Post as many problems as you can for others to try.

Please don't post problems from past year RMO papers.

Have FUN!

Here is a problem to start with :

Let \(a, b, c, d, e, f\) be real numbers such that the polynomial equation

\[x^{8} - 4x^{7} +7x^{6} +ax^{5}+ bx^{4} + cx^{3} + dx^{2} + ex + f = 0\]

has eight positive real roots. Determine all possible values of \(f\)

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## Comments

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TopNewestSince we were given that the roots were positive reals,we get motivated and tempted to use,\(R.M.S\geq A.M\geq G.M\geq H.M\),also since we have been given,\(\sum \alpha\) and \(\sum \alpha\beta\) we try to find the value of \(\sum \alpha^2=4^2-2\times 7=2\).We apply \(R.M.S \geq A.M\) \[\sqrt{\dfrac{2}{8}} \geq \dfrac{4}{8}\\ \Longrightarrow \dfrac{1}{2} \geq \dfrac{1}{2}\],hence equality occurs,meaning that all the roots are equal.Now,that means,\(A.M=G.M\),hence\[\dfrac{4}{8}=\sqrt[8]{f}\\ \Longrightarrow \dfrac{1}{256}=f\],hence there is only one value of \(f=\dfrac{1}{256}\).Is this correct Harsh?

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In your solution you need prove that \[(\sum_{cyclic}ω)^{2}-\sum_{cyclic}ω_{1}ω_{2}=\sum_{cyclic}ω^{2}\]

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That's an identity that is true for all numbers, real and imaginary. And there should be a 2 infront of that second summation.

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If \(x,y,z\) are positive real numbers, prove that,

\[\large (x+y+z)^2 (yz+zx+xy)^2 \le 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)\]

Give me as much different solutions please .

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U may use the concept of fermat pt to solve this one..then it will be very easy...suppose three line segments AF,BF,CF of length x,y,z meet each other at120 degrees and now use cosine rule to find out the lengths of the triangle and see that each term on the rhs of the inequality is actually squares of the sides...then express xy+yz+zx as \(\frac{4}{\sqrt{3}}T\) where T is the area of the triangle....then use \(T=\frac{abc}{4R}\) and see that the inequality reduces to 3R greater than or equal to (x+y+z) which is obvious since F is the fermat pt

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You have a set of consecutive positive integers,\(\{1,2,3,...,199,200\}\),if you choose \(101\) numbers from these at random,then prove that there at-least two numbers of which one divides the other.

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Find all positive integer solutions \(a\), \(b\) and \(c\) such that

\[\dfrac {7}{a^2} + \dfrac {11}{b^2} = \dfrac {c}{77}.\]

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\(\text{THIS IS ONLY AN OUTLINE OF THE SOLUTION, NOT A COMPLETE SOLUTION.}\)

\[\dfrac {7 }{a^2} + \dfrac {11 }{b^2} = \dfrac{c}{77}.\]

Case 1:\(c>1\)This forces \(a^{2} < 7 \text{ and } b^{2} < 11\)

Now it is easy to check that \( (a,b,c) = (1,1,1386) \text{ and } (a,b,c) = (3,3,154)\) satisfy the given conditions.

Case 2:\(c \leq 77\)Since \(c \leq 77\) this implies \(\dfrac{c}{77} \leq 1\).

Let \(\dfrac{c}{77} = \dfrac{1}{k}\) for some integer \(k \geq 1\).

\(\implies c = \dfrac{77}{k}\)

Now since \(c\) is a positive integer , this forces

\(k \text{ | } 77\).Possible values for \(k = 1,7,11,77\).

Now plugging the possible values of \(c\) in the given equation and using Simon's Favourite Factoring Trick, we can easily solve for \(a,b\)

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If \(x\) and \(y\) are integers such that \(y^2+3(xy)^2=30x^2+517\), Prove that \(3(xy)^2=588\)

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Let \(y^{2} = b\) and \(x^{2} = a\).

\(b(1 + 3a) = 30a +517 \)

\(\implies b(1+3a) -10(3a + 1) = 507 \)

\(\implies (b-10)(3a+1) = 507 \times 1 = 13 \times 39 = 3 \times 169\)

Since a , b are integers, after checking through all 6 cases, we get only admissible values of \((x ,y) = ( \pm 2, \pm 7)\).

This forces \(3(xy)^{2}= 588\).

\(\QED \)

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Let \(x,y\) and \(z\) be positive real numbers such that \(x+y+z=1\). Prove that \(\frac{yz}{1+x}+\frac{zx}{1+y}+\frac{xy}{1+z} \le \frac{1}{4}\).

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I got this. This is simple AM-HM inequality. Write 1+x=(1-y)+(1-z). Similarly for others

Apply AM-HM and then manipulate on the RHS.

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x, y, z are positive real numbers such that x^2+y^2+z^2=3 Prove that x+y+z >= (xy)^2+(yz)^2+(zx)^2

Sorry for not using latex as I was in a hurry.

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Can u send the solution of this question ? I have been trying. No progress ....

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I didn't get it.

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Find all primes \(p\) such that for all prime \(q\), the remainder of \(p\) upon divison by \(q\) is squarefree.

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I think that 3 is the only prime number. I used the fact that all primes>3 is of the form \(\pm 1 \pmod{6}\).

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In a triangle \(ABC\) , D is the midpoint of side AB and E is the point of trisection of side BC nearer to C.

Given that \(\angle ADC = \angle BAE \), find \(\angle BAC\).

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Is the answer angle BAC = 90 degrees ? I hope so My proof : Apply Menelaus theorem on triangle BCD with transversal EA. Obtain that F is the midpoint of CD. Then its simple angle chasing :)

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Comment deleted Dec 04, 2015

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Yes , Menelaus is indeed a good approach for this problem. For shortening the solution: Once you get F is the midpoint of CD , you can directly say that \(\angle BAC = 90^\circ\) , do you see why?

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Yup I also got 90 degrees.

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Shrihari,we get that \(CF=2DF\)

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(CF/FD)(DA/AB)=1Log in to reply

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Let \(p>3\) be a prime number. Suppose \(\displaystyle \sum_{k=1}^{p-1}{\frac{1}{k}}=\frac{a}{b}\) where \(\gcd(a,b)=1\). Prove that \(a\) is divisible by \(p^{2}\).

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I am able to prove that a is divisible by p. But not able to prove divisible by p^2. Could u send the solution ?

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Please post the proof, thanks!

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I have the solution only till \(a\) is divisible by \(p\). To prove that \(a\) is divisible by \(p^{2}\) is given below the solution as a challenge.

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If \(a,b \ge 0\). Prove that \(\sqrt{2}(\sqrt{a(a+b)^{3}}+b \sqrt{a^{2}+b^{2}}) \le 3(a^{2}+b^{2})\).

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Let \(a,b,c\) be positive reals satisfying \((a+b)(b+c)(c+a) = 1\).

Prove that \(ab+bc+ca \leq \frac{3}{4}\)

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Using am gm

(a + b) + (b + c) + (c + a) > 3

a + b + c > 3/2

now using cauchy,

\(3(a^2 + b^2 + c^2) > (a + b + c)^2\)

so

\(a^2 + b^2 + c^2 > 3/4\)

using cauchy again,

\((a^2 + b^2 + c^2)^2 > (ab + bc + ca)^2\)

3/4 > ab + bc + ca

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Dev,your solution is incorrect. Check the solution again,you will find your flaw.

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Nice solution. You could also have used the fact that \((a+b+c)^{2} \ge 3(ab+bc+ca)\)

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You got the following two inequalities,\[a^2+b^2+c^2>ab+bc+ca\\ and\\ a^2+b^2+c^2>\dfrac{3}{3}\],what next?How can you conclude that,\[\dfrac{3}{4}>ab+bc+ca\]?

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(sum(a^2))(sum(b^2)) > (sum(ab))^2

sum is cyclic

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Let \(a,b,c,d \in \mathbb{N}\) such that \(a \ge b \ge c \ge d \) and \(d \neq 0\). Show that the equation \(x^4 - ax^3 - bx^2 - cx -d = 0\) has no integer solution in \(x\).

Please post a "

complete solution" like you would do in RMO.Log in to reply

I'll have a crack at it. As \(d \neq 0\), \(x \neq 0\).

Denote \(y = \frac{1}{x}\).

We have

\[x^4 = ax^3 + bx^2 + cx + d \quad \quad\]

\[1 = ay + by^2 + cy^3 + dy^4 \quad \quad (1)\]

Now, we separate this into 2 cases:

Case 1:\(x = -n\) where \(n \in \mathbb{N}\), then \(y=-\dfrac{1}{n}\). Substituting this into \((1)\), we get

\[\dfrac {-a}{n} + \dfrac {b}{n^2} + \dfrac {-c}{n^3} + \dfrac {d}{n^4} = 1\]

\[\dfrac {-an+b}{n^2}+\dfrac{-cn+d}{n^4}=1\]

Here, the LHS is non-positive since \(a \geq b\) and \(c \geq d\), whereas the RHS is positive. Thus no solutions exist in this case.

Case 2:\(x = n\) where \(n \in \mathbb{N}\), then \(y=\dfrac{1}{n}\). Substituting this into \((1)\), we get

\[\dfrac {a}{n} + \dfrac {b}{n^2} + \dfrac {c}{n^3} + \dfrac {d}{n^4} = 1\]

This implies that \(n \neq 1\), and

\[\dfrac {a}{n} < 1\]

\[1 \leq \dfrac {a}{n} + \dfrac {a}{n^2} + \dfrac {a}{n^3} + \dfrac {a}{n^4} < \displaystyle \sum_{i=1}^{\infty} \dfrac{a}{n^i} = \dfrac {a}{n-1}\]

These two equations imply that

\[\dfrac {n}{a} > 1\]

\[\dfrac {n-1}{a} < 1\]

There are no such \(n\) that satisfy the above two inequalities. Thus, no solutions exist.

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Isn't this from INMO 2013 ??

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Yes it is.

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Let \(p\) and \(q\) be distinct primes. Show that

\(\displaystyle \sum_{j=1}^{(p-1)/2}[\frac{qj}{p}]+ \sum_{j=1}^{(q-1)/2}[\frac{pj}{q}]=\dfrac{(p-1)(q-1)}{4}\).

Note that \([ \ ]\) represents the greatest integer function.

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Let \(S={(x,y)|x,y \in N, \quad 1 \le x \le (p-1)/2, \quad 1 \le y \le (q-1)/2}\).

We note that the set \(S\) can be partitioned into two sets \(S_1\) and \(S_2\) according as \(qx>py\) or \(qx<py\). Note that there are no pairs in such \(S\) such that \(qx=py\).

The set \(S_1\) can be described as the set of all pairs \((x,y)\) satisfying \(1 \le x \le (p-1)/2, \quad 1 \le y < qx/p\). Then \(S_1\) has \(\displaystyle \sum_{j=1}^{(p-1)/2}{\frac{qj}{p}}\) elements.

Similarly, \(S_2\) has \(\displaystyle \sum_{i=1}^{(q-1)/2}{\frac{pi}{q}}\). Hence, we get the required result.

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Find all primes \(x,y\) such that \(x^{y}-y^{x}=xy^{2}-19\).Enjoy solving this.

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Using FLT, we have

\[x^y \equiv x \pmod{y} \Rightarrow x \equiv -19 \pmod {y} \Rightarrow x+19 \equiv 0 \pmod{y}\]

\[y^x \equiv y \pmod{x} \Rightarrow -y \equiv -19 \pmod {x} \Rightarrow 19-y \equiv 0 \pmod{x}\]

From here, it is clear that either \(x=2\) or \(y=2\). We now separate this into 2 cases:

Case 1:\(y=2\)When \(y=2\), we have \(19 - 2 = 17 \equiv 0 \pmod{x}\), which implies \(x=17\). The solution in this case is \((17, 2)\). Checking, however, we find that this clearly doesn't work. So, there are no solutions in this case.

Case 2:\(x=2\)When \(x=2\), we have \(19+2 = 21 \equiv 0 \pmod {y}\), which implies \(y=3\) or \(y=7\). Checking, we find both these solutions work. Thus, 2 solutions exist in this case: \((2, 3)\) and \((2,7)\).

Therefore, only 2 solutions exist: \((2, 3)\) and \((2,7)\).

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would you please explain how x = -19 mody

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Using FLT and modulo \(x\) and \(y\) we get that,\(x+19 \equiv 0 \pmod{y},y-19 \equiv 0 \pmod{x}\),now just use parity(as suggested by @Sharky Kesa).

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If \(a,b,c\) are positive integers such that \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}<1\). Prove that the maximum value of \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) is \(\dfrac{41}{42}\)

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If none of \(a,b,c=2\), the maximum value of the expression will be \(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}<\frac{41}{42}\). So, one of them has to \(2\) say \(a\). So we have to prove that the maximum value of \(\frac{1}{b}+\frac{1}{c}\) is \(\frac{10}{21}\). If none of \(b,c=3\), the maximum of the expression will be \(\frac{1}{4}+\frac{1}{5}=\frac{9}{20}<\frac{10}{21}\). So, one them has to be \(3\) say \(b\). Therefore, \(\frac{1}{c} \le \frac{1}{7}\).Therefore the maximum value of \(c\) is \(7\) and the maximum value of \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) is \(\frac{41}{42}\).

Not sure if this is the best method.

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Find all reals \(x, y, z\) in (1,∞) such that

\[x + y + z + \dfrac{3}{x-1}+ \dfrac{3}{y-1} + \dfrac{3}{z-1} = 2(\sqrt{x+2} + \sqrt{y+ 2} + \sqrt{z+2})\]

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https://brilliant.org/problems/three-variables-one-equation/?group=4bQcwX9pLM2R#!/solution-comments/106741/

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Let \(x,y\) be real numbers such that \(x,y > 1\)

Prove that \(\dfrac{x^{2}}{y-1} + \dfrac{y^{2}}{ x-1} \geq 8\)

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Just use AM-GM inequality directly on the expression in the LHS and then use the fact that (a-2)^2>0 Then rearrange a bit. You get the required answer

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Alternatively we can substitute p = x-1 and q = y-1.

Then it will be easy manipulate the expression.

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Nice idea!

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Since the expression is symmetric, let's assume without loss of generality that \(x \ge y\).

\(\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge \frac{y^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 4+\frac{y^{2}}{x-1} \implies \frac{x^{2}}{y-1} \ge 4\). Minimum value of \(\frac{x^{2}}{y-1}\) is \(4\) and it occurs if and only if \(x=y=2\).

Therefore, \(\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 8\).

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Hmm you have not proven that min value of \(\frac{x^{2}}{y-1}\) is 4.

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Everyone here who have posted questions on this board and have not got any solutions please post the solutions now as RMO is only 42 hours from now. Its better that we understand the methods. Sorry that I cannot send the solution of my geometry question as I myself have been trying that question but could not arrive at the solution.

And please post some more questions if possible :)

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Done :)

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Thanks

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\(1991^{k}|(1990^{1991^{1992}}+1992^{1991^{1990}})\)

\(max(k)=?\)

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1991

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Yes..... Can you post solution ?

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How ? Could u give some rough outline of your solution ?

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https://brilliant.org/problems/triangular-grid-problem/?group=U3Rx70QSpE0z&ref_id=1044906

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Try this set guys.

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Here is a geometry question. Waiting for numerous solutions. In a triangle ABC, excircle opposite to A touches BC at point Q. Let M be the midpoint of BC. Incircle of triangle ABC touches BC at point P. Prove that PM=MQ. I have been trying this question but I am stuck. So please solve this question.

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Its easy , note that \(BP=QC=s-b\) where \(s\) is the semiperimeter of the triangle and \(b\) is length of side opposite to \( \angle B\) respectively. Since \(M\) is mid point of \(BC\) , the result follows.

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Oh nice !

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Are protractors allowed for RMO? Nothing about it is mentioned in the website and the hall ticket.

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Let d(k) denote the sum of the digits of k in base10. Find all natural numbers n such that n, d(n), d(d(n)) sum to 2015.

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I am getting no solution. Is that right ? My proof : n,d(n),d(d(n)) are congruent to the same number say 'x' (mod 9). So their sum is congruent to 3x (mod 9). Therefore 2015 is congruent to 3x ( mod 9) which is impossible as 3 does not divide 2015

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Find all pairs \((m,n)\) such that \(2^m+3^n\) is a perfect square. (A popular NT problem)

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Its easy.Only one solution (2,2)

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Wrong. Check again.

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If \(m\) is a positive real number and satisfies the equation \[2m=1+\frac{1}{m}+\sqrt{m^{2}-\frac{1}{4}}\] Find \(m\).

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Could you send the solution of this question. I tried expanding and landed up with a fourth degree polynomial whose roots I am unable to guess

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Prove that

\[ 2 < (\dfrac{n+2}{n+1}) ^ {(n+1)}\]for all integers \(n > 0\).Log in to reply

Well its easy just use Bernoulli's inequality that (1+x)^n > 1+nx

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Prove that

\(1<\frac { 1 }{ 1001 } +\frac { 1 }{ 1002 } +.........+\frac { 1 }{ 3001 } <\frac { 4 }{ 3 } \)

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It is easy. I got a calculus and a non calculus answer

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Since Rmo is a precalculus olympiad, answer without calculus will be better:P

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Alternate text

This is an image file :D

this is a copy paste from this note on moderators

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Show that \(n^{5}+n^{4}+1\) is not prime for \(n>1\).

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\(n^{2} + n + 1\) is factor of the given expression.Hence the result is obvious.

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If \(a,b,c,d,e\) are real numbers, prove that the roots of \(x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e=0\) cannot all be real if \(2a^{2}<5b\).

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Let roots be \(p,q,r,s,t\).

Sum of roots = -a & Product of roots taken two at a time = b.

Therefore Sum of squares of roots = \(a^{2} - 2b\)

If all roots are real, then by Titu's Lemma,

\(a^{2} -2b \geq \frac{a^{2}}{5} \)

\(\implies 2a^{2} \geq 5b\)

Therefore the polynomial cannot have all roots real if \( 5b > 2a^{2}\).

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Let \(a,b \in R, a \neq 0\). Show that \(a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{b}{a} \ge \sqrt{3}\).

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Well the question becomes very easy if u make it a quadratic in b and then prove that the Discriminant is less than 0. I did it that way.

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An easy INMO geometry problem:

Let \(ABC\) be a triangle with circumcircle \(\Gamma\). Let \(M\) be a point in the interior of triangle \( ABC\) which is also on the bisector of \(\angle A\). Let \(AM, BM, CM\) meet \( \Gamma\) in \( A_{1}, B_{1}, C_{1}\) respectively. Suppose \(P\) is the point of intersection of \( A_{1}C_{1}\) with \(AB\) and \(Q\) is the point of intersection of \(A_{1}B_{1}\) with \(AC\). Prove that \(PQ\) is parallel to \(BC\) .

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Let \(ABC\) be an acute-angled tiangle .Let O denote its circumcenter.Let \(\Pi\) be the circle passing through the points A,O,B.Lines CA and CB meet \(\Pi\) again at P and Q respectively.Prove that PQ is perpendicular to the line CO.

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Harsh,could you please add a figure?

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Hint:Let \(\overrightarrow{QP} \cap \overleftrightarrow{CO} = \{R\}\) and \(\overleftrightarrow{CO} \cap \Pi = \{S\}\) and then it will suffice to prove that \(\Delta CRP \sim \Delta CAS\)Log in to reply

Find the least positive integer n such that 2549 divides \(n^{2545} - 2541\)

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Please post the solution , Thanks!

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Let \(ABC\) be a triangle and \(P\) be a point inside it.Let \(x,y,z\) denote the lengths of \(PA\),\(PB\),\(PC\).Let \(a,b,c\) be the lengths of sides \(BC\),\(CA\),\(AB\).Find all points \(P\) such that \(axy+byz+czx=abc\).

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If a circle goes through point \(A\) of a parallelogram \(ABCD\), cuts the two sides \(AB,AD\) and the diagonal \(AC\) at points \(P,R\) and \(Q\) respectively. Prove that \((AP×AB)+(AR×AD)=AQ×AC\).

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I hope you all have drawn the diagram.(Consider all the segments below as vectors in the given direction.)

Let \(AS\) be the diameter of given circle.

\(AB+AD=AC\)

Therefore taking dot product with\(AS\)

\(AB.AS+AD.AS=AC.AS \)

Using, \(a.b=|a|*|b|*cos(\theta) \) [\( \theta\) is angle between the vectors]

\(|AB|*|AP|+|AR|*|AD|=|AQ|*|AC| \)

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Comment deleted Dec 04, 2015

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You can also use Ptolemy's theorem to the quadrilateral \(PQRA\) and then replace the sides of \(\Delta PQR\) by the corresponding sides of the

similartriangle \(CBA\).( \(\Delta PQR \sim \Delta CBA\) by \(AAA\) similarity.)Log in to reply

Given seven arbitrary distinct real numbers, show that there exist two numbers \(x\) and \(y\) such thst \(0<\dfrac{x-y}{1+xy}<\dfrac{1}{\sqrt{3}}\).

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The question is easy. Consider x and y to be some tan(A) and tan(B) respectively and consider A and B to vary from 0 to pi so that all reals are covered. now divide it into six equal portions spaced by pi/6. Now u have 6 pigeons and 7 holes and you are done. Then that expression is simply tan(A-B).

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There are given \(2^{500}\) points on a circle labeled \(1, 2, . . . , 2^{500}\)in some order. Prove that one can choose 100 pairwise disjoint chords joining some of these points so that the 100 sums of the pairs of numbers at the endpoints of the chosen chords are equal.

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In the first question (sum(ω))^2>3sum(ω1*ω2) fails to hold? why?

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@Harsh Shrivastava @Dev Sharma @Adarsh Kumar @Svatejas Shivakumar @Saarthak Marathe@Nihar Mahajan Where are u guys ?????? Please be active on the INMO practise board as well like u were on this board.... we need a lot of questions and a lot of solutions .....

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Could someone please post some nice RMO-INMO level Geometry problems ? I have nothing to work on. Please not from the previous years.

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See the problems of this set. Can you suggest me some good books for Geometry? My geometry is horrible.

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Thanks for that set ! For geometry I have been solving only Challenges and Thrills ... I don't have something exclusively for geometry....

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@Nihar Mahajan @Saarthak Marathe @Svatejas Shivakumar @Harsh Shrivastava @Adarsh Kumar @Dev Sharma have your respective region's RMO results been declared ? Mumbai RMO results are declared and I am selected for INMO which is on January 17. Less than a month !!!! I think we should get back to work again. Lets start an INMO practise board just like this one as it helped me a lot in my RMO preparation. Hoping for a quick response

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BTW how many did you solve?

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I didn't qualify :(

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Hey !! That doesn't stop you from taking part in the discussions on INMO. Its okay ... Ur just in 8th standard, I came to know about RMO in 10th ...so u are at an advantage. And I am quite sure u will make it to the IMO by your 11-12th standard. I just don't know how u r able to solve olympiad questions being in 8th standard. So now bro .... u r targeting for IMO and thus u r participating in the INMO board ...

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Congratulations!! The results have not been declared from my region (I don't think that I have much chances of qualifying this time. I did many silly mistakes :(

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Congo dude! I am a bit busy right now!Sorry I can't take part in the discussion now!

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Congo!

Btw my region results are not out yet.

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Thanks. Which is your region ? And what is the approx result date for the past few years ?

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All the best to everyone writing RMO!

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All the best everyone ! :)

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All the best everyone.!!

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All the best everyone! :)

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Prove that the function \(f(n)=\lfloor n+\sqrt{n}+\frac{1}{2} \rfloor\) where \(n\) is a positive integer, contains all numbers except perfect squares.

\((\lfloor \rfloor)\) represents the greatest integer function.

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I have been trying this question but I have got no hint of even how to start. Could u give some starting line ?

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Is n a natural number or something like that ? Because if n is any real then put n=0.5 . You will get f(0.5)=1 which is a perfect square.

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yes n is a natural number. Sorry forgot to mention it

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There should be no perfect squares in the domain or the range?

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Range.

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For any integer \(a\), prove that \(a^{37} \equiv a \pmod{1729}\).

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Its easy just apply keep applying FLT for various factors of 1729

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Write the sum \(S_n= \displaystyle \sum_{k=0}^n{\frac{(-1)^{k}{n \choose{k}}}{k^{3}+ 9k^{2}+26k+24}}\) in the form \(\frac{p(n)}{q(n)}\) where \(p(n),q(n)\) are polynomials with integer coefficients.

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We first note that \(k^{3}+9k^{2}+26k+24=(k+2)(k+3)(k+4)\).Hence, \[S_n=\displaystyle \sum_{k=0}^n{\frac{(-1)^{k}{n \choose{k}}}{(k+2)(k+3)(k+4)}}\] \[=\displaystyle \sum_{k=0}^n{\frac{(-1)^{k}n!}{(n-k)k!(k+2)(k+3)(k+4)}}\] \[=\displaystyle \sum_{k=0}^n{\frac{(-1)^{k}(n+4)!}{(n-k!)(k+4!)}}\frac{(k+1)}{(n+1)(n+2)(n+3)(n+4)}\]

Let \(T_n=S_n(n+1)(n+2)(n+3)(n+4)=\small \displaystyle \sum_{k=0}^n{(-1)^{k}{n+4 \choose{k+4}}(k+1)}\).

We note that \(\small \displaystyle \sum_{k=0}^n{(-1)^{k}{n \choose{k}}}\).Now,

\(\displaystyle \sum_{k=0}^n{(-1)^{k}{n \choose{k}}k}\)\(=\displaystyle \sum_{k=1}^n{(-1)^{k}n{n-1 \choose{k-1}}+0}\)\(=-n \displaystyle \sum_{j=0}^{n-1}{(-1)^{j}{n-1 \choose{j}}=0}\).

\[T_n=\displaystyle \sum_{k=0}^n{(-1)^{k}{n+4 \choose{k+4}}(k+1)}\]. \[=\displaystyle \sum_{k=0}^n{(-1)^{k+4}{n+4 \choose{k+4}}(k+1)}\]. \[=\displaystyle \sum_{k=-4}^n{(-1)^{k+4}{n+4 \choose{k+4}}(k+1)-((-3)+2(n+4)-{n+4 \choose{2}}})\].

Substituting \(j=k+4\);

\[T_n=\displaystyle \sum_{j=0}^{n+4}{(-1)^{j}{n+4 \choose{j}}(j-3)-((-3)+2(n+4)-{n+4 \choose{2}}})\].

\(T_n=\displaystyle \sum_{j=0}^{n+4}{(-1)^{j}{n+4 \choose{j}}}-3 \displaystyle \sum_{j=0}^{n+4}{(-1)^{j}{n+4 \choose{j}}(j-3)-((-3)+2(n+4)-{n+4 \choose{2}}})\).

The first two terms are zero, hence

\(T_n={n+4 \choose{j}}-2n-8+3=\frac{(n+4)(n+3)}{2}-2n-5\).

Hence, \(T_n=\frac{(n+1)(n+2)}{2}\) and \(S_n=\frac{1}{2(n+3)(n+4)}\).

Phew!!

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Nicely done. There are a few mistakes in the solution but one may understand the correct part as he reads it.

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A barrel contains \(2n\) balls, numbered \(1\) to \(2n\). Choose three balls at random, one after the other, and with the balls replaced after each draw.

What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, is repeated?

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1/2

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Correct. Can you show your steps?

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\(a,b,c\) are real numbers such that \(a+b+c=0\) and \(a^2+b^2+c^2=1\). Prove that \(a^{2}b^{2}c^{2}≤\frac{1}{54}\).

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See my note

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Sorry didn't see this post but since I found it, here you go - 65 bugs are placed at different squares of a 9*9 square board. Abug in each moves to a horizontal or vertical adjacent square. No bug makes 2 horizontal or vertical moves in succession. Show that after some moves, there will be at least 2 bugs in the same square.

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