Category Archives: IITJEE Advanced Mathematics

Triangle inequality and max-min

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


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.


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.


More later,

Nalin Pithwa


Maxima and minima — using calculus — for IITJEE Advanced Mathematics


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.


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+nwhere 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


A Trigonometry problem for IITJEE Advanced Mathematics or Mathematics Olympiad

Ref: Titu Andreescu and Zumin Feng.


Let x, y, z be positive real numbers such that x+y+z=1. Determine the minimum value of



An application of CauchySchwarz inequality makes this as a one step problem. Nevertheless, we present a proof which involves only the easier inequality x^{2}+y^{2} \geq 2xy for real numbers x and y by setting first x=\tan {b} and y=2\tan{b} and second x=\tan{a} and y=\cot{a}.

Clearly, z is a real number in the interval [0,1]. Hence, there is an angle a such that z=\sin^{2} {a}. Then, x+y=1-\sin^{2}{a}=\cos^{2}{a}, or \frac{x}{\cos^{2}{a}}+\frac{y}{\cos^{2}{a}}=1. For an angle b, we have \cos^{2}{b}+\sin^{2}{b}=1. Hence, we can set x=\cos^{2}{a}\cos^{2}{b}, and y=\cos^{2}{a}\sin^{2}{b} for some b, it suffices to find the minimum value of

P = \sec^{2}{a} \sec^{2}{b}+4\sec^{2}{a}\csc^{2}{b}+9\csc^{2}{a}


P = (\tan^{2}{a}+1)(\tan^{2}{b}+1)+4(\tan^{2}{a}+1)(\cot^{2}{b}+1)+9(\cot^{2}{a}+1)

Expanding the right hand side gives

P = 14 + 5\tan^{2}{a}+9\cot^{2}{a}+(\tan^{2}{b}+4\cot^{b})(1+\tan^{2}{a})

\geq 14 + 5\tan^{2}{a}+9\cot^{2}{a}+2\tan{b}.2\cot{b}(1+\tan^{2}{a})

= 18 + 9(\tan^{2}{a}+\cot^{2}{a}) \geq 18 + 9.2\tan{a}\cot{a}=36

Equality holds when \tan{a}=\cot{a} and \tan{b}=2\cot{b}, which implies that \cos^{2}{a}=\sin^{2}{a} and 2\cos^{2}{b}=\sin^{b}. Because \sin^{2}{\theta}+\cos^{2}{\theta}=1, equality holds when \cos^{2}{a}=\frac{1}{2} and \cos^{2}{b}=\frac{1}{3}, that is, x=\frac{1}{6}, y=\frac{1}{3}, z=\frac{1}{2}.

More later.

Nalin Pithwa


Inclusion Exclusion Principle theorem and examples

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


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 <j \leq n}|A_{i} \bigcup A_{j}|- \ldots + \{ (-1)^{k} \sum_{1 \leq i_{1} < i_{2}< \ldots < i_{k} \leq n}|A_{i_{1} \bigcup A_{i_{2}}} \ldots \bigcup A_{i_{k}}|\}+ \ldots+ (-1)^{n}|A_{1} \bigcup A_{2} \ldots \bigcup A_{n}|.


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










Solution to another S L Loney problem for IITJEE Advanced Mathematics

Exercise XXVII. Problem #32. The sides of a triangle are in AP and the greatest and least angles are \theta and \phi. Prove that



Let a, b, c be in AP. Hence, a<b<c, which in turn implies c is greatest and a is the least.

Hence, \theta=\angle{C} and \angle{A}.





(b-a)^{2}=(b-c)^{2}. Hence,








Now, we have 1-\cos{A}=1-\frac{b^{2}+c^{2}-a^{2}}{2bc} and this is equal to the following:


= \frac{2bc-b^{2}+2ab-2bc}{2bc}=\frac{2ab-b^{2}}{2bc}.

Also, similarly, we have the following:



= \frac{b^{2}-2ab+2bc}{2bc}+\frac{b^{2}-2bc+2ab}{2ab}




= \frac{ab+bc-2a^{2}-2c^{2}+4ac}{2ac}





but it is given that b-a=c-b, hence, a+c=2b, a+c-b=b. So the above expression changes to

= \frac{(b+a-c)(c-b+c)}{ac}

= \frac{(2c-b)(a+b-c)}{ac}

= \frac{(2c-b)(a+b-\overline{2b-a})}{ac}

= \frac{(2c-b)(a-b+c)}{ac}

= \frac{(2c-b)(2a-b)}{ac}

= \frac{4ac-2bc-2ab+b^{2}}{ac}

= 4(1-\cos{A})(1-\cos{C}).


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.


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.


which is equal to


which is equal to


which is equal to


which in turn equals


From the above, consider only the expression, given below. We will see what it simplifies to:




=\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.




= \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:


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


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.


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!!

Well, here’s your cryptic complex problem to cudgel your brains!


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}



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


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

More later,

Nalin Pithwa

Announcement: A Full Scholarship Program

We are Mathematics Hothouse, Bangalore, We are pleased to announce that henceforth, every academic year, we will be admitting 5 students with full scholarship or 100% discount, from any part of India, who are talented, deserving or needy, to our program for RMO and INMO coaching. The coaching will be via on-line, live, video interactive Skype sessions mimicking traditional classroom or just classroom coaching or even correspondence course.

If you wish to apply, please write to


Nalin Pithwa

Some properties of continuous functions : continued

We now turn to the question whether f(x_{n}) approximates f(x) when x_{n} approximates x. This is the same as asking: suppose (x_{n})_{n=1}^{\infty} is a sequence of real numbers converging to x, does (f(x_{n}))_{n=1}^{\infty} converges to f(x)?


If f: \Re \rightarrow \Re is continuous at x \in \Re if and only if (f(x_{n}))_{n=1}^{\infty} converges to f(x) whenever (x_{n})_{n=1}^{\infty} converges to x, that is,

\lim_{n \rightarrow \infty}f(x_{n})=f(\lim_{n \rightarrow \infty} x_{n}).


Suppose f is continuous at x and (x_{n})_{n=1}^{\infty} converges to x. By continuity, for every \varepsilon > 0, there exists \delta >0 such that |f(x)-f(y)|<\varepsilon whenever |x-y|<\delta. Since (x_{n})_{n=1}^{\infty} converges to x, for this \delta >0, we can find an n_{0} such that |x-x_{0}|<\delta for n > n_{0}.

So |f(x)-f(x_{n})|<\varepsilon for n > n_{0} as |x_{n}-x|<\delta.

Conversely, suppose (f(x_{n}))_{n=1}^{\infty} converges to f(x) when (x_{n})_{n=1}^{\infty} converges to x. We have to show that f is continuous at x. Suppose f is not continuous at x. That is to say, there is an \varepsilon >0 such that however small \delta we may choose, there will be a y satisfying |x-y|<\delta yet |f(x)-f(y)|\geq \varepsilon. So for every n, let x_{n} be such a number for which |x-x_{n}|<(1/n) and |f(x)-f(x_{n})| \geq \varepsilon. Now, we see that the sequence (x_{n})_{n=1}^{\infty} converges to x. But (f(x_{n}))_{n=1}^{\infty} does not converge to f(x) violating our hypothesis. So f must be continuous at x. QED.

More later,

Nalin Pithwa



Quick Review of Trigonometric Optimization Methods

Let us review together the four general methods we can use for triangular optimization.

I) Trigonometric Method: 

The essence of this method is the observation that the cosine of an angle is at most one and that it equals 1 only when the angle is zero. This fact is applied to the difference between two of the angles A, B and C, holding the third angle fixed, to show that unless those two angles are equal, the objective functionn can be increased (or decreased as the case may be). Consequently, at an optimal solution, these two angles must be equal. If the objective function is symmetric (as is the case, in almost all problems of triangular optimization), then every two of A, B and C must be equal to each other and hence, the triangle ABC must be equilateral.

This method is elementary and easy to apply. Even when the objective function is only partially symmetric, that is, symmetric in two but not in all the three variables, it can be applied to those two variables, holding the third variable fixed. Suppose, for example, that we want to maximize f(A,B,C)=\cos{A}+2\cos{B}+\cos{C}. This is symmetric in A and C. So, by the same reasoning as for maximizing \cos{A}+\cos{B}+\cos{C}, which at an optimal solution we must have A=C. Then, B=\pi-2A, which makes f effectively a function of just one variable, viz., \cos{A}+2\cos(\pi-2A)+\cos{A}, which equals 2\cos{A}-4\cos^{2}{A}+2. This can be maximized as a quadratic in \cos{A} either by completing the square or using calculus. The maximum occurs when A=\cos^{-1}{1/4}. Thus, the maximum value of \cos{A}+2\cos{B}+\cos{C} for a triangle ABC is 1/4. The method, of course, fails if the function is not even partially symmetric. This is not surprising. Basically, in a triangular optimization problem, we are dealing with a function f(A,B,C) of three variables. Because of the constraint A+B+C=\pi, any one of the variables can be expressed in terms of the other two. This effectively makes f a function of two variables. Optimization of functions of several variables requires advanced methods. It is only when f satisfies some other conditions such as partial symmetry that we can hope to reduce the number of variables further so that elementary methods can be applied.

II) Algebraic Method: 

The essence of this method is to reduce the optimization problem to some inequality using suitable trigonometric formulae or identities. The inequality is then established using some standard inequality such as the AM-GM-HM inequality, or Jensen’s inequality, or sometimes, by doing some more basic work. The fundamental ideas are very simple, viz., (a) the square of any real number is non-negative and is zero only when that number is zero, and (b) the sum of two or more non-negative numbers is non-negative and vanishes if and only if each of the term is zero. When this method works, it works elegantly. But it is not always easy to come up with the right algebraic manipulations. Sometimes, certain simplifying substitutions have to be used. Still, it is an elementary method and deserves to be tried.

III) Jensen’s inequality:

This is a relatively advanced method. It is directly applicable when the objective function is, or can be recast, in a certain form, viz., h(A)+h(B)+h(C), where h is a function of one variable whose second derivative maintains the same sign over a suitable interval. But, even when h fails to do so, the method can sometimes be applied with a suitable conversion of the problem.

IV) Lagrange’s Multipliers:

This is a highly advanced method based on the calculus of functions of several variables. It is applicable to all types of objective functions, not just those that are symmetric or partially symmetric. When applied to triangular optimization problems with symmetric objective functions, the optimal solution is either degenerate or an equilateral triangle.

Naturally, for a particular given problem, some of these methods may work better than others. The method of Lagrange’s multipliers is the surest but the most mechanical of all the four. The algebraic method is artistic and sometimes gives the answer very fast. Jensen’s inequality also works fast once you are able to cast the objective function in a certain form. Such a recasting may involve some ingenuity sometimes. The trouble is that both these methods work only in the case of an  optimization problem where the objective function is symmetric. And, in such cases, the method of Lagrange’s multipliers makes a mincemeat of the problem. From an examination point of view, this is a boon if a question about triangular optimization is asked in a “fill in the blanks” or “multiple choice” form, where you don’t have to show any reasoning. if the objective function is symmetric, then the optimal solution is either degenerate or an equilateral triangle. But, degenerate triangles are often excluded from the very definition of a triangle because of the requirement that the three vertices of a triangle must be distinct and non-collinear and, in any case, such absurdities are unlikely to be asked in an examination! So, it is a safe bet to simply assume that the optimal solution is an equilateral triangle and proceed with further work (namely, calculating the value of the objective function for an equilateral triangle). This saves you a lot of time.

More later,

Nalin Pithwa