Algebra contest

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

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

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

  3. Please make a substantial comment.

  4. 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.

  5. 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.

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

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

  8. Format your solution in LaTeX\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.

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

  10. 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)

  11. Handwritten picture solution are accepted only when they are easily understandable!!

Note by Prince Loomba
3 years, 2 months ago

No vote yet
1 vote

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PROBLEM 1

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

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

Here ii=1\sqrt {-1}

Prince Loomba - 3 years, 2 months ago

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SOLUTION TO PROBLEM 1 let a+bi=eixa+bi=e^{ix} then e2002ix=eixe^{2002ix}=e^{-ix} and e2003ix=1e^{2003ix}=1 so all te possible a+bia+bi are 2003th roots of unity . we know from here that a+bia+bi has 2003 diffent solutions and for each solution we get a different ordered pair for (a,b)(a,b) and hence the number of ordered pairs (a,b)(a,b) is 2003\boxed{2003}

Aareyan Manzoor - 3 years, 2 months ago

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Right. Post next

Prince Loomba - 3 years, 2 months ago

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@Prince Loomba No, answer should be 2004 as (0,0) is also a solution.

Archit Agrawal - 3 years, 2 months ago

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@Archit Agrawal yes yes.....trivial one

Abhisek Mohanty - 2 years, 5 months ago

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@Archit Agrawal Classic division by zero mistake.

Jesse Nieminen - 2 years, 5 months ago

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Problem 22

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

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

a3+b3+c3(a2+b2+c2)32+3abca^3+b^3+c^3 \leq (a^2+b^2+c^2)^{\frac32} + 3abc

Sharky Kesa - 3 years, 1 month ago

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(a3+b3+c33abc)2=[(a+b+c)(a2+b2+c2abbcca)]2=(a2+b2+c2+2ab+2bc+2ca)(a2+b2+c2abbcca)2(a2+b2+c2)3    a3+b3+c3(a2+b2+c2)32+3abc.\begin{aligned} \left(a^3 + b^3 + c^3 - 3abc\right)^2 &= \left[\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right]^2 \\ &= \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 \\ &\implies a^3 + b^3 + c^3 \leq \left(a^2 + b^2 + c^2\right)^{\frac 32} + 3abc. \square \end{aligned}

(a2+b2+c2+2ab+2bc+2ca)(a2+b2+c2abbcca)2(a2+b2+c2)3\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=a2+b2+c2,y=ab+bc+cax = a^2 + b^2 + c^2, y = ab + bc + ca, we get (x+2y)(xy)2x32y3x2ab+2bc+2ca3a2+3b2+3c2a2+b2+c2+(ab)2+(bc)2+(ca)20\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 - 3 years, 1 month ago

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PROBLEM 9

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

f(x+a)=12+f(x)f(x)2f (x+a)=\frac {1}{2}+\sqrt {f (x)-{f (x)}^{2}}

Prove that f(x)f (x) is periodic and give an example for f(x)f (x) when a=1a=1.

Prince Loomba - 3 years, 2 months ago

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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.

Archit Agrawal - 3 years, 2 months ago

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Absolutely right. Noww post next

Prince Loomba - 3 years, 2 months ago

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Problem 17

Find real numbers aa, bb and cc such that

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

for all real numbers xx, yy and zz.

Sharky Kesa - 3 years, 2 months ago

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Unordered triplets are (1,0,0) and (-1,0,0).

Archit Agrawal - 3 years, 1 month ago

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Post solution too

Prince Loomba - 3 years, 1 month ago

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@Prince Loomba 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.

Archit Agrawal - 3 years, 1 month ago

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@Archit Agrawal Yes! You have got the answer!

Sharky Kesa - 3 years, 1 month ago

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problem 2 given the roots of y8y7+y61=0y^8-y^7+y^6-1=0 are y1,y2,...,y7,y8y_1,y_2,...,y_7,y_8 find the value of 1y19+1y29+...+1y79+1y89\dfrac{1}{y_1^9}+\dfrac{1}{y_2^9}+...+\dfrac{1}{y_7^9}+\dfrac{1}{y_8^9}

Aareyan Manzoor - 3 years, 2 months ago

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Solution to Problem 2

y8y7+y61=0y^8 - y^7 + y^6 - 1 = 0

Now we have to form a polynomial whose roots are 1y1,.....\frac{1}{y_{1}}, .....

Now, let x=1yx = \frac{1}{y}

then y=1xy = \frac{1}{x}

Putting it in the above equation, we get

x8+x2x+1=0x^8 + x^2 - x + 1 = 0

x8=xx21x^8 = x - x^2 - 1

x9=x2x3x...(1)x^9 = x^2 - x^3 - x ...(1)

Putting x=1y1,...,1y8x = \frac{1}{y_{1}},..., \frac{1}{y_{8}} repeatedly in the equation and add them all and using bit Newton Sum (S1=S2=S3=0S_{1} = S_{2} = S_{3} = 0)

And answer is 00.

Prince Loomba - 3 years, 2 months ago

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Post the next question.

Harsh Shrivastava - 3 years, 2 months ago

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you are right.

Aareyan Manzoor - 3 years, 2 months ago

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Is thE Answer 0?

Kaustubh Miglani - 3 years, 2 months ago

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Problem 3

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

If f(f(x))=x2+2f (f (x))=x^{2}+2, find the value of f(3)f (3)

Prince Loomba - 3 years, 2 months ago

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Harsh Shrivastava - 3 years, 2 months ago

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Its right. Post next

Prince Loomba - 3 years, 2 months ago

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Post next soon

Prince Loomba - 3 years, 2 months ago

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Problem 4

Let a,ba,b lie in first quadrant and let xx be an acute angle.

Given that (acosx)2+(bsinx)2=[(sin2x)(a+b2)]2acosx)^2 +(bsinx)^2 = [(sin2x)(\dfrac{a+b}{2} )]^2

Find the value of tanx.

Harsh Shrivastava - 3 years, 2 months ago

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If we denote the three vertices by A(a,b),B(acscx,0),C(0,bsecx)A(a,b),B(a\csc x,0),C(0,b\sec x) then observe that AB2=a2cot2x+b2,BC2=a2+b2tan2x,CA2=a2csc2x+b2sec2x\displaystyle |AB|^2=a^2\cot^2 x+b^2,|BC|^2=a^2+b^2\tan^2 x,|CA|^2=a^2\csc^2 x +b^2\sec^2 x

which in turn shows AB2+BC2=CA2\displaystyle |AB|^2+|BC|^2=|CA|^2 and so B=π2\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 (M,N)=(acscx2,bsecx2)\displaystyle (M,N)=(\frac{a\csc x}{2},\frac{b\sec x}{2})

From the relation we have , a2cos2x+b2sin2x=sin2xcos2x(a+b)2\displaystyle a^2\cos^2 x+b^2\sin^2 x=\sin^2 x\cos^2 x(a+b)^2

    (acos2x)2+(bsin2x)22absin2xcos2x=0    acos2x=bsin2x\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 , cscx=a+ba,secx=a+bb\displaystyle \csc x = \sqrt{\frac{a+b}{a}} , \sec x=\sqrt{\frac{a+b}{b}}

So the circumcenter is : (a2+ab2,b2+ab2)\displaystyle (\frac{\sqrt{a^2+ab}}{2},\frac{\sqrt{b^2+ab}}{2})

Aditya Narayan Sharma - 3 years, 2 months ago

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Correct @Aditya Sharma but sorry I rephrased the problem so @Prince Loomba will decide who will post the next problem.

Harsh Shrivastava - 3 years, 2 months ago

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You post the next problem. The original was this only

Prince Loomba - 3 years, 2 months ago

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Wait @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 - 3 years, 2 months ago

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You post the next prob.

Harsh Shrivastava - 3 years, 2 months ago

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Post the next

Prince Loomba - 3 years, 2 months ago

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

Prince Loomba - 3 years, 2 months ago

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Corrections done.

If I tell the name of the topic, the problem will be trivial :)

Harsh Shrivastava - 3 years, 2 months ago

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I got tanxtanx=a/b\sqrt {a/b}. Idk how to find circumcentre

Prince Loomba - 3 years, 2 months ago

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Lemme rephrase the problem.

Harsh Shrivastava - 3 years, 2 months ago

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Done post the solution.

Harsh Shrivastava - 3 years, 2 months ago

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@Harsh Shrivastava

Prince Loomba - 3 years, 2 months ago

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Problem 5:\text{Problem 5:}

Find all functions f:ZZf:{\rm Z}\to{\rm Z} such that f(m+n)+f(m)f(n)=f(mn+1)\displaystyle f(m+n)+f(m)f(n)=f(mn+1) where m,nm,n are integers.

Aditya Narayan Sharma - 3 years, 2 months ago

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Is the answer only 1? (f(x) = 0)

Harsh Shrivastava - 3 years, 2 months ago

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NO, f(x)=0f(x)=0 is trivial, And more functions are there.

Aditya Narayan Sharma - 3 years, 2 months ago

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x-1 is also

Prince Loomba - 3 years, 2 months ago

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@Prince Loomba Yes one of them, you need to show it other than trial and error.

Aditya Narayan Sharma - 3 years, 2 months ago

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I got a weird thing. All of the functions of the type x2n1x^{2n}-1 are satisfying according to me. Please check.

Prince Loomba - 3 years, 2 months ago

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hmm ri8 but there are some complicated functions involving mod also. anyways post the solution otherways

Aditya Narayan Sharma - 3 years, 2 months ago

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@Aditya Narayan Sharma Mod function: (Alternatively the trig function) f(x)=0f(x) = 0 if x1,3(mod4)x\equiv 1, 3 \pmod{4}, f(x)=1f(x) = 1 if x2(mod4)x \equiv 2 \pmod{4}, f(x)=1f(x) = -1 if x0(mod4)x \equiv 0 \pmod{4}.

Sharky Kesa - 3 years, 2 months ago

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@Sharky Kesa ya those are the solutions exactly. you may post the solution

Aditya Narayan Sharma - 3 years, 2 months ago

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@Aditya Narayan Sharma I am just saying this. I did not say these are all.

Prince Loomba - 3 years, 2 months ago

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Two functions f (x)=0 and f (x)=x-1

Prince Loomba - 3 years, 2 months ago

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There are 6, excluding the zero function

Aditya Narayan Sharma - 3 years, 2 months ago

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@Aditya Narayan Sharma x21x^{2}-1

Prince Loomba - 3 years, 2 months ago

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@Prince Loomba 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 there

Aditya Narayan Sharma - 3 years, 2 months ago

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@Aditya Narayan Sharma |x|-1 also satisfies. I am just checking by graph which can hold and verifying my answer

Prince Loomba - 3 years, 2 months ago

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f(x)=-cos((2n+1)×pi×x/2)

f (x)=sin (n×pi×x)

Prince Loomba - 3 years, 2 months ago

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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 solution

Aditya Narayan Sharma - 3 years, 2 months ago

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One more f (x)={x} fractional part. 6 are done

Prince Loomba - 3 years, 2 months ago

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f (x)=[x]-1 (floor function)

Prince Loomba - 3 years, 2 months ago

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There are infinite functions I can make

Prince Loomba - 3 years, 2 months ago

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Prince Loomba - 3 years, 2 months ago

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yes it will work , i wll ppst other functions later . for now you may post the next question

Aditya Narayan Sharma - 3 years, 2 months ago

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Is there any mistakes I made?

Prince Loomba - 3 years, 2 months ago

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Problem 26

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

Prove for positive xx, yy and zz

x2+y2xy+y2+z2yzz2+x2+zx\sqrt{x^2+y^2-xy}+\sqrt{y^2+z^2-yz} \geq \sqrt{z^2+x^2+zx}

Sharky Kesa - 3 years, 1 month ago

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Post easy, I will solve haha

Prince Loomba - 3 years, 1 month ago

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

Sharky Kesa - 3 years, 1 month ago

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Take a triangle ABC and construct a point P inside the triangle such that angleAPB = angleBPC = angleCPA= 120°.

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 - 3 years, 1 month ago

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Problem 6

Let S1,S2,S3S_{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

n=2S3S1S1S2S2S3S12S2+S3.\large n=\frac {2S_{3}S_{1}-S_{1}S_{2}-S_{2}S_{3}}{S_{1}-2S_{2}+S_{3}}.

Prince Loomba - 3 years, 2 months ago

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I would like to cut the algebraic bash out of my solution, which I have, but it is quite a standard expansion.

S1=n2(2a+(n1)d1)S2=n2(2a+(n1)d2)S3=n2(2a+(n1)d3)d1=1xd2=1x+kd3=1x+2ka=1    S1=n+n12rS2=n+n12r+2kS3=n+n12r+4k\begin{aligned} 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{aligned}

Substituting back into the original expression, after much painful expansion, you get the fraction to simplify to nn.


For those who want the expansion:

2(n+n12r)(n+n14k+2r)+(n+n12r)(n+n12r+2k)+(n+n12r+2k)(n+n12r+4k)(n+n12r)2(n+n12r+2k)+(n+n12r+4k)=2(n+n12r)(n+n14k+2r)+(n+n12r)(n+n12r+2k)+(n+n12r+2k)(n+n12r+4k)k2(n1)r(k+r)(2k+r)=r(2(n+n12r)(n+n14k+2r)+(n+n12r)(n+n12r+2k)+(n+n12r+2k)(n+n12r+4k))(k+r)(k+2r)k2(n1)=r(k2n(n1)r(k+r)(2k+r))(k+r)(k+2r)k2(n1)=k2n(n1)k2(n1)=n\begin{aligned} & \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{aligned}

Sharky Kesa - 3 years, 2 months ago

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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 have

Prince Loomba - 3 years, 2 months ago

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@Prince Loomba Can you please post the easy solution. @Prince Loomba

Harsh Shrivastava - 3 years, 2 months ago

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Problem 7

Let Hn=11+12++1nH_n = \frac{1}{1} + \frac {1}{2} + \ldots + \frac {1}{n}. Prove that for n2n \geq 2

n(n+1)1n<n+Hnn(n+1)^{\frac {1}{n}} < n + H_n

Sharky Kesa - 3 years, 2 months ago

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Note that n+Hn=(1+11)+(1+12)++(1+1n)n+H_n = (1+\frac {1}{1}) + (1+\frac{1}{2}) + \ldots + (1+\frac {1}{n}) (These brackets are all distinct terms). Now, by AM-GM, we have

i=1n(1+1i)n>i=1n(1+1i)nHn+nn>n+1nHn+n>n(n+1)1n\begin{aligned} \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{aligned}

Thus, proven.

Sharky Kesa - 3 years, 2 months ago

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Problem 8

Let nn be a positive integer. Prove (x2+x)2n+1(x^2+x)^{2^n} + 1 is irreducible.

Sharky Kesa - 3 years, 2 months ago

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look at the polynomial mod 2

its equivalent to (x2+x+1)2n(x^{2}+x+1)^{2^{n}}

so if (x2+x)2n+1(x^{2}+x)^{2^{n}+1} \equiv f(x) g(x) mod 2

f(x)=(x2+x+1)mf(x)=(x^{2}+x+1)^{m} mod 2

g(x)=(x2+x+1)(2nm)g(x) = (x^{2}+x+1)^{(2^{n}-m)} mod 2

f(x)=(x2+x+1)m+2P(x)f(x)=(x^{2}+x+1)^{m} + 2P(x)

g(x)=(x2+x+1)(2nm)+2Q(x)g(x)=(x^{2}+x+1)^{(2^{n}-m)} + 2Q(x)

f(x)g(x)=((x2+x+1)m+2P(x))((x2+x+1)(2nm)+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)

(w2+w)2n+1=2P(w)2Q(w)(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 - 3 years, 2 months ago

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Let aIa\in I be a root of it. (a2+a)2n=1(a^{2}+a)^{2^{n}}=-1. Clearly for n>=1, LHS is a perfect square with range >=0, so inequality cant hold

Thus no integer can be its root.

Hence proved

Prince Loomba - 3 years, 2 months ago

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Ya same way About to post it But u did it first

Kaustubh Miglani - 3 years, 2 months ago

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@Kaustubh Miglani But it is incorrect.

Sharky Kesa - 3 years, 2 months ago

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@Sharky Kesa Yeah U are right.Thanks for explaining

Kaustubh Miglani - 3 years, 2 months ago

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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.

Sharky Kesa - 3 years, 2 months ago

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It sufficies to consider the equivalent rational polynomial

(x21)2n+42n (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 - 3 years, 2 months ago

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Problem 10

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

{1,2,...91,2,...9}

that are divisible by x2x+1x^{2}-x+1

Prince Loomba - 3 years, 2 months ago

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Let P(x)=ax5+bx4+cx3+dx2+ex+fP(x)=ax^5 + bx^4 + cx^3 + dx^2 + ex + f. Note that the roots of x2x+1x^2 - x + 1 are eiπ3e^{\frac{i\pi}{3}} and eiπ3e^{\frac {-i\pi}{3}}. Thus, P(x)P(x) is divisible by x2x+1x^2 - x + 1 if P(eiπ3)=0P(e^{\frac{i\pi}{3}})=0 and P(eiπ3)=0P(e^{\frac{-i\pi}{3}}) = 0. From this, we gather that a+b=d+ea+b=d+e and a2b2cd2+e2+f=0\frac {a}{2} - \frac {b}{2} - c - \frac {d}{2} + \frac {e}{2} + f = 0, which implies e+2f+a=b+2c+de + 2f + a = b + 2c + d. Also, since a+b=d+ea+b=d+e, ad=eb=cfa-d=e-b=c-f.

We now consider P(x)=(k+l)x5+9x4+(m+l)x3+kx2+(9+l)x+mP(x)= (k+l)x^5 + 9x^4 + (m+l)x^3 + kx^2 + (9+l)x + m, with l>0l > 0 and k9mk \leq 9 \leq m. For a given ll, there are (9k3)\dbinom{9-k}{3} ways of choosing kk, mm and nn such that l+m9l+m \leq 9. Now since the coefficients are distinct, we must subtract any combination with the difference in kk, mm and nn being ll.

There are 92l9-2l ways to select two numbers differing by ll and 7l7-l ways to select the remaining numbers. This implies that for a given ll, there are (9l3)(92l)(7l)+93l\dbinom{9-l}{3} - (9-2l)(7-l)+9-3l polynomials satisfying, which adds up to 5353. However, we take into account the rearrangements of the coefficients (there are 12 for each) so there are 53×12=63653 \times 12 = 636 polynomials satisfying.

Sharky Kesa - 3 years, 2 months ago

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Problem 11

Let ff be a function f:ZZf:\mathbb{Z} \to \mathbb{Z} that satisfies below:

f(y+f(x))=f(xy)+y(f(y+1)f(y1))f(y+f(x))=f(x-y)+y(f(y+1)-f(y-1))

Find all functions that satisfy.

Sharky Kesa - 3 years, 2 months ago

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Put y=0. f (f (x))=f (x)

This implies either f (x) is a constant function or f (x)=x.

Prince Loomba - 3 years, 2 months ago

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Its a constant function

Kaustubh Miglani - 3 years, 2 months ago

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And cannot be f(x)=x

Kaustubh Miglani - 3 years, 2 months ago

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@Kaustubh Miglani Bro LHS becomes y+f (x)=y+x

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 - 3 years, 2 months ago

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@Prince Loomba Sorry calc error

Kaustubh Miglani - 3 years, 2 months ago

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@Kaustubh Miglani No problem

Prince Loomba - 3 years, 2 months ago

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Not necessarily.

Sharky Kesa - 3 years, 2 months ago

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@Sharky Kesa Then you tell. Idk the solutions of f (f(x))=f (x)

Prince Loomba - 3 years, 2 months ago

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One more: f (x)=|x|, one more f (x)=-|x|

Prince Loomba - 3 years, 2 months ago

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Here are the solutions:

1) f(x)=af(x)=a if xx is even and aa is even, and f(x)=bf(x)=b if xx is odd and bb is odd.

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

3)f(x)=xxf (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 - 3 years, 2 months ago

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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?)

Wen Z - 3 years, 2 months ago

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Ok the final answer

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 - 3 years, 2 months ago

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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!

Prince Loomba - 3 years, 2 months ago

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Sorry, I was busy, I'll post solution ASAP.

Sharky Kesa - 3 years, 2 months ago

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PROBLEM 12

For the equation x2+bx+cx^{2}+bx+c, as x approches \infty, it also approaches \infty.

For the equation x2+bx+c|-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 - 3 years, 2 months ago

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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.

Archit Agrawal - 3 years, 2 months ago

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Right post next

Prince Loomba - 3 years, 2 months ago

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Problem 13

In problem 12, if c approaches -\infty, such that when x tends to \infty, x2+bx+cx^{2}+bx+c tends to 0.

Which of the following can be the value of x2+bx+c|-x^{2}+bx+c|?(b0)b\neq 0)

A) 0

B) \infty

C) -\infty

D)20162015\sqrt{2016^{2015}}

Prince Loomba - 3 years, 2 months ago

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B

Archit Agrawal - 3 years, 2 months ago

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Wrong. Think again

Prince Loomba - 3 years, 2 months ago

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

Aman Rajput - 3 years, 2 months ago

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

Prince Loomba - 3 years, 2 months ago

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And are you guys kidding? Why three contests are going on simultaneously?

Aman Rajput - 3 years, 2 months ago

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

Prince Loomba - 3 years, 2 months ago

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

Prince Loomba - 3 years, 2 months ago

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PROBLEM 14

Which 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.

Question

You have to explain why all the others are wrong and this is right (Your answer)

Prince Loomba - 3 years, 2 months ago

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Firstly, in A), we have the the set of all living children is a subset of the set of all living humans, so they can be compared.

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 - 3 years, 2 months ago

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Problem 15

Find the least real number rr such that for each triangle with side lengths aa, bb, cc,

max(a,b,c)a3+b3+c3+3abc3<r\dfrac {\max(a, b, c)}{\sqrt[3]{a^3 + b^3 + c^3 + 3abc}} < r

Sharky Kesa - 3 years, 2 months ago

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WLOG, it can be assumed that abc \displaystyle a \ge b \ge c

Using the triangle inequality, and some trivially true statements:

b+ca>0b + c - a > 0

b2+c2+a2+ab+acbc>0b^2 + c^2 + a^2 + ab + ac - bc > 0

b3+c3a3+3abc>0b^3 + c^3 - a^3 + 3abc > 0

a3+b3+c3+3abc>2a3a^3 + b^3 + c^3 + 3abc > 2a^3

The rest should follow after rearranging...

a3a3+b3+c3+3abc<12\dfrac {a^3}{a^3+b^3+c^3+3abc} < \dfrac {1}{2}

aa3+b3+c3+3abc3<123\dfrac {a}{\sqrt[3]{a^3+b^3+c^3+3abc}} < \dfrac {1}{\sqrt[3]{2}}

Thus, r=123\displaystyle r=\frac {1}{\sqrt[3]{2}}.

Jack Lam - 3 years, 2 months ago

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Problem 16

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

Assume θ is such that the sum converges.

Jack Lam - 3 years, 2 months ago

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is it ? i replaced theta with x 12.2eixcosxcosxe2ixcos2xeixcosxcosxe2ix+eix\frac12.\frac{2e^{ix}-\cos x - \cos x e^{2ix}}{\cos^2 x e^{ix} - \cos x - \cos x e^{2ix}+e^{ix}}

Aman Rajput - 3 years, 2 months ago

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

Jack Lam - 3 years, 2 months ago

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I will prove the fraction simplifies to 2. We are required to prove:

2eixcosxcosxe2ixcos2xeixcosxcosxe2ix+eix=22eixcosxcosxe2ix=2cos2xeix2cosx2cosxe2ix+2eixcosxe2ix+cosx=2cos2xeixcosx(2cos2x+2isinxcosx)=2cos2x(cosx+isinx)2cos2x(cosx+isinx)=2cos2x(cosx+isinx)\begin{aligned} \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{aligned}

which is clearly true. Thus, the original fraction must be equal to 2, so the value of the summation is 1.

Sharky Kesa - 3 years, 2 months ago

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Problem 18

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

Consider the sequence

an=21n2+n4+14,n1a_n = 2 - \dfrac {1}{n^2 + \sqrt{n^4 + \frac {1}{4}}}, \quad n \geq 1

Prove that a1+a2++a119\sqrt{a_1}+\sqrt{a_2}+\ldots+\sqrt{a_{119}} is an integer and find the value of it.

Sharky Kesa - 3 years, 1 month ago

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168\boxed{168}

Aman Rajput - 3 years, 1 month ago

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Can you prove it? That was the question asked.

Sharky Kesa - 3 years, 1 month ago

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Here is the solution:

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

an=21n2+n4+14=2n2n4+14n4(n4+14)=2+4n224n4+1\begin{aligned} 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{aligned}