|=|z_{2}|=|z_{3}|$.

Method II:

Using the relations between the roots and the coefficients, it follows that $z_{1}$, $z_{2}$, $z_{3}$ are the roots of polynomial $z^{3}-p$, where $p=z_{1}z_{2}z_{3}$. Hence, $z_{1}^{3}-p=z_{2}^{3}-p=z_{3}^{3}-p=0$ implying $z_{1}^{3}=z_{2}^{3}=z_{3}^{3}$, and the conclusion follows.

More complex fun later,

Nalin Pithwa

### How deep is the well?

In one episode of the television series Time Team, the indefatigable archaeologists want to measure the depth of a mediaeval well. They drop something into it and time its fall, which takes an amazingly long six seconds. You hear it clattering its way down for ages. They come dangerously close to calculating the depth using Newton’s laws of motion, but cop out at the last moment and use three very long tape measures joined together instead.

The formula they very nearly state is

$s=\frac{1}{2}gt^{2}$

where s is the distance travelled under gravity, falling from rest, and g is the acceleration due to gravity. It applies when air resistance can be ignored. This formula was discovered experimentally by Galileo Galilei and later generalized by Isaac Newton to describe motion under the influence of any force.

Taking $g= 10 m s^{-2}$how deep is the well?

You’ve got three days to do it.

Ref: Prof Ian Stewart’s Cabinet of Mathematical Curiosities.

More treasures coming from Prof Stewart’s cabinet for us later,

Nalin Pithwa

### Triangle inequality and max-min

Here is a problem I culled from Prof. Titu Andreescu’s literature (Geometric maxima and minima) for IITJEE Mathematics:

Problem:

Find the greatest real number k such that for any triple of positive numbers a, b, c such that $kabc > a^{3}+b^{3}+c^{3}$, there exists a triangle with side lengths a, b, c.

Solution:

We have to find the greatest real number k such that for any a, b, $c > 0$ with $a + b \leq c$, we have $kabc \leq a^{3}+b^{3}+c^{3}$. First take $b=a$ and $c=2a$. Then, $2ka^{3} \leq 10a^{3}$, that is, $k \leq 5$. Conversely, let $k=5$. Set $c=a+b+x$, where $x \geq 0$. Then,

$a^{3}+b^{3}+c^{3}-5abc =2(a+b)(a-b)^{2}+(ab+3a^{2}+3b^{2})x+3(a+b)x^{2}+x^{3} \geq 0$.

QED.

More later,

Nalin Pithwa

### Weird Arithmetic

“No, Henry, you can’t do that,” said the teacher, pointing to Henry’s exercise book, where he had written:

$\frac{1}{4} \times \frac{8}{5}=\frac{18}{45}$

“Sorry, sir<” said Henry. “What’s wrong? I checked it on my calculator and it seemed to work.

“Well, Henry, the answer is right, I guess,” the teacher admitted. “Though you should probably cancel the 9’s to get $\frac{2}{5}$, which is simpler. What’s wrong is —”

Explain the mistake to Henry. Then, find all the sums, with single non-zero  digits in the first two fractions, that are correct.

More later,

Nalin Pithwa

Ref: Professor Ian Stewart’s Cabinet of Mathematical Curiosities.

### The Surprise Examination

This paradox is so famous that I nearly left it out. It raises some intriguing issues.

Teacher tells the class that there will be a test one day next week (Monday to Friday), and that it will be a surprise. This seems reasonable: the teacher can choose any day out of five, and there is no way that the students can know which day it will be. But, the students don’t see things that way at all. They reason that the test can’t be on Friday —- because if it was, then as soon as Thursday passed without a test, they’d know it had to be Friday, so no surprise. And, once they have ruled out Friday, they apply the same reasoning to the remaining four days of the week, so the test can’t be on Thursday, either. In which case, it can’t be on Wednesday, so it can’t be on Tuesday, so it can’t be on Monday. Apparently, no surprise test is possible.

That’s all very well, but if the teacher decides to set the test on Wednesday, there seems to be no way that the students could actually know the day ahead of time! Is this a genuine paradox or not?

Reference: Professor Ian Stewart’s Cabinet of Mathematical Curiosities.

Thanks for Prof Stewart for entertaining us! 🙂

More later,

Nalin Pithwa

### Maxima and minima — using calculus — for IITJEE Advanced Mathematics

Problem:

The length of the edge of the cube $ABCDA_{1}B_{1}C_{1}D_{1}$ is 1. Two points M and N move along the line segments AB and $A_{1}D_{1}$, respectively, in such a way that at any time t $(0 \leq t < \infty)$ we have $BM = |\sin{t}|$ and $D_{1}N=|\sin(\sqrt{2}t)|$. Show that MN has no maximum.

Proof:

Clearly, $MN \geq MA_{1} \geq AA_{1} =1$. If $MN=1$ for some t, then $M=A$ and $N=A_{1}$, which is equivalent to $|\sin{t}|=1$ and $|\sin{\sqrt{2}t}|=1$. Consequently, $t=\frac{\pi}{2}+k\pi$ and $\sqrt{2}t=\frac{\pi}{2}+n\pi$ for some integers k and n, which implies $\sqrt{2}=\frac{2n+1}{2k+1}$, a contradiction since $\sqrt{2}$ is irrational. This is why $MN >1$ for any t.

We will now show that MN can be made arbitrarily close to 1. For any integer k, set $t_{k}=\frac{\pi}{2}+k\pi$. Then, $|\sin{t_{k}}|=1$, so at any time $t_{k}$, the point M is at A. To show that N can be arbitrarily close to $A_{1}$ at times $t_{k}$, it is enough to show that $|\sin(\sqrt{2}t_{k})|$ can be arbitrarily close to 1 for appropriate choices of k.

We are now going to use Kronecker’s theorem:

If $\alpha$ is an irrational number, then the set of numbers of the form $ma+n$where m is a positive integer while n is an arbitrary integer, is dense in the set of all real numbers. The latter means that every non-empty open interval (regardless of how small it is) contains a number of the form $ma+n$.

Since $\sqrt{2}$ is irrational, we can use Kronecker’s theorem with $\alpha=\sqrt{2}$. Then, for $x=\frac{1-\sqrt{2}}{2}$, and any $\delta>0$, there exist integer $k \geq 1$ and $n_{k}$ such  that $k\sqrt{2}-n_{k} \in (x-\delta, x+\delta)$. That is, for $\in_{k}=\sqrt{2}k+\frac{\sqrt{2}}{2}-\frac{1}{2}-n_{k}$ we have $|\in_{k}|<\delta$. Since $\sqrt{2}(k+\frac{1}{2})=\frac{1}{2}+n_{k}+\in_{k}$, we have

$|\sin(\sqrt{2}t_{k})|=|\sin \pi \sqrt{2}(k+\frac{1}{2})|=|\sin(\frac{\pi}{2}+n_{k}\pi+\in_{k}\pi)|=|\cos{(\pi \in_{k})}|$.

It remains to note that $|\cos{(\delta\pi)}|$ tends to 1 as $\delta$ tends to 0.

Hence, MN can be made arbitrarily close to 1.

Ref: Geometric Problems on Maxima and Minima by Titu Andreescu, Oleg Mushkarov, and Luchezar Stoyanov.

Thanks to Prof Andreescu, et al !

Nalin Pithwa

It’s envy, and the problem is to avoid it.

Stefan Banach and Bromslaw Knaster extended Steinhaus’s method of fair cake division to any number of people, and simplified it for three people. Their work pretty much summed up  the whole area until a subtle flaw emerged: the procedure may be fair, but it takes no account of envy. A method is envy-free if no one thinks that anyone else has got a bigger share than they have. Every envy-free method is fair, but a fair method need not be envy free. And, neither Steinhaus’s method, nor that of Banach and Knaster is envy-free.

For example, Belinda may think that Arthur’s division is fair. Then, Steinhaus’s method stops after step 3, and both Arthur and Belinda consider all three pieces to be of size $1/3$. Charlie must think that his own piece is at least $1/3$, so the allocation is proportional. But, if Charlie sees Arthur’s piece as $1/6$ and Belinda’s as $1/2$, then he will envy Belinda, because Belinda got first crack at a piece that Charlie thinks is bigger than his.

Can you find an envy-free method for dividing a cake among three people?

Ref: Professor Ian Stewart’s Cabinet of Mathematical Curiosities

We will see from treasures from Professor Stewart’s cabinet soon,

-Nalin Pithwa

### Fair Shares

In 1944, as the Russian army fought to reclaim Poland from the Germans, the mathematician Hugo Steinhaus, trapped in the city of Lvov, sought distraction in a puzzle. As you do.

The puzzle was this. Several people want to share a cake (by all means, replace that by a pizza if you wish). And, they want the procedure to be fair, in the sense that no one will feel that they have got less than their fair share.

Steinhaus knew that for two people there is a simple method: one person cuts the cake in two pieces, and the other chooses which one they want. The second person can’t complain, because they made the choice. The first person also can’t complain — if they do, it was their fault for cutting the cake wrongly.

How can three people divide a cake fairly?

Ref: Professor Ian Stewart’s Cabinet of Mathematical Curiosities

More fun coming,

Nalin Pithwa

### Geometric maxima minima — a little question for IITJEE Mains

Problem:

Of all triangles with a given perimeter, find the one with maximum area.

Solution:

Consider an arbitrary triangle with side lengths a, b, c and perimeter $2s=a+b+c$. By Heron’s formula, its area F is given by

$F = \sqrt{s(s-a)(s-b)(s-c)}$

Now, the arithmetic mean geometric mean inequality gives

$\sqrt[3]{s(s-a)(s-b)(s-c)} \leq \frac{(s-a)+(s-b)+(s-c)}{3}=\frac{s}{3}$

Therefore, $F \leq \sqrt{s(\frac{s}{3})^{3}}=s^{2}\frac{\sqrt{3}}{9}$

where inequality holds if and only if $s-a=s-b=s-c$, that is, when $a=b=c$.

Thus, the area of any triangle with perimeter 2s does not exceed $\frac{s^{2}\sqrt{3}}{9}$ and is equal to $\frac{s^{2}\sqrt{3}}{9}$ only for an equilateral triangle. QED.

More later,

Nalin Pithwa

PS: Ref: Geometric Problems on Maxima and Minima by Titu Andreescu et al.

In mathematical logic, a paradox is a self-contradictory statement. The best known is “This sentence is a lie.” Another is Bertrand Russell’s “barber paradox”. In a village, there is a barber who shaves everyone who  does not shave themselves. So, who shaves the barber? Neither ‘the barber’ nor ‘someone else’ is logically acceptable. If it is the barber, then he shaves himself — but we are told that he doesn’t. But, if it is someone else, then the barber does not shave himself…but we are told that he shaves all such people, so he does shave himself !

In the real world. there are plenty of get-outs (are we talking about shaving beards here, or legs, or what? Is the barber a woman? Can such a barber actually exist anyway?) But in mathematics, a more carefully stated version of Russell’s paradox ruined the life’s work of Gottlob Frege, who attempted to base the whole of mathematics on set theory — the study of collections of objects, and how these can be combined to form other collections.

Protagoras was a Greek lawyer who lived and taught in the fifth century BC. He had a student, and it was agreed that the student would pay for his teaching after he had won his first case. But, the student did not get any clients, and eventually Protagoras threatened to sue him. Protagoras reckoned that he would win either way: if the court upheld his case, the student would be required to pay up, but if Protagoras lost, then by their agreement the student would have to pay anyway. The student argued exactly the other way round: if Protagoras won, then by their agreement the student would not have to pay, but if Protagoras lost, the court would have ruled that the student did not have to pay.

Is this a genuine logical paradox or not?

Ref; Professor Ian Stewart’s Cabinet of Mathematical Curiosities

More later,

Nalin Pithwa