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For 6 - something to be less than or equal to 6, we need something to be non-negative

Let $z_1 = \cos \theta_1+ i \sin \theta_1; z_2 = \cos \theta_2+ i \sin \theta_2; z_3 = \cos \theta_3+ i \sin \theta_3$

Then

$\sum_{cyc} |z_1-z_2|^2 = \sum_{cyc}[\cos \theta_1-\cos \theta_2)^2+(\sin \theta_1-\sin \theta_2)^2] = \sum_{cyc} 2 - 2 \cos (\theta_1-\theta_2) \\ \\<br/>= 6 - 2\sum_{cyc}\cos (\theta_1-\theta_2)$

Now, let us look at the summation term

$\sum_{cyc}\cos (\theta_1-\theta_2) = \cos (\theta_1-\theta_2) +\cos (\theta_2-\theta_3) + \cos (\theta_3-\theta_1) \\ \\ = \cos (\theta_1-\theta_2) +\cos (\theta_3-\theta_2) + \cos (\theta_1-\theta_3)$

Now suppose we fix $\theta_1-\theta_2$

Notice that $\theta_3-\theta_2+ \theta_1-\theta_3 = \theta_1-\theta_2$

That means $\cos (\theta_3-\theta_2)+ \cos( \theta_1-\theta_3)$ attains a minimum when $\theta_3-\theta_2 = \theta_1-\theta_3 = \frac{\theta_1-\theta_2}{2}$

Hence let, $\theta_1-\theta_2 = \theta$

Hence the minimum of the expression is $\cos \theta + 2 \cos \frac{\theta}{2} = 2 \cos^2 \frac{\theta}{2} + 2 \cos \frac{\theta}{2} -1 = 2 [(\cos \frac{\theta}{2}+\frac{1}{2})^2 - \frac{3}{4}) \ge -\frac{3}{2}$

Hence $-2 \sum_{cyc} \cos (\theta_1-\theta_2) \le 3$

That means the minimum of the expression is 9 attained when the three points are spread out at angles of 120⁰ each.

@ravi Could I request you for the source, please? Really nice problem!

Some clarifications [I dont want to go through the trouble of editing the post and then removing all the line breaks :( ]

1. In the 3rd para I missed writing f'(x) >0 for all x as the the second possibility.

2.In the lemma that inequality holds when f(x) is concave. The opposite is true for a convex function.

This is a nice question.

We are given f(x) f

The first important thing to note here is this immediately means $f(x) \ne 0$ . This isnt a trivial fact as combined with the fact that f(x) is differentiable everywhere and hence continuous everywhere this implies that either $f(x) > 0 \ \forall x$ or $f(x) < 0 \ \forall x$ .i.e. it retains the same sign throughout the real line. This is a consequence of the Mean Value Theorem as if f(x) changes sign in some interval we will have some point where f(x) = 0 and that violates the given condition.

Similarly since and f"(x) exists at all x, f'(x) is continuous and hence this again tells us that either or

Thus we have only two cases to deal with. I will demonstrate one and you can easily extend this to the other case.

Case I: f(x)>0 everywhere

Now, we have to prove that f'(x)>0 everywhere.

Let us assume to the contrary that f'(x)<0 for all x.

From the given condition this tells us that f"(x) <0 for all x.

The structure of the proof is now easy to see as f"(x)<0 means that f(x) is concave everywhere. (To get an idea of the shape sin x is a concave function in the interval $(0,\frac{\pi}{2})$ ). If you draw such a curve you will easily see that its got to intersect the x-axis at some large x. And we will prove precisely this.

Now for a lemma:

If we have three points x_1, x_2, x_3 such that x_1 < x_2 < x_3 , then the following relation holds:

$\frac{f(x_1) - f(x_2)}{x_2-x_1} < \frac{f(x_1)-f(x_3)}{x_3-x_1}$

This too can be seen geometrically, but we will supply an easy analytic proof.

First we must note that any concave function has the property that $\omega_1f(x_1)+\omega_2 f(x_2) \le f(\omega_1 x_1+\omega_2 x_2)$ where $\omega_1+ \omega_2 = 1$ (this is, of course, Jensen's Inequality)

So now we can write

$\frac{x_3-x_2}{x_3-x_1} f(x_1)+\frac{x_2-x_1}{x_3-x_1} f(x_3) \le f \left(\frac{x_3-x_2}{x_3-x_1}x_1+\frac{x_2-x_1}{x_3-x_1} x_3} \right) = f(x_2)$

And with a bit of bashing about, we can see that it is equivalent to the statement given above.

Now, let us fix x_1 = a; x_2 = b

So now we have $\frac{f(a)-f(b)}{b-a} < \frac{f(a) - f(x_0)}{x_0-a}$

or $f(x_0) < f(a) - \frac{f(a)-f(b)}{b-a} (x_0-a)$

The only variable here is the quantity (x₀- a) and this can be made arbitrarily large and since f(a) is finite we will have for some large enough x₀, f(x₀) <0 which contradicts our original stipulation that f(x)>0 for all x.

What remains is to prove that f'(x)>0 is feasible.

That would mean that f"(x)>0 i.e. a convex function and all the inequalities reverse and it is easy to see that we will have f(x)>0 for all x and hence there is not contradiction here.

Thus f'(x)>0 for all x is the only possibility

Case 2: f(x) <0 for all x. The argument proceeds on very similar lines proving that f"(x)>0 again leads to f(x)>0 at some x so we must f(x)>0 for some x

Hence we must have f'(x)>0 for all x

Fantastic work jishnu and this forum owes you for this. I had heard about this technique but I had never encountered it before. So, personally, thanks from me.

I guess decoder is alluding to the result that 1^k+2^k+3^k+...+n^k is divisible by $\frac{n(n+1)}{2}$ when k is odd.

Consider the function:

$f(x) = a_1(x-\lambda_2)(x-\lambda_3) + a_2(x-\lambda_3)(x-\lambda_1) + a_3 (x-\lambda_1)(x-\lambda_2)$

We can see that the following are true

$f(\lambda_1) = a_1 (\lambda_1-\lambda_2)(\lambda_1-\lambda_3)>0; \ f(\lambda_2) = a_2 (\lambda_2-\lambda_3)(\lambda_3-\lambda_1)<0; \\ \\ f(\lambda_3) = a_3 (\lambda_3-\lambda_2)(\lambda_3-\lambda_1)>0$

Thus we see that f(x) has one root each in the intervals $(\lambda_1, \lambda_2), (\lambda_2, \lambda_3)$

Since none of are $\lambda_1, \lambda_2, \lambda_3$ roots, we can divide f(x) by $(x-\lambda_1) (x- \lambda_2)(x- \lambda_3)$ and hence the claim that the given expression has roots in the intervals $(\lambda_1, \lambda_2), (\lambda_2, \lambda_3)$

Just my tuppence worth on this discussion.

While considering the question whether there are naturals a and b such $a \ne b$ and a^b = b^a ,

it may help to rewrite it as $a^{\frac{1}{a}} = b^{\frac{1}{b}}$

It is well known that the the sequence $a_n = n^{\frac{1}{n}}$ behaves as a_1 <a_2<a_3>a_4>a_5>a_6>....

i.e. it monotonically increases up to n = 3 and then monotonically decreases. We know that a₂ = a₄ and further a_i>1 for all i>1. Hence a=2 and b = 4 is the only ordered solution set.

Moving to rationals, $f(x) = x^{\frac{1}{x}}$ monotonically increases up to x = e and is monotonically decreasing for x>e. Which means infinitely many rationals in the interval (1,e) get mapped to rationals in the interval $(e, \infty)$ . So that means infinitely many solutions exist among rationals and it looks quite dense to be covered by a closed form.

It is an adaptation of a quite well known method

$(\sqrt{1996}+44) (\sqrt{1996} - 44) =1996-1936 = 60$ .

Further $(\sqrt{1996}+44) >60 \Rightarrow \sqrt{1996}-44<1$

Let $(\sqrt{1996}+44)^{100} = I+f$ (Obviously LHS is not an integer so f>0)

Obviously $(\sqrt{1996}-44)^{100} = f$

Now consider $(\sqrt{1996}+44)^{100}+(\sqrt{1996}-44)^{100} = I+f+f$

That the LHS is an even integer is easily proved.

That means f+f' = An Integer

and that integer can only be 1 (as f,f'>0).

Thus I+1 is an even integer which makes I an odd integer

It is similar to http://goiit.com/posts/list/algebra-4-diffrnt-objcts-1-2-3-4-r-dstributd-at-random-on-4-82895.htm#397618

Following the arguments there the required number of mappings would be

5! - 5 .4!+⁵C₂ . 3! - ⁵C₃.2!+ ⁵C₄-1 = 44

That was an application of factor theorem, not a short-cut. And you can be sure that the question wasnt given so that you exapnd and arrive at the results.

@rajatsen: In the 1st qn - what allowed you to assume that f is differentiable?

In the same way for the second question you need not take the trouble of expanding the determinant

You will notice that putting a= b the determinant is zero. Similarly setting b=c and c = a too we get det = 0

Hence the triangle need be isoceles for the det = 0 (that covers equilateral too)

$A (x) = \left | \begin{array} {ccc} a^2+x & ab & ac \\ ab & b^2+x & bc \\ ac & bc & c^2+x \end{array} \right |$

$A (0) = \left | \begin{array} {ccc} a^2 & ab & ac \\ ab & b^2 & bc \\ ac & bc & c^2 \end{array} \right | = abc \left | \begin{array} {ccc} a & b & c \\ a & b & c \\ a & b & c \end{array} \right | = 0$

Hence we get

Thus x = 0 is a root of A(x) and A'(x). Hence x is a repeated root of A(x) and is repeated at least twice.

This implies x²|A(x)

I really dont know how to answer this. Although you may be in 12th, your instructors seem to be giving you problems without paying heed to that! In the past you have even brought IMO problems to this forum.

Abel Summation is just another way of writing out the LHS that can easily be adopted in this problem. No special pyrotechnics involved.This will definitely look unfamilar in a JEE forum because we dont do much of inequalities in JEE. So you ask an olympiad type of question and ask a JEE type answer, there is little that can be done. But in an olympiad forum, you give this majorization condition, it becomes easy meat as the Abel Summation technique is well known and it will be done by them without having to exercise their brain at all.( so solving this problem doesnt mean one is a genius. It only means you happened to be familiar with the technique)

Yet again I want to stress, for JEE oriented students, you dont really have to go deep into inequalities, Number Theory etc. Just remember, JEE is looking for Engineers and not mathletes

For Inequalities - practice AM-GM (most useful) and Cauchy Schwarz (very basic). Know well the inequalities concerrning bounds of sin A + sin B + sin C, tan A + tan B + tan C etc. One more is the inequality that arises in Riemann Integral [m(b-a) <= I <= M(b-a) stuff]. This one had come in JEE last year and its easy to spot too.

For Number Theory - Well, Unique Factorisation to Fermat's Little theorem a^p-a is divisible by p is about the range. Knowledge of Wilson's Theorem [(p-1)!+1 is divisible by p ] will not harm you. Get good practice in stuff concerning Euclid's lemma like a square leaves a remainder of 0 or 1 on division by 3, odd squares are always of the form 4k+1 and 8k+1 etc.. These do help in some problems. If you could learn congruences your life will be easier in problems that need you to find remainders. But be thorough with dealing with rational and irrational numbers. You will almost certainly not encounter a problem that deals with NT exclusively, but it will be woven into a problem.

A case in point: A circle is drawn around the point $(0, \sqrt 2)$ . Call a point (x,y) rational if both x and y are rational. Prove that there are at most two rational points on the circle.

(I was told this was a problem that appeared in one of the JEEs, but I cannot confirm that.).

What they will be looking for in-depth is: Coordinate Geometry (just see last year's JEE), Complex Numbers, Calculus (Diff. Equations) and Vectors. These are essential toolkits for any good engineer and thats where they will be looking to test you.

I said this so that all of you can just check your flight paths and if required effect a course correction.

PS: Please dont interpret me to mean that I am trying to dampen your enthusiasm. I am cautioning you guys because I know of enough students who made it big in the olympiad scene but didnt do so well in JEE. If you plan to specialize in a pure sciences subject it is quite alright that you study that subject more. But if you are looking for a engineering career, then be pragmatic and balance your studying so that you are well prepared in M, P and C and particularly in the fields they are looking for.

Which operator in calculus calculates rate of change of y with respect to x?