**This contest has ended on 3rd september**

**TOPICS**

Relation and function, matrix and determinants, vector, system of equations, sequence and series, permutation and combination, binomial theorem, complex numbers, polynomials, inequalities .

**Rules**

I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

Please make a substantial comment.

Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 24 hours. Then, you must post the solution and you have the right to post a new problem.

If the one who solves the last problem does not post a new problem in 2 hours, the creator of the previous problem has the right to post another problem.

It is NOT compulsory to post original problems. But make sure it has not been posted on Brilliant.

If a diagram is involved in your problem please make sure it is drawn by a computer program.

Format your solution in \(\LaTeX\), picture solution will be accepted but only picture will not be, i.e. you can use picture for diagram or something, but not for the complete solution. Also make sure your solution is detailed and make sure to proof all claims.

Do not post problems on same topic frequently like 3 continuous inequality problem are not allowed. 2 are sufficient.

**For those who are on slack**Please post in #general that new problem is up along with problem number and the link to contest (https://brilliant.org/discussions/thread/algebra-contest/?sort=new)Handwritten picture solution are accepted only when they are easily understandable!!

## Comments

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TopNewestProblem 22Passed on from Jesse Nieminen, since he didn't wish to post.

Let \(a, b, c \in \mathbb{R}\) be real numbers such that \(a+b+c>0\). Prove the inequality

\[a^3+b^3+c^3 \leq (a^2+b^2+c^2)^{\frac32} + 3abc\] – Sharky Kesa · 6 months ago

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\(\left(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\right)\left(a^2 + b^2 + c^2 - ab - bc - ca\right)^2 \leq \left(a^2 + b^2 + c^2\right)^3\)

is true because if we substitute \(x = a^2 + b^2 + c^2, y = ab + bc + ca\), we get \(\left(x+2y\right)\left(x-y\right)^2 \leq x^3 \Leftrightarrow 2y \leq 3x \Leftrightarrow 2ab + 2bc + 2ca \leq 3a^2 + 3b^2 + 3c^2 \Leftrightarrow a^2 + b^2 + c^2 + \left(a-b\right)^2 + \left(b-c\right)^2 + \left(c-a\right)^2 \geq 0\) – Jesse Nieminen · 6 months ago

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PROBLEM 1Find the number of ordered pairs of real numbers (a,b) such that

\((a+bi)^{2002}=a-bi\)

Here \(i\)=\(\sqrt {-1}\) – Prince Loomba · 6 months, 2 weeks ago

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SOLUTION TO PROBLEM 1let \[a+bi=e^{ix}\] then \[e^{2002ix}=e^{-ix}\] and \[e^{2003ix}=1\] so all te possible \(a+bi\) are 2003th roots of unity . we know from here that \(a+bi\) has 2003 diffent solutions and for each solution we get a different ordered pair for \((a,b)\) and hence the number of ordered pairs \((a,b)\) is \[\boxed{2003}\] – Aareyan Manzoor · 6 months, 2 weeks agoLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

Right. Post nextLog in to reply

– Archit Agrawal · 6 months, 2 weeks ago

No, answer should be 2004 as (0,0) is also a solution.Log in to reply

Problem 17Find real numbers \(a\), \(b\) and \(c\) such that

\[|ax + by + cz| + |bx + cy + az| + |cx + ay + bz| = |x| + |y| + |z|\]

for all real numbers \(x\), \(y\) and \(z\). – Sharky Kesa · 6 months, 1 week ago

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– Archit Agrawal · 6 months, 1 week ago

Unordered triplets are (1,0,0) and (-1,0,0).Log in to reply

– Prince Loomba · 6 months, 1 week ago

Post solution tooLog in to reply

– Archit Agrawal · 6 months, 1 week ago

Put x,y=0, z=1 from this we get |a|+|b|+|c|=1. Then put x=y=z from this we get |a+b+c|=1. So a,b,c must be if same sign or 0. 1st case none of them is 0. Take x a very large negative number and x,y be very small positive number, the LHS mod will be opened with negative sign. The equation must be true for all x,y,z which is not happening. Do same in case of 1 of them being 0, you will see it is also not possible. So 2 of them must be 0 and other must be 1 or -1.Log in to reply

– Sharky Kesa · 6 months, 1 week ago

Yes! You have got the answer!Log in to reply

PROBLEM 9Let \(a>0\) be a real number and \(f (x)\) a real function defined on all R satisfying for all x belong to R

\(f (x+a)=\frac {1}{2}+\sqrt {f (x)-{f (x)}^{2}}\)

Prove that \(f (x)\) is periodic and give an example for \(f (x)\) when \(a=1\). – Prince Loomba · 6 months, 2 weeks ago

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– Archit Agrawal · 6 months, 2 weeks ago

Replace x by x+a, 2 times you will find that f(x+3a)=f(x+a). Hence it is periodic with period 2a. Eg. f(x)=(2+√6)/4.Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

Absolutely right. Noww post nextLog in to reply

Problem 26Can someone please post a solution to my problems!!!

Prove for positive \(x\), \(y\) and \(z\)

\[\sqrt{x^2+y^2-xy}+\sqrt{y^2+z^2-yz} \geq \sqrt{z^2+x^2+zx}\] – Sharky Kesa · 5 months, 3 weeks ago

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Now by law of cosines, we can let AP = x,BP = y and CP = z.

The terms in the inequality represents side lengths of triangle ABC and thus the inequality is true by triangle inequality. – Harsh Shrivastava · 5 months, 3 weeks ago

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– Prince Loomba · 5 months, 3 weeks ago

Post easy, I will solve hahaLog in to reply

– Sharky Kesa · 5 months, 3 weeks ago

This is very, very simple. You just have to think outside the box, and you will get it.Log in to reply

\(\text{Problem 5:}\)

Find all functions \(f:{\rm Z}\to{\rm Z}\) such that \(\displaystyle f(m+n)+f(m)f(n)=f(mn+1)\) where \(m,n\) are integers. – Aditya Narayan Sharma · 6 months, 2 weeks ago

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– Prince Loomba · 6 months, 2 weeks ago

I got a weird thing. All of the functions of the type \(x^{2n}-1\) are satisfying according to me. Please check.Log in to reply

– Aditya Narayan Sharma · 6 months, 2 weeks ago

hmm ri8 but there are some complicated functions involving mod also. anyways post the solution otherwaysLog in to reply

– Sharky Kesa · 6 months, 2 weeks ago

Mod function: (Alternatively the trig function) \(f(x) = 0\) if \(x\equiv 1, 3 \pmod{4}\), \(f(x) = 1\) if \(x \equiv 2 \pmod{4}\), \(f(x) = -1\) if \(x \equiv 0 \pmod{4}\).Log in to reply

– Aditya Narayan Sharma · 6 months, 2 weeks ago

ya those are the solutions exactly. you may post the solutionLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

I am just saying this. I did not say these are all.Log in to reply

– Harsh Shrivastava · 6 months, 2 weeks ago

Is the answer only 1? (f(x) = 0)Log in to reply

– Aditya Narayan Sharma · 6 months, 2 weeks ago

NO, \(f(x)=0\) is trivial, And more functions are there.Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

x-1 is alsoLog in to reply

– Aditya Narayan Sharma · 6 months, 2 weeks ago

Yes one of them, you need to show it other than trial and error.Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

Is there any mistakes I made?Log in to reply

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– Aditya Narayan Sharma · 6 months, 2 weeks ago

yes it will work , i wll ppst other functions later . for now you may post the next questionLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

There are infinite functions I can makeLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

f (x)=[x]-1 (floor function)Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

One more f (x)={x} fractional part. 6 are doneLog in to reply

f (x)=sin (n×pi×x) – Prince Loomba · 6 months, 2 weeks ago

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– Aditya Narayan Sharma · 6 months, 2 weeks ago

Firstly the floor is meaningless here since it's a function from integers to integers. And moreover you are not supposed to guess functions, you have to show the derivation. Otherwise it's not a solutionLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

Two functions f (x)=0 and f (x)=x-1Log in to reply

– Aditya Narayan Sharma · 6 months, 2 weeks ago

There are 6, excluding the zero functionLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

\(x^{2}-1\)Log in to reply

– Aditya Narayan Sharma · 6 months, 2 weeks ago

see, dont check your answers then the problem is meaningless . i would suggest that if you hve gor the functions post your solution and i wll cmmnt thereLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

|x|-1 also satisfies. I am just checking by graph which can hold and verifying my answerLog in to reply

Problem 4Let \(a,b\) lie in first quadrant and let \(x\) be an acute angle.

Given that (\(acosx)^2 +(bsinx)^2 = [(sin2x)(\dfrac{a+b}{2} )]^2 \)

Find the value of tanx. – Harsh Shrivastava · 6 months, 2 weeks ago

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which in turn shows \(\displaystyle |AB|^2+|BC|^2=|CA|^2\) and so \(\displaystyle B=\frac{\pi}{2}\).

Now, the circumcentre of a right angled triangle lies at the mid-point of it's hypotenuse and so the co-ordinates of the circumcenter is \(\displaystyle (M,N)=(\frac{a\csc x}{2},\frac{b\sec x}{2}) \)

From the relation we have , \(\displaystyle a^2\cos^2 x+b^2\sin^2 x=\sin^2 x\cos^2 x(a+b)^2\)

\(\displaystyle \implies (a\cos^2 x)^2+(b\sin^2 x)^2-2ab\sin^2 x\cos^2 x =0 \implies a\cos^2 x=b\sin^2 x\)

So we have , \(\displaystyle \csc x = \sqrt{\frac{a+b}{a}} , \sec x=\sqrt{\frac{a+b}{b}}\)

So the circumcenter is : \(\displaystyle (\frac{\sqrt{a^2+ab}}{2},\frac{\sqrt{b^2+ab}}{2}) \) – Aditya Narayan Sharma · 6 months, 2 weeks ago

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@Aditya Sharma but sorry I rephrased the problem so @Prince Loomba will decide who will post the next problem. – Harsh Shrivastava · 6 months, 2 weeks ago

CorrectLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

You post the next problem. The original was this onlyLog in to reply

@Harsh Shrivastava , you can't change the problem for the reason someone can't do or have problems with that, the solution is done. Let the problem be back to it's original state – Aditya Narayan Sharma · 6 months, 2 weeks ago

WaitLog in to reply

– Harsh Shrivastava · 6 months, 2 weeks ago

You post the next prob.Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

Post the nextLog in to reply

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– Prince Loomba · 6 months, 2 weeks ago

Meaning of circumvented?Log in to reply

– Harsh Shrivastava · 6 months, 2 weeks ago

I meant circumvented damn autocorrect.Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

I got \(tanx\)=\(\sqrt {a/b}\). Idk how to find circumcentreLog in to reply

– Harsh Shrivastava · 6 months, 2 weeks ago

Done post the solution.Log in to reply

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– Harsh Shrivastava · 6 months, 2 weeks ago

Lemme rephrase the problem.Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

Circumvented? And is this in the topic list at the starting of note?Log in to reply

If I tell the name of the topic, the problem will be trivial :) – Harsh Shrivastava · 6 months, 2 weeks ago

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Problem 3Let \(a,b,x,f (x)\) be positive integers such that when \(a>b\), \(f (a)>f (b)\).

If \(f (f (x))=x^{2}+2\), find the value of \(f (3)\) – Prince Loomba · 6 months, 2 weeks ago

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– Prince Loomba · 6 months, 2 weeks ago

Post next soonLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

Its right. Post nextLog in to reply

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– Prince Loomba · 6 months, 2 weeks ago

I want solution.Log in to reply

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– Prince Loomba · 6 months, 2 weeks ago

Yeah absolutely correctLog in to reply

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– Prince Loomba · 6 months, 2 weeks ago

Are you posting?Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

Read point 11 in the noteLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

Handwritten will work. The only condition is it should be understandable!Log in to reply

problem 2given the roots of \[y^8-y^7+y^6-1=0\] are \(y_1,y_2,...,y_7,y_8\) find the value of \[\dfrac{1}{y_1^9}+\dfrac{1}{y_2^9}+...+\dfrac{1}{y_7^9}+\dfrac{1}{y_8^9}\] – Aareyan Manzoor · 6 months, 2 weeks agoLog in to reply

Solution to Problem 2\[y^8 - y^7 + y^6 - 1 = 0\]

Now we have to form a polynomial whose roots are \(\frac{1}{y_{1}}, .....\)

Now, let \(x = \frac{1}{y}\)

then \(y = \frac{1}{x}\)

Putting it in the above equation, we get

\[x^8 + x^2 - x + 1 = 0\]

\[x^8 = x - x^2 - 1\]

\[x^9 = x^2 - x^3 - x ...(1)\]

Putting \(x = \frac{1}{y_{1}},..., \frac{1}{y_{8}}\) repeatedly in the equation and add them all and using bit Newton Sum (\(S_{1} = S_{2} = S_{3} = 0\))

And answer is \(0\). – Prince Loomba · 6 months, 2 weeks ago

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– Aareyan Manzoor · 6 months, 2 weeks ago

you are right.Log in to reply

– Harsh Shrivastava · 6 months, 2 weeks ago

Post the next question.Log in to reply

– Kaustubh Miglani · 6 months, 2 weeks ago

Is thE Answer 0?Log in to reply

Problem 25Prove \((a_1+b_1)(a_2+b_2)\ldots (a_n+b_n) \leq (a_1+b_{\sigma(1)})(a_2+b_{\sigma(2)})\ldots (a_n+b_{\sigma(n)})\) where \(\{a\}\) and \(\{b\}\) are similarly ordered sequences and \(\sigma\) denotes a permutation. – Sharky Kesa · 5 months, 4 weeks ago

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– Prince Loomba · 5 months, 3 weeks ago

Time is up bro..Log in to reply

– Prince Loomba · 5 months, 4 weeks ago

Permutations are always greater than or equal to 1 so each bracket of lhs is smaller than corresponding bracket of rhsLog in to reply

Furthermore, even if permutation is defined as what you think it was, you're wrong. What if \(b\) was negative? – Sharky Kesa · 5 months, 4 weeks ago

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Problem 24Prove \(2^{1/x} < \dfrac {x+1}{x}\) for \(x>1\). – Sharky Kesa · 6 months ago

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\[\begin{align} 2^{1/x} &< \dfrac {x+1}{x}\\ 2x^x &< (x+1)^x\\ 2x^x &< x^x + x \cdot x^{x-1} + \{\text{ Stuff }\}\\ 0 &< \{\text{Stuff}\} \end{align}\]

Since the value of \(\text{Stuff}\geq 1\), the final inequality must be true, so the original inequality is true. Thus, proven. – Sharky Kesa · 5 months, 4 weeks ago

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– Jesse Nieminen · 5 months, 4 weeks ago

I was waiting for someone else to post a solution. I solved this in first 5 minutes.Log in to reply

– Jesse Nieminen · 6 months ago

Counter-example: \(x=50\).Log in to reply

– Sharky Kesa · 6 months ago

LOL, sorry, the other way!Log in to reply

– Prince Loomba · 5 months, 4 weeks ago

Time elapsed post nextLog in to reply

Problem 23Since Jesse Nieminen didn't post, my turn again!

Let \(a,b,c,d\in\mathbb{R}^{+}\) and \(a+b+c+d+abcd=5\). Prove

\[\dfrac {1}{a} + \dfrac {1}{b} + \dfrac {1}{c} + \dfrac {1}{d} \geq 4\] – Sharky Kesa · 6 months ago

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– Prince Loomba · 6 months ago

Post next. Time elapsedLog in to reply

– Jesse Nieminen · 6 months ago

I have a solution using Lagrange Multipliers :PLog in to reply

– Sharky Kesa · 6 months ago

Just write it. My solution is Lagrange Multipliers as wellLog in to reply

– Jesse Nieminen · 6 months ago

I found a solution using AM-GM inequality. I will post it in few minutes if my battery doesn't run out.Log in to reply

@Sharky Kesa :D – Jesse Nieminen · 6 months ago

This competition is now me solving or not solving problems byLog in to reply

Problem 21Let \(a,b,c\) be positive real numbers such that \(27ab + 2bc + 18ca = 81\). Prove that

\[\dfrac1a + \dfrac2b + \dfrac3c \geq 3\] – Jesse Nieminen · 6 months, 1 week ago

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– Jack Lam · 6 months ago

According to Mathematica, the inequality is stricter than the one you have provided us with. Mathematica states that the expression is greater than or equal to \( \frac{7\sqrt{7}}{3\sqrt{3}} \), which is approximately 3.56Log in to reply

– Prince Loomba · 6 months ago

You can post nextLog in to reply

Here's the solution:\(\begin{align}\dfrac 1a + \dfrac 2b + \dfrac 3c &= \dfrac 1{2a} + \dfrac 1{2b} + \dfrac 3{2b} + \dfrac 9{4c} + \dfrac 3{4c} + \dfrac 1{2a} \\ &\geq 2\sqrt{\dfrac 1{2a} \cdot \dfrac 1{2b}} + 2\sqrt{\dfrac 3{2b} \cdot \dfrac 9{4c}} + 2\sqrt{\dfrac 3{4c} \cdot \dfrac 1{2a}} \\ &= \sqrt{\dfrac 1{ab}} + \sqrt{\dfrac {27}{2bc}} + \sqrt{\dfrac{3}{2ca}} \\ &\geq 3 \cdot \sqrt[3]{\sqrt{\dfrac 1{ab}} \cdot \sqrt{\dfrac {27}{2bc}} \cdot \sqrt{\dfrac{3}{2ca} }} \\ &= 3 \cdot \sqrt{\sqrt[3]{\dfrac 1{ab}} \cdot \sqrt[3]{\dfrac {27}{2bc}} \cdot \sqrt[3]{\dfrac{3}{2ca}}} \\ &\geq 3 \cdot \sqrt{\dfrac{3}{ab + \dfrac{2bc}{27} + \dfrac{2ca}{3} }} \\ &= 3 \cdot \sqrt{\dfrac{81}{27ab + 2bc + 18ca }} \\ &= 3. \ \square \end{align}\) – Jesse Nieminen · 6 months ago

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– Sharky Kesa · 6 months ago

I did not see that approach at all! Great job!Log in to reply

– Prince Loomba · 6 months ago

Please post nextLog in to reply

– Jesse Nieminen · 6 months ago

I give my right to post to everyone.Log in to reply

Problem 20Let \(x_1, x_2, \ldots, x_n\) be the roots of the polynomial \(X^n + X^{n-1} + \ldots + X + 1\). Prove that

\[\displaystyle \sum_{k=1}^n \dfrac {1}{1-x_k} = \dfrac {n}{2}\] – Sharky Kesa · 6 months, 1 week ago

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After multiplying out the products we get \(\displaystyle \dfrac{n \times 1 - \left(n-1\right)\times\left(x_1 + \cdots + x_n\right) + \cdots - \left(-1\right)^n \times1\times\left(x_1x_2\cdots x_{n-1} + \cdots + x_2\cdots x_n\right)}{1 - \left(x_1+\cdots+x_n\right) + ... + \left(-1\right)^n \times \left(x_1x_2\cdots x_n\right)}\)

After using Vieta's formulas we get \(\dfrac{\dfrac{n\left(n+1\right)}2}{n+1} = \dfrac n2\).

\(Q.E.D\) – Jesse Nieminen · 6 months, 1 week ago

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– Sharky Kesa · 6 months, 1 week ago

Good solution. You can post the next problem, but see if you can get a simpler, non-expansive solution.Log in to reply

Post the next problem. – Harsh Shrivastava · 6 months, 1 week ago

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Problem 19Let \(a, b, c > 0\) be real parameters. Solve the system of equations:

\[\begin{align} ax - by + \frac {1}{xy} &= c\\ bz - cx + \frac {1}{zx} &= a\\ cy - az + \frac {1}{yz} &= b \end{align}\] – Sharky Kesa · 6 months, 1 week ago

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Consider \(a\), \(b\) and \(c\) as the unknowns instead, so we get a set of linear equations. If we multiply the first equation by \(x\), and then \(y\) and substitute \(cx\) in the second question and \(cy\) in the last one, we obtain

\[\begin{align} a(x^2+1)-b(xy+z)&=\frac{1}{xz}-\frac{1}{y} \quad (i)\\ a(xy-z)-b(y^2+1)&=- \frac {1}{x} - \frac {1}{yz} \quad (ii) \end{align}\]

Multiply equation \((i)\) by \(y^2+1\) and equation \((ii)\) by \(-(xy+z)\), and add them up. It follows that

\[a(x^2+y^2+z^2+1)=\dfrac{x^2+y^2+z^2+1}{xz}\]

so \(a=\frac {1}{xz}\). Similarly, \(b=\frac{1}{yz}\) and \(c=\frac{1}{zx}\). Thus, \(abc=\frac{1}{(xyz)^2}\), so \(xyz=\pm \frac{1}{\sqrt{abc}}\). From this, we can attain the solutions of the system are

\[\left (\dfrac {b}{\sqrt{abc}}, \dfrac {a}{\sqrt{abc}}, \dfrac {c}{\sqrt{abc}} \right ), \left (\dfrac {-b}{\sqrt{abc}}, \dfrac {-a}{\sqrt{abc}}, \dfrac {-c}{\sqrt{abc}} \right ) \] – Sharky Kesa · 6 months, 1 week ago

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– Prince Loomba · 6 months, 1 week ago

I made a silly mistake while calculating by matrices... Now I got it hahaLog in to reply

Problem 18Since @Archit Agrawal's time ran out, I'll post next question.

Consider the sequence

\[a_n = 2 - \dfrac {1}{n^2 + \sqrt{n^4 + \frac {1}{4}}}, \quad n \geq 1\]

Prove that \(\sqrt{a_1}+\sqrt{a_2}+\ldots+\sqrt{a_{119}}\) is an integer and find the value of it. – Sharky Kesa · 6 months, 1 week ago

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We will take the conjugate of the expression to get as follows:

\[\begin{align} a_n &= 2 - \dfrac {1}{n^2 + \sqrt{n^4+\frac {1}{4}}}\\ &= 2 - \dfrac{n^2 - \sqrt{n^4 + \frac {1}{4}}}{n^4 - \left ( n^4 + \frac {1}{4} \right )}\\ &= 2 + 4n^2 - 2 \sqrt{4n^4 + 1} \end{align}\]

We can factorise \(4n^4 + 1\) as \((2n^2 - 2n + 1)(2n^2 + 2n + 1)\). Note that \((2n^2 - 2n + 1) + (2n^2 + 2n + 1) = 4n^2 +2\). Thus, this expression is the perfect square identity. We have

\[\begin{align} a_n &= 4n^2 + 2 - 2\sqrt{4n^4 + 1}\\ &= (\sqrt{2n^2 + 2n + 1})^2 + (\sqrt{2n^2 - 2n + 1})^2 - 2 \times \sqrt{2n^2 + 2n + 1} \times \sqrt{2n^2 - 2n + 1}\\ &= (\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1})^2\\ \sqrt{a_n} &= \sqrt{2n^2 + 2n + 1} - \sqrt{2n^2-2n +1} \end{align}\]

Note that this is a telescoping series so almost all terms cancel out in the, so we are left with \(\sqrt{2\times 119^2 + 2 \times 119 + 1} - \sqrt{2\times 1^2 - 2 \times 1 + 1} = 169 - 1 = 168\), which is an integer. Thus proven. – Sharky Kesa · 6 months, 1 week ago

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– Aman Rajput · 6 months, 1 week ago

\[\boxed{168}\]Log in to reply

– Sharky Kesa · 6 months, 1 week ago

Can you prove it? That was the question asked.Log in to reply

Problem 16

Evaluate: \( \displaystyle \sum_{n=0}^{\infty} \cos^n{\theta} \cos{n \theta} \)

Assume θ is such that the sum converges. – Jack Lam · 6 months, 1 week ago

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– Aman Rajput · 6 months, 1 week ago

is it ? i replaced theta with x \[\frac12.\frac{2e^{ix}-\cos x - \cos x e^{2ix}}{\cos^2 x e^{ix} - \cos x - \cos x e^{2ix}+e^{ix}}\]Log in to reply

\[\begin{align} \dfrac{2e^{ix}-\cos x - \cos x e^{2ix}}{\cos^2 x e^{ix} - \cos x - \cos x e^{2ix}+e^{ix}}&= 2\\ 2e^{ix} - \cos x - \cos x e^{2ix} &= 2\cos^2 x e^{ix} - 2\cos x - 2\cos x e^{2ix} + 2e^{ix}\\ \cos x e^{2ix} + \cos x &= 2 \cos^2 x e^{ix}\\ \cos x (2 \cos^2 x + 2i \sin x \cos x) &= 2 \cos^2 x (\cos x + i \sin x)\\ 2 \cos^2 x (\cos x + i \sin x) &= 2 \cos^2 x (\cos x + i \sin x) \end{align}\]

which is clearly true. Thus, the original fraction must be equal to 2, so the value of the summation is 1. – Sharky Kesa · 6 months, 1 week ago

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– Jack Lam · 6 months, 1 week ago

It is assumed that I want the answer in the simplest possible form...Log in to reply

Problem 15Find the least real number \(r\) such that for each triangle with side lengths \(a\), \(b\), \(c\),

\[\dfrac {\max(a, b, c)}{\sqrt[3]{a^3 + b^3 + c^3 + 3abc}} < r\] – Sharky Kesa · 6 months, 1 week ago

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Using the triangle inequality, and some trivially true statements:

\[b + c - a > 0\]

\[b^2 + c^2 + a^2 + ab + ac - bc > 0\]

\[b^3 + c^3 - a^3 + 3abc > 0\]

\[a^3 + b^3 + c^3 + 3abc > 2a^3\]

The rest should follow after rearranging...

\[\dfrac {a^3}{a^3+b^3+c^3+3abc} < \dfrac {1}{2}\]

\[\dfrac {a}{\sqrt[3]{a^3+b^3+c^3+3abc}} < \dfrac {1}{\sqrt[3]{2}}\]

Thus, \(\displaystyle r=\frac {1}{\sqrt[3]{2}}\). – Jack Lam · 6 months, 1 week ago

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PROBLEM 14Which of the following sets are incomparable

A). Set of all living children, set of all living humans

B). Set of regular polygons, Set of irregular polygons

C). Set of all the triangles in a plane, set of all the squares in a plane

D). Set of all the things in this universe,set of all the chairs.

QuestionYou have to explain why all the others are wrong and this is right (Your answer) – Prince Loomba · 6 months, 1 week ago

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Secondly, in C), we have the set of all triangles, which is an uncountable infinite, versus the set of all squares, which is also an uncountable infinite, and we cannot compare these two sets.

Thirdly, in D), we has the set of all chairs is a subset of the set of all things in the universe, so they can be compared.

Finally, in B), we have the set of regular polygons is disjoint from the set of irregular polygons. However, note that the probability an arbitrary polygon is regular is 0 (not impossible, 0), so it is almost guaranteed that an arbitrary polygon is irregular. Thus, we must have the set of all regular polygons to be less than the set of irregular polygons.

My answer is C). – Sharky Kesa · 6 months, 1 week ago

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Problem 13In

problem 12, if c approaches \(-\infty\), such that when x tends to \(\infty\), \(x^{2}+bx+c\) tends to 0.Which of the following can be the value of \(|-x^{2}+bx+c|\)?(\(b\neq 0)\)

A) 0

B) \(\infty\)

C) \(-\infty\)

D)\(\sqrt{2016^{2015}}\) – Prince Loomba · 6 months, 1 week ago

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– Aman Rajput · 6 months, 1 week ago

And are you guys kidding? Why three contests are going on simultaneously?Log in to reply

– Prince Loomba · 6 months, 1 week ago

Your time limit to create a question is up. Read point 5.Log in to reply

– Prince Loomba · 6 months, 1 week ago

Geometry one is stucked, JEE algebra one I told Rajdeep not to start but he started!Log in to reply

– Aman Rajput · 6 months, 1 week ago

B) and D) both can be correct. Check your question again unless it is a multiple options correct question.Log in to reply

– Prince Loomba · 6 months, 1 week ago

Aman you are right. Post the next. I created this question as multicorrect only!Log in to reply

– Archit Agrawal · 6 months, 1 week ago

BLog in to reply

– Prince Loomba · 6 months, 1 week ago

Wrong. Think againLog in to reply

PROBLEM 12For the equation \(x^{2}+bx+c\), as x approches \(\infty\), it also approaches \(\infty\).

For the equation \(|-x^{2}+bx+c|\), as x approaches \(\infty\), it also approaches \(\infty\)

Can we compare the 2 infinities? If yes, which is greater? – Prince Loomba · 6 months, 1 week ago

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– Archit Agrawal · 6 months, 1 week ago

Yes, we can compare the two infinities. If b is +ve then x^2+bx+c is greater and if b is -ve then |-x^2+bx+c|is greater.Log in to reply

– Prince Loomba · 6 months, 1 week ago

Right post nextLog in to reply

Problem 11Let \(f\) be a function \(f:\mathbb{Z} \to \mathbb{Z}\) that satisfies below:

\[f(y+f(x))=f(x-y)+y(f(y+1)-f(y-1))\]

Find all functions that satisfy. – Sharky Kesa · 6 months, 2 weeks ago

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– Prince Loomba · 6 months, 2 weeks ago

Your time is up. 24 hours were given to solve the problem and 2 hours to post the next one for you. 26 hours are over, so by point number 5 since I am creator of problem 10, I can post problem 12!Log in to reply

– Sharky Kesa · 6 months, 1 week ago

Sorry, I was busy, I'll post solution ASAP.Log in to reply

As I have wrote f (f (x))=f (x)

The following satisfy

Constant function

f is odd when x is odd and f is even when x is even

f (x)=x – Prince Loomba · 6 months, 2 weeks ago

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1) \(f(x)=a\) if \(x\) is even and \(a\) is even, and \(f(x)=b\) if \(x\) is odd and \(b\) is odd.

2) \(f (x)=c \forall c\)

3)\(f (x)=x \forall x\)

Sharky discussed some time back this with me so it would be unfair for me to post the solution. – Zoha Asif · 6 months, 2 weeks ago

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– Wen Z · 6 months, 2 weeks ago

Well I found these solutions but couldn't find a way to prove that these were the only ones, so I gave up. I'm just waiting for the solution now. (is it some kind of induction?)Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

One more: f (x)=|x|, one more f (x)=-|x|Log in to reply

This implies either f (x) is a constant function or f (x)=x. – Prince Loomba · 6 months, 2 weeks ago

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– Sharky Kesa · 6 months, 2 weeks ago

Not necessarily.Log in to reply

– Prince Loomba · 6 months, 2 weeks ago

Then you tell. Idk the solutions of f (f(x))=f (x)Log in to reply

– Kaustubh Miglani · 6 months, 2 weeks ago

And cannot be f(x)=xLog in to reply

RHS becomes x-y+y (y+1-y+1)=x+y

Clearly f (x)=x satisfies

Alternatively take inverse of my equation, you will get f (x)=x – Prince Loomba · 6 months, 2 weeks ago

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– Kaustubh Miglani · 6 months, 2 weeks ago

Sorry calc errorLog in to reply

– Prince Loomba · 6 months, 2 weeks ago

No problemLog in to reply

– Kaustubh Miglani · 6 months, 2 weeks ago

Its a constant functionLog in to reply

Problem 10Find the number of polynomials of degree 5 with distinct coefficient from the set

{\(1,2,...9\)}

that are divisible by \(x^{2}-x+1\) – Prince Loomba · 6 months, 2 weeks ago

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We now consider \(P(x)= (k+l)x^5 + 9x^4 + (m+l)x^3 + kx^2 + (9+l)x + m\), with \(l > 0\) and \(k \leq 9 \leq m\). For a given \(l\), there are \(\dbinom{9-k}{3}\) ways of choosing \(k\), \(m\) and \(n\) such that \(l+m \leq 9\). Now since the coefficients are distinct, we must subtract any combination with the difference in \(k\), \(m\) and \(n\) being \(l\).

There are \(9-2l\) ways to select two numbers differing by \(l\) and \(7-l\) ways to select the remaining numbers. This implies that for a given \(l\), there are \(\dbinom{9-l}{3} - (9-2l)(7-l)+9-3l\) polynomials satisfying, which adds up to \(53\). However, we take into account the rearrangements of the coefficients (there are 12 for each) so there are \(53 \times 12 = 636\) polynomials satisfying. – Sharky Kesa · 6 months, 2 weeks ago

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Problem 8Let \(n\) be a positive integer. Prove \((x^2+x)^{2^n} + 1\) is irreducible. – Sharky Kesa · 6 months, 2 weeks ago

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its equivalent to \((x^{2}+x+1)^{2^{n}}\)

so if \((x^{2}+x)^{2^{n}+1}\) \(\equiv\) f(x) g(x) mod 2

\(f(x)=(x^{2}+x+1)^{m}\) mod 2

\(g(x) = (x^{2}+x+1)^{(2^{n}-m)}\) mod 2

\(f(x)=(x^{2}+x+1)^{m} + 2P(x)\)

\(g(x)=(x^{2}+x+1)^{(2^{n}-m)} + 2Q(x)\)

\(f(x)*g(x)= ((x^{2}+x+1)^m + 2P(x))((x^{2}+x+1)^{(2^n-m)} + 2Q(x))\)

Let x=w (complex cube root of unity)

\((w^{2}+w)^{2^{n}} + 1 = 2P(w)*2Q(w)\)

2 = 4P(w)Q(w)

We got P (w)Q (w) as rational numbers but they were integer polynomials.

Hence a contradiction! – Prince Loomba · 6 months, 2 weeks ago

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\( (x^2 -1)^{2^n} + 4^{2^n} \)

It is known that every non-trivial binomial coefficient in the rows of Pascal's Triangle which are powers of 2 are all even.

Thus, in the expasion of the above polynomial, only the leading co-efficient and the constant are not even.

The leading coefficient is clearly not divisible by 2.

The constant term is clearly not divisible by 4.

Therefore, Eisenstein's Criterion holds and the polynomial is irreducible.

Hence, the original polynomial is also irreducible. – Jack Lam · 6 months, 2 weeks ago

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Thus no integer can be its root.

Hence proved – Prince Loomba · 6 months, 2 weeks ago

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– Sharky Kesa · 6 months, 2 weeks ago

Not quite. Your method is incorrect since irreducible just means that it cannot be factored. A property of irreducible polynomials (degree > 1) is that they have no integral roots, but the converse isn't true.Log in to reply

– Kaustubh Miglani · 6 months, 2 weeks ago

Ya same way About to post it But u did it firstLog in to reply

– Sharky Kesa · 6 months, 2 weeks ago

But it is incorrect.Log in to reply

– Kaustubh Miglani · 6 months, 2 weeks ago

Yeah U are right.Thanks for explainingLog in to reply

Problem 7Let \(H_n = \frac{1}{1} + \frac {1}{2} + \ldots + \frac {1}{n}\). Prove that for \(n \geq 2\)

\[n(n+1)^{\frac {1}{n}} < n + H_n\] – Sharky Kesa · 6 months, 2 weeks ago

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\[\begin{align} \dfrac {\displaystyle \sum_{i=1}^n \left ( 1+\frac{1}{i} \right ) }{n} &> \sqrt[n]{\displaystyle \prod_{i=1}^n \left ( 1+\frac{1}{i} \right )}\\ \dfrac {H_n + n}{n} &> \sqrt[n]{n+1}\\ H_n+n &> n(n+1)^{\frac{1}{n}} \end{align}\]

Thus, proven. – Sharky Kesa · 6 months, 2 weeks ago

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Problem 6Let \(S_{1}, S_{2}, S_{3}\) denote the sum of n terms of 3 arithmetic progressions with first terms unity and their common differences are in harmonic progression. Prove that

\[\large n=\frac {2S_{3}S_{1}-S_{1}S_{2}-S_{2}S_{3}}{S_{1}-2S_{2}+S_{3}}.\] – Prince Loomba · 6 months, 2 weeks ago

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\[\begin{align} S_1 &= \frac {n}{2} (2a+(n-1)d_1)\\ S_2 &= \frac {n}{2} (2a+(n-1)d_2)\\ S_3 &= \frac {n}{2} (2a+(n-1)d_3)\\ d_1 &= \frac{1}{x}\\ d_2 &= \frac {1}{x+k}\\ d_3 &= \frac {1}{x+2k}\\ a &= 1\\ \implies S_1 &= n + \frac{n-1}{2r}\\ S_2 &= n + \frac{n-1}{2r+2k}\\ S_3 &= n + \frac{n-1}{2r+4k} \end{align}\]

Substituting back into the original expression, after much painful expansion, you get the fraction to simplify to \(n\).

For those who want the expansion:

\[\begin{align} & \dfrac {2 \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{4k+2r} \right ) + \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+2k} \right ) \left (n + \frac {n-1}{2r+4k} \right )}{\left (n + \frac {n-1}{2r} \right ) - 2 \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+4k} \right )}\\ & = \dfrac{2 \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{4k+2r} \right ) + \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+2k} \right ) \left (n + \frac {n-1}{2r+4k} \right )}{\frac {k^2 (n-1)}{r(k+r)(2k+r)}}\\ & = \dfrac{r \left ( 2 \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{4k+2r} \right ) + \left (n + \frac {n-1}{2r} \right ) \left (n + \frac {n-1}{2r+2k} \right ) + \left (n + \frac {n-1}{2r+2k} \right ) \left (n + \frac {n-1}{2r+4k} \right ) \right )(k+r)(k+2r)}{k^2 (n-1)}\\ &=\dfrac{r \left (\frac {k^2 n(n-1)}{r(k+r)(2k+r)} \right )(k+r)(k+2r)}{k^2 (n-1)}\\ &= \dfrac {k^2 n (n-1)}{k^2 (n-1)}\\ &= n \end{align}\] – Sharky Kesa · 6 months, 2 weeks ago

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– Prince Loomba · 6 months, 2 weeks ago

This is not the solution. The solution is 10 times easier. I will consider it as solution either you solve full or take out the shortcut that I haveLog in to reply

@Prince Loomba – Harsh Shrivastava · 6 months, 2 weeks ago

Can you please post the easy solution.Log in to reply