**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!!

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## Comments

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

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

Here \(i\)=\(\sqrt {-1}\)

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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 \[\boxed{2003}\]Log in to reply

Right. Post next

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Problem 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

\[a^3+b^3+c^3 \leq (a^2+b^2+c^2)^{\frac32} + 3abc\]

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\(\begin{align} \left(a^3 + b^3 + c^3 - 3abc\right)^2 &= \left[\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right]^2 \\ &= \left(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca\right)\left(a^2 + b^2 + c^2 - ab - bc - ca\right)^2 \\ &\leq \left(a^2 + b^2 + c^2\right)^3 \\ &\implies a^3 + b^3 + c^3 \leq \left(a^2 + b^2 + c^2\right)^{\frac 32} + 3abc. \square \end{align}\)

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

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

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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 \(a=1\).

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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.

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Absolutely right. Noww post next

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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 \(z\).

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

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Post solution too

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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 \[\dfrac{1}{y_1^9}+\dfrac{1}{y_2^9}+...+\dfrac{1}{y_7^9}+\dfrac{1}{y_8^9}\]Log 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 \(0\).

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Post the next question.

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you are right.

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Is thE Answer 0?

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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 \(f (3)\)

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Its right. Post next

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Post next soon

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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.

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If we denote the three vertices by \(A(a,b),B(a\csc x,0),C(0,b\sec x)\) then observe that \(\displaystyle |AB|^2=a^2\cot^2 x+b^2,|BC|^2=a^2+b^2\tan^2 x,|CA|^2=a^2\csc^2 x +b^2\sec^2 x\)

which in turn shows \(\displaystyle |AB|^2+|BC|^2=|CA|^2\) 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 : \(\displaystyle (\frac{\sqrt{a^2+ab}}{2},\frac{\sqrt{b^2+ab}}{2}) \)

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

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You post the next problem. The original was this only

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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

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You post the next prob.

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Post the next

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

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Corrections done.

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

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

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Lemme rephrase the problem.

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Done post the solution.

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\(\text{Problem 5:}\)

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.

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

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

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x-1 is also

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

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

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

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There are 6, excluding the zero function

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f(x)=-cos((2n+1)×pi×x/2)

f (x)=sin (n×pi×x)

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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

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

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

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There are infinite functions I can make

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

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Is there any mistakes I made?

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

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

\[\sqrt{x^2+y^2-xy}+\sqrt{y^2+z^2-yz} \geq \sqrt{z^2+x^2+zx}\]

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Post easy, I will solve haha

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

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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.

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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

\[\large n=\frac {2S_{3}S_{1}-S_{1}S_{2}-S_{2}S_{3}}{S_{1}-2S_{2}+S_{3}}.\]

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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.

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

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

For those who want the expansion:

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

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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

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@Prince Loomba

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

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

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

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

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

Thus, proven.

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Problem 8Let \(n\) be a positive integer. Prove \((x^2+x)^{2^n} + 1\) is irreducible.

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look at the polynomial mod 2

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!

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Let \(a\in I\) be a root of it. \((a^{2}+a)^{2^{n}}=-1\). 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

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

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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.

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It sufficies to consider the equivalent rational polynomial

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

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.

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

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

that are divisible by \(x^{2}-x+1\)

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Let \(P(x)=ax^5 + bx^4 + cx^3 + dx^2 + ex + f\). Note that the roots of \(x^2 - x + 1\) are \(e^{\frac{i\pi}{3}}\) and \(e^{\frac {-i\pi}{3}}\). Thus, \(P(x)\) is divisible by \(x^2 - x + 1\) if \(P(e^{\frac{i\pi}{3}})=0\) and \(P(e^{\frac{-i\pi}{3}}) = 0\). From this, we gather that \(a+b=d+e\) and \(\frac {a}{2} - \frac {b}{2} - c - \frac {d}{2} + \frac {e}{2} + f = 0\), which implies \(e + 2f + a = b + 2c + d\). Also, since \(a+b=d+e\), \(a-d=e-b=c-f\).

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.

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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.

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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.

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Its a constant function

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And cannot be f(x)=x

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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

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Not necessarily.

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

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Here are the solutions:

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.

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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?)

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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

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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!

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

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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?

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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.

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Right post next

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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)\(\sqrt{2016^{2015}}\)

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B

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Wrong. Think again

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

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

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

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

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

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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)

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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).

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

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

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

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, \(\displaystyle r=\frac {1}{\sqrt[3]{2}}\).

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Problem 16

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

Assume θ is such that the sum converges.

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

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

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

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

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

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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.

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\[\boxed{168}\]

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Can you prove it? That was the question asked.

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Here is the solution:

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.

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

\[\begin{align} ax - by + \frac {1}{xy} &= c\\ bz - cx + \frac {1}{zx} &= a\\ cy - az + \frac {1}{yz} &= b \end{align}\]

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This solution is a bit messy, but is quite quick as opposed to the other possible solutions (I think).

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

\[\left (\dfrac {b}{\sqrt{abc}}, \dfrac {a}{\sqrt{abc}}, \dfrac {c}{\sqrt{abc}} \right ), \left (\dfrac {-b}{\sqrt{abc}}, \dfrac {-a}{\sqrt{abc}}, \dfrac {-c}{\sqrt{abc}} \right ) \]

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I made a silly mistake while calculating by matrices... Now I got it haha

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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

\[\displaystyle \sum_{k=1}^n \dfrac {1}{1-x_k} = \dfrac {n}{2}\]

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\(\displaystyle \sum_{k=1}^n \dfrac {1}{1-x_k} = \dfrac{\left[\left(1-x_1\right)\left(1-x_2\right)\cdots\left(1-x_{n-1}\right)\right]+\cdots+ \left[\left(1-x_2\right)\left(1-x_3\right)\cdots\left(1-x_{n}\right) \right]}{\left(1-x_1\right)\left(1-x_2\right)\cdots\left(1-x_{n}\right)}\) (Numerator has total of \(n\) addends, each missing different \(1 - x_k\))

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 \(\dfrac{\dfrac{n\left(n+1\right)}2}{n+1} = \dfrac n2\).

\(Q.E.D\)

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Nice solution.

Post the next problem.

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

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

\[\dfrac1a + \dfrac2b + \dfrac3c \geq 3\]

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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 3.56

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You can post next

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Please post next

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I give my right to post to everyone.

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

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I did not see that approach at all! Great job!

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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

\[\dfrac {1}{a} + \dfrac {1}{b} + \dfrac {1}{c} + \dfrac {1}{d} \geq 4\]

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Post next. Time elapsed

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This competition is now me solving or not solving problems by @Sharky Kesa :D

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I have a solution using Lagrange Multipliers :P

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Problem 24Prove \(2^{1/x} < \dfrac {x+1}{x}\) for \(x>1\).

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Counter-example: \(x=50\).

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LOL, sorry, the other way!

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Why did no one give the quick and simple solution?!

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

Since the value of \(\text{Stuff}\geq 1\), the final inequality must be true, so the original inequality is true. Thus, proven.

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

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Time elapsed post next

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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.

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

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No, what is meant is that the permutation is a permutation of the set \({1,2,3,4,...,n}\). e.g. if \(n=3\), we can assume a permutation of it to be \({2,1,3}\), so \(\sigma(1)=2\), \(\sigma(2)=1\) and \(\sigma(3)=3\). Note that the above must hold true for any permutation.

Furthermore, even if permutation is defined as what you think it was, you're wrong. What if \(b\) was negative?

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Time is up bro..

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