## Category Archives: RMO

### Announcement: Scholarships for RMO Training

Mathematics Hothouse.

### Logicalympics — 100 meters!!!

Just as you go to the gym daily and increase your physical stamina, so also, you should go to the “mental gym” of solving hard math or logical puzzles daily to increase your mental stamina. You should start with a laser-like focus (or, concentrate like Shiva’s third eye, as is famous in Hindu mythology/scriptures!!) for 15-30 min daily and sustain that pace for a month at least. Give yourself a chance. Start with the following:

The logicalympics take place every year in a very quiet setting so that the competitors can concentrate on their events — not so much the events themselves, but the results. At the logicalympics every event ends in a tie so that no one goes home disappointed 🙂 There were five entries in the room, so they held five races in order that each competitor could win, and so that each competitor could also take his/her turn in 2nd, 3rd, 4th, and 5th place. The final results showed that each competitor had duly taken taken their turn in finishing in each of the five positions. Given the following information, what were the results of each of the five races?

The five competitors were A, B, C, D and E. C didn’t win the fourth race. In the first race A finished before C who in turn finished after B. A finished in a better position in the fourth race than in the second race. E didn’t win the second race. E finished two places  behind C in the first race. D lost the fourth race. A finished ahead of B in the fourth race, but B finished before A and C in the third race. A had already finished before C in the second race who in turn finished after B again. B was not first in the first race and D was not last. D finished in a better position in the second race than in the first race and finished before B. A wasn’t second in the second race and also finished before B.

So, is your brain racing now to finish this puzzle?

Cheers,

Nalin Pithwa.

PS: Many of the puzzles on my blog(s) are from famous literature/books/sources, but I would not like to reveal them as I feel that students gain the most when they really try these questions on their own rather than quickly give up and ask for help or look up solutions. Students have finally to stand on their own feet! (I do not claim creating these questions or puzzles; I am only a math tutor and sometimes, a tutor on the web.) I feel that even a “wrong” attempt is a “partial” attempt; if u can see where your own reasoning has failed, that is also partial success!

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

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.

### Inclusion Exclusion Principle theorem and examples

Reference: Combinatorial Techniques by Sharad Sane, Hindustan Book Agency.

Theorem:

The inclusion-exclusion principle: Let X be a finite set and let and let $P_{i}: i = 1, 2, \ldots n$  be a set of n properties satisfied by (s0me of) the elements of X. Let $A_{i}$ denote the set of those elements of X that satisfy the property $P_{i}$ . Then, the size of the set $\overline{A_{1}} \bigcup \overline{A_{2}} \bigcup \ldots \bigcup \overline{A_{n}}$ of all those elements that do not satisfy any one of these properties is given by $\overline{A_{1}} \bigcup \overline{A_{2}} \bigcup \ldots \bigcup \overline{A_{n}} = |X| - \sum_{i=1}^{n}|A_{n}|+ \sum_{1 \leq i .

Proof:

The proof will show  that every object in the set X is counted the same number of times on both the sides. Suppose $x \in X$ and assume that x is an element of the set on the left hand side of above equation. Then, x has none of the properties $P_{i}$. We need to show that in this case, x is counted only once on the right hand side. This is obvious since x is not in any of the $A_{i}$ and $x \in X$. Thus, X is counted only once in the first summand and is not counted in any other summand since $x \notin A_{i}$ for all i. Now let x have k properties say $P_{i_{1}}$, $P_{i_{2}}$, $\ldots$, $P_{i_{k}}$ (and no  others). Then x is counted once in X. In the next sum, x occurs ${k \choose 2}$ times and so on. Thus, on the right hand side, x is counted precisely, ${k \choose 0}-{k \choose 1}+{k \choose 2}+ \ldots + (-1)^{k}{k \choose k}$

times. Using the binomial theorem, this sum is $(1-1)^{k}$ which is 0 and hence, x is not counted on the right hand side. This completes the proof. QED.

More later,

Nalin Pithwa

### A solution to a S L Loney Part I trig problem for IITJEE Advanced Math

Exercise XXVII. Problem 30.

If a, b, c are in AP, prove that $\cos{A}\cot{A/2}$, $\cos{B}\cot{B/2}$, $\cos{C}\cot{C/2}$ are in AP.

Proof:

Given that $b-a=c-b$

TPT: $\cos{B}\cot{B/2}-\cos{A}\cot{A/2}=\cot{C/2}\cos{C}-\cos{B}\cot{B/2}$. —— Equation 1

Let us try to utilize the following formulae: $\cos{2\theta}=2\cos^{2}{\theta}-1$ which implies the following: $\cos{B}=2\cos^{2}(B/2)-1$ and $\cos{A}=2\cos^{2}(A/2)-1$

Our strategy will be reduce LHS and RHS of Equation I to a common expression/value. $LHS=(\frac{2s(s-b)}{ac}-1)(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-\cos{A}\cot{(A/2)}$

which is equal to $(\frac{2s(s-b)}{ac}-1))(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-(2\cos^{2}(A/2)-1)\frac{\cos{A/2}}{\sin{A/2}}$

which is equal to $(\frac{2s(s-b)}{ac}-1))(\frac{\sqrt{\frac{(s)(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}})-(\frac{2s(s-a)}{bc}-1)\frac{\sqrt{\frac{s(s-a)}{bc}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}}$

which is equal to $(\frac{2s(s-b)}{ac}-1))\sqrt{s(s-b)}{(s-a)(s-c)}-(\frac{2s(s-a)}{bc}-1)\sqrt{\frac{s(s-a)}{(s-b)(s-c)}}$

which in turn equals $\sqrt{\frac{s}{(s-a)(s-b)(s-c)}}((\frac{2s(s-b)}{ac}-1)(s-b)-(\frac{2s(s-a)}{bc}-1)(s-a))$

From the above, consider only the expression, given below. We will see what it simplifies to: $\frac{2s(s-b)^{2}}{ac}-(s-b)-\frac{2s(s-a)^{2}}{bc}+(s-a)$ $=\frac{2s(s-b)^{2}}{ac}-\frac{2s(s-a)^{2}}{bc}+b-a$ $=(\frac{2s}{c})(\frac{(s-b)^{2}}{a}-\frac{(s-a)^{2}}{b})+b-a$ $=\frac{2s(s-b)^{2}}{ca}-\frac{2s(s-a)^{2}}{bc}+c-b$ —- Equation II.

Now, consider RHS of Equation I. Let us see if it also boils down to the above expression after simplification. $RHS=\cot{(C/2)}\cos{C}-\cos{B}\cot{(B/2)}$ $=(2\cos^{2}{(C/2)}-1)\cot{(C/2)}-(2\cos^{2}({B/2})-1)\cot{(B/2)}$ $=(\frac{2s(s-c)}{ab}-1)\frac{\sqrt{\frac{s(s-c)}{ab}}}{\sqrt{\frac{(s-b)(s-a)}{ab}}}-(\frac{2s(s-b)}{ac})\frac{\sqrt{\frac{s(s-b)}{ac}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}}$ $= \sqrt{\frac{s}{(s-a)(s-b)(s-c)}}((\frac{2s(s-c)}{ab}-1)(s-c)-(\frac{2s(s-b)}{ac}-1)(s-b))$

From equation II and above, what we want is given below: $\frac{2s(s-c)^{2}}{ab}-(s-c)-\frac{2s(s-b)^{2}}{ac}+(s-b)=\frac{2s(s-b)^{2}}{ac}-\frac{2s(s-a)^{2}}{bc}+c-b$

that is, want to prove that $c(s-c)^{2}+a(s-a)^{2}=2b(s-b)^{2}$

but, it is given that $a+c=2b$ and hence, $c=2b-a$, which means $a+c-b=b$ and $b-a=c-b$

that is, want to prove that $c(a+b-c)^{2}+a(b+c-a)^{2}=2b(a+c-b)^{2}=2b^{3}$

i.e., want: $c(a+b-c)^{2}+a(b+c-a)^{2}=2b^{3}$

i.e., want: $(2b-a)(a+b-2b+a)^{2}+a(b-a+2b-a)^{2}=2b^{3}$

i.e., want: $(2b-a)(2a-b)^{2}+a(3b-2a)^{2}=2b^{3}$

Now, in the above, $LHS=(2b-a)(4a^{2}+b^{2}-4ab)+a(9b^{2}+4a^{2}-12ab)$ $= 8a^{2}b+2b^{3}-8ab^{2}-4a^{3}-ab^{2}+4ab^{2}+9ab^{2}+4a^{3}-12a^{2}b$ $= 2b^{3}$.

Hence, $LHS+RHS$.

QED.

### Graphs of trig raised to trig

Question: Consider the function $y=f(x)=x^{x}$. Can you graph it? It is variable raised to variable. Send me your observations.

Now, consider the functions: $(\tan \theta)^{\tan \theta}$, $(\tan \theta)^{\cot \theta}$, $(\cot \theta)^{\tan {\theta}}$, $(\cot \theta)^{\cot \theta}$.

Can you graph these? What is the difference between these and the earlier generalized case?

Now, consider the function:

Let $0 \deg < \theta < 45 \deg$.

Arrange $t_{1}=(\tan \theta)^{\tan \theta}$, $t_{2}=(\tan \theta)^{\cot \theta}$ $t_{3}=(\cot \theta)^{\tan \theta}$ and $t_{4}=(\cot \theta)^{\cot \theta}$

in decreasing order.

More later,

Nalin Pithwa

### Are complex numbers complex ?

You  might perhaps think that complex numbers are complex to handle. Quite contrary. They are easily applied to various kinds of engineering problems and are easily handled in pure math concepts compared to real numbers. Which brings me to another point. Mathematicians are perhaps short of rich vocabulary so they name some object as a “ring”, which is not a wedding or engagement ring at all; there is a “field”, which is not a field of maize at all; then there is a “group”, which is just an abstract object and certainly not a group of people!!

Problem:

Prove the identity: $({n \choose 0}-{n \choose 2}+{n \choose 4}- \ldots)^{2}+({n \choose 1}-{n \choose 3}+{n \choose 5}-\ldots)^{2}=2^{n}$

Solution:

Denote $x_{n}={n \choose 0}-{n \choose 2}+{n \choose 4}- \ldots$ and $y_{n}={n \choose 1}-{n \choose 3}+{n \choose 5}-\ldots$

and observe that $(1+i)^{n}=x_{n}+i y_{n}$

Passing to the absolute value it follows that $|x_{n}+y_{n}i|=|(1+i)^{n}|=|1+i|^{n}=2^{n/2}$.

This is equivalent to $x_{n}^{2}+y_{n}^{2}=2^{n}$.

More later,

Nalin Pithwa

### A Cute Complex Problem

Question:

If $w=\cos{\frac{\pi}{n}}+i\sin{\frac{\pi}{n}}$, then find the value of $1+w+w^{2}+w^{3}+\ldots+w^{n-1}$.

Solution:

We have $S=1+w+w^{2}+w^{3}+\ldots+w^{n-1}=\frac{1-w^{n}}{1-w}$.

But, $w^{n}=\cos{\frac{n\pi}{n}}+i\sin{\frac{n\pi}{n}}=-1$

Thus, $S=\frac{2}{1-w}$

but, $1-w=1-\cos{\frac{\pi}{n}}-i\sin{\frac{\pi}{n}}$ which equals $2\sin^{2}{\frac{\pi}{2n}}-2i\sin{\frac{\pi}{2n}}\cos{\frac{\pi}{2n}}$

that is, $-2i\sin{\frac{\pi}{2n}}[\cos{\frac{\pi}{2n}}+i\sin{\frac{\pi}{2n}}]$.

Thus, $S=\frac{-2}{2i\sin{\frac{\pi}{2n}}}[\cos{\frac{\pi}{2n}}+i\sin{\frac{\pi}{2n}} ]^{-1}=1+i\cot{\frac{\pi}{2n}}$.

Hope you are finding it useful,

More later,

Nalin Pithwa

### Complex Numbers for you

If $iz^{3}+z^{2}-z+i=0$, then $|z|$ equals

(a) 4

(b) 3

(c) 2

(d) 1.

Solution.

We can write the given equation as $z^{3}+\frac{1}{i}z^{2}-\frac{1}{i}z+1=0$, or $z^{3}-iz^{2}+iz-i^{2}=0$ $\Longrightarrow z^{2}(z-i)+i(z-i)=0$ $\Longrightarrow (z^{2}+i)(z-i)=0 \Longrightarrow z^{2}=-i, z=i$ $\Longrightarrow |z|^{2}=|-i|$ and $|z|=|i|$ $\Longrightarrow |z|^{2}=1$ and $|z|=1$ $\Longrightarrow |z|=1$