Chebyshev plus one

 A puzzle inspired by a recent problem from the JEE 2026 advanced math paper.

Find the value of 

$$ \prod_{k=0}^{4} \left(1 - \sqrt{2} \cos \left(\frac{(2k+1)\pi}{11}\right)\right)^2 $$

Scroll down for a solution.

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The actual value is $$ \frac{(\sqrt{2} - 1)^2}{32}$$

We will prove the following identity:

$\textbf{Proposition:}$

$$1 + \cos ((2n+1) \theta) = 2^{2n} (1 + \cos \theta) \prod_{k=0}^{n-1} \left( \cos \theta - \cos \left(\frac{(2k+1)\pi}{2n+1}\right)\right)^2$$

Setting $\theta = \frac{\pi}{4}$ and $n = 5$ in the above solves the puzzle.

So let's try to prove the above identity.

$\textbf{Proof:}$

We first make the following claim:

$\textbf{Lemma:}$ There is a polynomial $T$ of degree $2n+1$ and leading coefficient $2^{2n}$ such that $T(\cos \theta) = \cos (2n+1)\theta$

$\textbf{Proof of Lemma:}$ This is easily proved using complex numbers and DeMoivre's formula. So we will skip it.

Now $\cos (2n+1)\theta = -1$ for $\theta = \frac{(2k+1)\pi}{2n+1}$, $k = 0 \dots n$

Differentiating $T(\cos \theta) = \cos (2n+1)\theta$ we get

$$T'(\cos \theta) \sin \theta = (2n+1) \sin (2n+1)\theta$$

Thus for $k = 0, \dots n-1$, we have that $T'\left(\cos \frac{(2k+1)\pi}{2n+1}\right) = 0$

Thus the $\cos \frac{(2k+1)\pi}{2n+1}$ are double roots of $T(x) + 1 = 0$, for $k = 0 \dots n-1$. Since the leading coefficient is $2^{2n}$, we have the factorization of $T + 1$

$$T(x) + 1= 2^{2n} (x + 1) \prod_{k=0}^{n-1} \left(x - \cos \left(\frac{(2k+1)\pi}{2n+1}\right) \right)^2$$

The $x+1$ factor corresponds to $k = n$, and the rest correspond to the double roots for $k=0, \dots, n-1$, totaling $2n+1$ in number, which is the degree of $T$.

Setting $x = \cos \theta$ gives us the identity we wanted to prove.

An arithmetic limit that gives the geometric mean

 Let $a_1, a_2, \dots a_k$ be positive reals.

Show that

$$ \lim_{n \to \infty} \left(\dfrac{a_1^{\frac{1}{n}} + \dots + a_k^{\frac{1}{n}}}{k}\right)^n = \sqrt[k]{a_1 a_2 \dots a_k}$$

Scroll down for a solution that does not use "calculus".

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A repeated use of (weighted) AM $\ge$ GM does it. 

Using weighted AM-GM we have that

$$\dfrac{x_1 \times \frac{1}{x_1} + \dots + x_k \times \frac{1}{x_k}}{x_1 + \dots + x_k} \ge \sqrt[x_1 + \dots + x_k] { \frac{1}{x_1^{x_1}} \dots \frac{1}{x_k^{x_k}}}$$

And so

$$ x_1^{\frac{x_1}{S}} \dots x_k^{\frac{x_k}{S}} \ge \dfrac{x_1 + \dots + x_k}{k}$$

Where $S = x_1 + \dots + x_k$.

And so we have the following (by adding the classic AM-GM on the right) and taking the $n^{th}$ powers

$$\left(x_1^{\frac{x_1}{S}} \dots x_k^{\frac{x_k}{S}}\right)^n \ge \left(\dfrac{x_1 + \dots + x_k}{k}\right)^n \ge \left(\sqrt[k]{x_1 \dots x_k}\right)^n$$

Now set $x_j = a_j^{\frac{1}{n}}$ (and so $x_j^n = a_j$)

Now as $n \to \infty, \frac{x_j}{S} \to \frac{1}{k}$.

Therefore 

$$\left(x_1^{\frac{x_1}{S}} \dots x_k^{\frac{x_k}{S}}\right)^n = a_1^{\frac{x_1}{S}} \dots a_k^{\frac{x_k}{S}} \to \sqrt[k]{a_1 \dots a_k}$$

We also have

$$\left(\sqrt[k]{x_1 \dots  x_2}\right)^n =  \sqrt[k]{a_1 \dots a_k}$$

Thus $$ \left(\dfrac{a_1^{\frac{1}{n}} + \dots + a_k^{\frac{1}{n}}}{k}\right)^n = \left(\dfrac{x_1 + \dots + x_k}{k}\right)^n \to \sqrt[k]{a_1 \dots a_k}$$

by the sandwich theorem and we are done.

Maximizing a trigonometric expression (IIT JEE prep question with cute solutions)

Found this cute problem in some IIT JEE prep material. The question as posed there, was a single answer multiple choice question, to find the maximum value. Anyway the puzzle here is:

For what $x \in [0, 2\pi]$ does the expression

$$ \cos x (\sin x + \sqrt{\sin^2 x + 1})$$

take the maximum value? What is the maximum value? No calculus allowed.

Scroll down for solutions.

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$\textbf{Method 1}$: One nice method is to use vector algebra. Consider the formula:

$$ u.v = |u | |v| \cos \theta$$

with $u = (\sin x, \cos x),  v= (\cos x, \sqrt{\sin^2 x + 1})$. 

The expression in question (say $f$) is $f = u.v$ with $|u| = 1$ and $|v| = \sqrt{2}$. Thus  $f \leq \sqrt{2}$. $f$ will be $ = \sqrt{2}$ iff $u$ and $v$ are parallel. And so we are looking for solutions of

$$\sin x \sqrt{\sin^2 x + 1} = \cos^2 x$$

Since $\cos x = 0$ cannot be a solution, we can divide by $\cos^2 x$ and rewrite as

$$\tan x \sqrt{\tan^2 x + \sec^2 x} = 1$$

If $t = \tan x$, then this is 

$$ t \sqrt{2t^2 + 1} = 1$$

Which is a quadractic in $t^2$ with the solution $ t  = \frac{1}{\sqrt{2}}$.

Thus $x$ must satisfy $\tan x = \frac{1}{\sqrt{2}}$. We can verify that it is indeed a solution.

$\textbf{Method 2}:$ This is also a nice solution.

Let $$f = \cos x (\sin x + \sqrt{\sin^2 x + 1})$$

Clearly $\cos x = 0$ cannot be it. So assume $\cos x \neq 0$.

Now this looks quite similar to formula for the root of a quadratic.

$$ f = \frac{ - ( -\sin x) + \sqrt{(-\sin x)^2 - 4 \left(\frac{\sec x}{2}\right) \left(\frac{-\cos x}{2}\right)}}{2 \times\frac{\sec x}{2}}$$

Thus $f$ is the root of $ax^2 + bx + c$ with $a = \frac{\sec x}{2}, b = -\sin x, c = \frac{-\cos x}{2}$.

Thus we have that

$$ \frac{\sec x}{2} f^2 - f \sin x - \frac{\cos x}{2} = 0$$

Multiplying by $2\sec x$ gives us

$$ \sec^2 x f^2 - 2f \tan x - 1 = 0$$

i.e

$$f^2 + \tan^2 x f^2 - 2f \tan x = 1$$

$$ f^2 + (f \tan x - 1)^2 = 2$$

Since $(f\tan x -1)^2 \ge 0$, we must have $f^2 \le 2$. For $f^2 = 2$ (and so $f = \sqrt{2}$) we need $f\tan x = 1$, ie $\tan x = \frac{1}{f} = \frac{1}{\sqrt{2}}$. This is easily verified.

Colouring grid points (problem by Edgar Dijkstra)

 A problem due to Edgar Dijkstra.

You have an $N \times N$ grid of squares  and you have some number of squares on the grid marked as "special" squares. Show that you can colour each of the special squares red or blue, such for any row or column, the number of red squares is within one of the number of blue squares.

Scroll down for a solution.

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This is a graph theory problem. 

You can view the grid as a bipartite graph, with the rows being the left set and the columns being the right set. There will be an edge between row $r$ and column $c$ iff the intersection of those is a special square. Thus it is enough to prove that we can colour the edges of a bipartite graph red or blue so that for each vertex the red degree is within one of the blue degree.

We prove by induction on the number of edges.  

Base cases are easy to verify.

If the graph is a tree (or a forest), pick any leaf node (vertex of degree one) and delete the edge that connects it to its tree. The rest can be coloured by induction. We can now add this edge back and colour it appropriately, without violating the constraint at the vertices where it is connected.

If the graph is not a tree/forest, it has a cycle. Since it is a bipartite graph, this cycle must be of even length. We alternately colour the edges of this cycle red and blue. Now colour the rest of the edges inductively.

A trigonometric problem from IIT JEE 1995

 The IIT JEE 1995 subjective maths paper had some nice problems. Already posted an integral, previously.

Here is a trigonometric one.

Find the smallest positive $\rho$ such that the equation in $x$ $$\cos (\rho \sin x) = \sin (\rho \cos x)$$ has a solution in $[0, 2\pi]$.

Scroll down for a solution.

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Let's first try to find some $\rho$. Notice that for $x = \frac{\pi}{4}$  to be a solution we need $$\cos \left(\frac{\rho}{\sqrt{2}}\right) = \sin\left(\frac{\rho}{\sqrt{2}}\right)$$

We can try $$\frac{\pi}{2} - \frac{\rho}{\sqrt{2}} =  \frac{\rho}{\sqrt{2}}$$

giving us $$\rho = \frac{\pi}{2\sqrt{2}}$$

 Thus the answer we seek must $ \leq \frac{\pi}{2\sqrt{2}}$.

Now if $\cos x = \cos y $ then we must have that $x = 2n\pi \pm y$ and so $\cos (\rho \sin x) = \sin (\rho \cos x)$ gives us

$$\rho \sin x = 2n\pi \pm \left(\frac{\pi}{2} - \rho \cos x\right)$$

i.e

$$\rho (\sin x \pm \cos x) = 2n \pi \pm \frac{\pi}{2}$$

i.e $$ \sqrt{2} \rho \sin (x \pm \frac{\pi}{4}) = 2n\pi \pm \frac{\pi}{2}$$

Since $0 < \rho \le \frac{\pi}{2\sqrt{2}}$  and $|\sin \theta| \leq 1$, we must have that $n = 0$ and

$$\sqrt{2} \rho |\sin (x \pm \frac{\pi}{4})| = \frac{\pi}{2}$$

This implies that $ \rho \geq \frac{\pi}{2\sqrt{2}}$.

Thus the least possible value is $\rho = \frac{\pi}{2\sqrt{2}}$.

An integral from IIT JEE 1995 entrance exam

 Back in the 20th century, the IIT JEE entrance exam used to have some pretty good subjective questions.

 Unfortunately, the format now is objective questions only, and these days it feels like the students in the coaching classes are taught to game the exams rather than learn the concepts: a regrettable outcome of removing the subjective questions. While one could argue that there are issues with having subjective questions, has it really been worthwhile?

Anyway, this is a cute integral problem from the 1995 JEE maths entrance exam.

Prove, using induction, that

$$\int_{0}^{\pi} \dfrac{1 - \cos mx} {1 - \cos x} \text {  dx} = m \pi$$

Scroll down for solutions.

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1) $\textbf{A fully elementary induction based solution:}$ Everything in this solution falls in the syllabus for IIT JEE and we could expect students to be able to come up with this during the exam.

Let $$I_m = \int_{0}^{\pi} \dfrac{1 - \cos mx} {1 - \cos x} \text {  dx}$$

It is easy to verify that $I_m = m \pi$ for $m = 0, 1$.

Now $$I_{m+1} - I_m = \int_{0}^{\pi} \dfrac{\cos mx - \cos (m+1)x}{1 - \cos x}$$

Now using the formula $\cos 2A - \cos 2B = 2 \sin(A+B)\sin(B-A)$ and $1 - \cos x = 2 \sin^2 \left(\frac{x}{2}\right)$we get

$$I_{m+1} - I_m = \int_{0}^{\pi} \dfrac{2\sin ((m + \frac{1}{2})x) \sin \frac{x}{2}}{2\sin^2 \left(\frac{x}{2}\right)} = \int_{0}^{\pi} \dfrac{\sin ((m + \frac{1}{2})x)}{\sin \frac{x}{2}}$$

[A more knowledgeable reader will recognize the Dirichlet Kernel and cut short the rest of the proof]

Making the substitution $x = 2u$ gives us

$$ I_{m+1} - I_m = 2 \int_{0}^{\frac{\pi}{2}} \dfrac{\sin ((2m+1)u)}{\sin u} \text{ du}$$

Let $$J_m = \int_{0}^{\frac{\pi}{2}} \dfrac{\sin ((2m+1)u)}{\sin u} \text{ du}$$

Note that $2J_0 = \pi$. 

Now we have that

$$J_{m} - J_{m-1}= \int_{0}^{\frac{\pi}{2}} \dfrac{\sin ((2m+1)u) - \sin ((2m-1)u)}{\sin u} \text{ du}$$

Now using $\sin(A+B) - \sin(A-B) = 2 \cos A \sin B$ gives us

$$J_{m} - J_{m-1} = \int_{0}^{\frac{\pi}{2}} \dfrac{2 \cos (2mu) \sin u}{\sin u} = \int_{0}^{\frac{\pi}{2}} 2 \cos (2mu) =   0$$

Thus $J_m = J_0$, and so $I_{m+1} - I_{m} = 2J_m = 2J_0 = \pi$ and so $I_{m} = m \pi$ by induction.

2) $\textbf{Elementary proof 2:}$  Show that $I_{m+1} + I_{m-1} = 2I_m$.

3) $\textbf{Elementary proof 3:}$ Show that:

$$ \frac{1 - \cos mx}{1 - \cos x} =  m + 2 \sum_{k=1}^{m} (m-k) \cos kx$$

This basically comes from the first proof (or even the second), or we can use complex numbers etc.

Note that this formula actually allows you to compute the indefinite integral.

4) $\textbf{Possibly advanced proof:}$ Looks like the function is just the Fejer Kernel (multiplied by $m$) and result is immediate: we are convolving it with $\eta(x) = 1$ and the result is the cesaro summation of $\eta$ evaluated at $0$. Of course, the proofs around the Fejer Kernel etc are taken for granted and this could possibly be a circular argument.