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 \(f\)

## 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? – Adarsh Kumar · 10 months, 1 week ago

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– Shivam Jadhav · 10 months, 1 week ago

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

– Trevor Arashiro · 9 months ago

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

– Nihar Mahajan · 9 months ago

Yes , its one of the short Newton identities.Log in to reply

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 . – Anish Harsha · 10 months ago

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– Souryajit Roy · 10 months ago

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 ptLog in to reply

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

\[\dfrac {7}{a^2} + \dfrac {11}{b^2} = \dfrac {c}{77}.\] – Sharky Kesa · 10 months ago

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

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 \(a,b\) – Harsh Shrivastava · 10 months ago

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If \(x\) and \(y\) are integers such that \(y^2+3(xy)^2=30x^2+517\), Prove that \(3(xy)^2=588\) – Lakshya Sinha · 10 months, 1 week ago

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

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

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

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

\(\QED \) – Harsh Shrivastava · 10 months ago

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Let \(x,y\) and \(z\) be positive real numbers such that \(x+y+z=1\). Prove that \(\frac{yz}{1+x}+\frac{zx}{1+y}+\frac{xy}{1+z} \le \frac{1}{4}\). – Svatejas Shivakumar · 10 months ago

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Apply AM-HM and then manipulate on the RHS. – Saarthak Marathe · 10 months ago

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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. – Saarthak Marathe · 10 months ago

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– Shrihari B · 9 months, 4 weeks ago

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

– Saarthak Marathe · 9 months, 3 weeks ago

I didn't get it.Log in to reply

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. – Adarsh Kumar · 10 months ago

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Find all primes \(p\) such that for all prime \(q\), the remainder of \(p\) upon divison by \(q\) is squarefree. – Dev Sharma · 10 months, 1 week ago

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– Svatejas Shivakumar · 9 months, 4 weeks ago

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

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 \(\angle BAC\). – Harsh Shrivastava · 9 months, 4 weeks ago

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– Shrihari B · 9 months, 4 weeks ago

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 :)Log in to reply

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– Shrihari B · 9 months, 3 weeks ago

Any solution is welcome. How did u solve using Trigonometry ?Log in to reply

– Nihar Mahajan · 9 months, 3 weeks ago

Let \(\angle FDA = \angle FAD = \alpha \ , \ \angle AEC=\gamma\).Let \(FD=FA=x \ , \ CE=y \ , \ BE=2y\). Using Sine Rule in \(\Delta ADF\) , we get \(AD=2x\cos(\alpha)\) and thus \(AB=4x\cos(\alpha)\). By angle chase we have \(\angle CFE= 180-2\alpha\) and again applying Sine Rule to \(\Delta CEF\) , we get \(FC=\dfrac{y\sin(\gamma)}{sin(2\alpha)}\). Now apply Sine Rule to \(\Delta ABE\) and knowing that \(\sin(180^\circ - \gamma)= \sin(\gamma)\) , we get \(x=\dfrac{y\sin(\gamma)}{sin(2\alpha)}= FC\) . This means \(F\) is midpoint of \(DC\) , and we get the desired result.Log in to reply

– Shrihari B · 9 months, 3 weeks ago

Nice work there !Log in to reply

– Nihar Mahajan · 9 months, 4 weeks ago

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?Log in to reply

– Shrihari B · 9 months, 4 weeks ago

FD=FC=FA . Great !Log in to reply

– Shrihari B · 9 months, 4 weeks ago

Do u have some problems for practice ? I need some number theory. I am not that good or in other words I am bad in itLog in to reply

– Harsh Shrivastava · 9 months, 4 weeks ago

I also need some NT probs...Log in to reply

– Nihar Mahajan · 9 months, 4 weeks ago

Even I need NT problems...Log in to reply

– Shrihari B · 9 months, 4 weeks ago

Are you going to appear for RMO from mumbai region or from the pune region ?Log in to reply

– Nihar Mahajan · 9 months, 4 weeks ago

Pune region.Log in to reply

– Harsh Shrivastava · 9 months, 4 weeks ago

Yup I also got 90 degrees.Log in to reply

– Saarthak Marathe · 9 months, 4 weeks ago

Shrihari,we get that \(CF=2DF\)Log in to reply

– Shrihari B · 9 months, 4 weeks ago

I don't think so check again.Log in to reply

(CF/FD)(DA/AB)=1 – Saarthak Marathe · 9 months, 4 weeks agoLog in to reply

– Shrihari B · 9 months, 4 weeks ago

BE/EC is 2 and DA/AB is 1/2. So CF/FD is 1Log in to reply

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 \(p^{2}\). – Svatejas Shivakumar · 10 months ago

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– Shrihari B · 9 months, 4 weeks ago

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 ?Log in to reply

– Harsh Shrivastava · 9 months, 3 weeks ago

Please post the proof, thanks!Log in to reply

– Shrihari B · 9 months, 3 weeks ago

My proof is incomplete I could not prove that a is divisible by p^2Log in to reply

– Harsh Shrivastava · 9 months, 3 weeks ago

Post the proof that a is divisible by p, Thanks!Log in to reply

– Shrihari B · 9 months, 3 weeks ago

Take LCM on the LHS and try proving that each term in the numerator leaves a different residue modulo p. Do it by contradiction. Once you are able to prove that, the rest is clear. If it is difficult, first take a smaller example say p=7. Then work it outLog in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

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.Log in to reply

If \(a,b \ge 0\). Prove that \(\sqrt{2}(\sqrt{a(a+b)^{3}}+b \sqrt{a^{2}+b^{2}}) \le 3(a^{2}+b^{2})\). – Svatejas Shivakumar · 10 months ago

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

Prove that \(ab+bc+ca \leq \frac{3}{4}\) – Harsh Shrivastava · 10 months ago

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(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 – Dev Sharma · 10 months ago

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– Svatejas Shivakumar · 10 months ago

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

– Saarthak Marathe · 10 months ago

Dev,your solution is incorrect. Check the solution again,you will find your flaw.Log in to reply

– Adarsh Kumar · 10 months ago

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,\[\dfrac{3}{4}>ab+bc+ca\]?Log in to reply

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

sum is cyclic – Dev Sharma · 10 months ago

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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. – Nihar Mahajan · 10 months agoLog in to reply

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. – Sharky Kesa · 10 months ago

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– Shrihari B · 9 months ago

Isn't this from INMO 2013 ??Log in to reply

– Nihar Mahajan · 9 months ago

Yes it is.Log in to reply

– Shrihari B · 9 months ago

Hey why aren't u participating in the INMO practice board ?Log in to reply

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. – Svatejas Shivakumar · 10 months ago

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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. – Svatejas Shivakumar · 9 months, 4 weeks ago

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Find all primes \(x,y\) such that \(x^{y}-y^{x}=xy^{2}-19\).Enjoy solving this. – Souryajit Roy · 10 months ago

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

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

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 \((2,7)\). – Sharky Kesa · 10 months ago

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– Dev Sharma · 10 months ago

would you please explain how x = -19 modyLog in to reply

– Sharky Kesa · 10 months ago

Putting \(x^y-y^x=xy^2-19\) in \(\pmod{y}\), we get \(x=-19 \pmod{y}\).Log in to reply

@Sharky Kesa). – Adarsh Kumar · 10 months ago

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 byLog in to reply

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 \(\dfrac{41}{42}\) – Svatejas Shivakumar · 10 months ago

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Not sure if this is the best method. – Svatejas Shivakumar · 10 months ago

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

\[x + y + z + \dfrac{3}{x-1}+ \dfrac{3}{y-1} + \dfrac{3}{z-1} = 2(\sqrt{x+2} + \sqrt{y+ 2} + \sqrt{z+2})\] – Harsh Shrivastava · 10 months ago

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– Surya Prakash · 10 months ago

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

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

Prove that \(\dfrac{x^{2}}{y-1} + \dfrac{y^{2}}{ x-1} \geq 8\) – Harsh Shrivastava · 9 months, 3 weeks ago

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– Shrihari B · 9 months, 3 weeks ago

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 answerLog in to reply

Then it will be easy manipulate the expression. – Harsh Shrivastava · 9 months, 3 weeks ago

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– Nihar Mahajan · 9 months, 3 weeks ago

Nice idea!Log in to reply

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

Therefore, \(\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 8\). – Svatejas Shivakumar · 9 months, 3 weeks ago

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– Harsh Shrivastava · 9 months, 3 weeks ago

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

– Svatejas Shivakumar · 9 months, 3 weeks ago

Do you mean that I haven't proven that minimum value of \(\frac{y^{2}}{y-1}\) is \(4\)?Log in to reply

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 :) – Shrihari B · 9 months, 4 weeks ago

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– Svatejas Shivakumar · 9 months, 4 weeks ago

Done :)Log in to reply

– Shrihari B · 9 months, 4 weeks ago

ThanksLog in to reply

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

\(max(k)=?\)

– Akshat Sharda · 9 months, 4 weeks agoDetails:Log in to reply

– Saarthak Marathe · 9 months, 3 weeks ago

1991Log in to reply

– Akshat Sharda · 9 months, 3 weeks ago

Yes..... Can you post solution ?Log in to reply

– Shrihari B · 9 months, 3 weeks ago

How ? Could u give some rough outline of your solution ?Log in to reply

https://brilliant.org/problems/triangular-grid-problem/?group=U3Rx70QSpE0z&ref_id=1044906 – Saarthak Marathe · 9 months, 3 weeks ago

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– Shrihari B · 9 months, 3 weeks ago

Hey Saarthak I haven't heard of Catalan numbers before. You need to elaborate a bit Could you give a proof for the number of ways ? It will help me understand better. Thanks in advanceLog in to reply

Try this set guys. – Akshat Sharda · 9 months, 4 weeks ago

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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. – Shrihari B · 10 months ago

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– Nihar Mahajan · 9 months, 4 weeks ago

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.Log in to reply

– Shrihari B · 9 months, 4 weeks ago

Oh nice !Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

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

– Nihar Mahajan · 9 months, 4 weeks ago

In the instructions given in question paper , it is said that protractors are not allowed.Log in to reply

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. – Rakhi Bhattacharyya · 10 months ago

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– Shrihari B · 10 months ago

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 2015Log in to reply

Find all pairs \((m,n)\) such that \(2^m+3^n\) is a perfect square. (A popular NT problem) – Nihar Mahajan · 10 months ago

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– Saarthak Marathe · 9 months, 3 weeks ago

Its easy.Only one solution (2,2)Log in to reply

– Nihar Mahajan · 9 months, 3 weeks ago

Wrong. Check again.Log in to reply

– Svatejas Shivakumar · 9 months, 3 weeks ago

I think he means (4,2).Log in to reply

– Saarthak Marathe · 9 months, 3 weeks ago

Ya I meant (4,2). Sorry,typing error. :PLog in to reply

– Nihar Mahajan · 9 months, 3 weeks ago

Its okay , please post your method. :)Log in to reply

– Saarthak Marathe · 9 months, 3 weeks ago

Taking mod 4 we get that m,n should be even. Then using \( gcd({2}^{m},3^{n})=1 \), we use the general formula for pythagorean triplets. Taking various cases,we get that only \( m=4,n=2\) are the solutions.Log in to reply

If \(m\) is a positive real number and satisfies the equation \[2m=1+\frac{1}{m}+\sqrt{m^{2}-\frac{1}{4}}\] Find \(m\). – Shivam Jadhav · 10 months ago

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– Shrihari B · 9 months, 4 weeks ago

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 guessLog in to reply

Prove that

\[ 2 < (\dfrac{n+2}{n+1}) ^ {(n+1)}\]for all integers \(n > 0\). – Harsh Shrivastava · 10 months agoLog in to reply

– Shrihari B · 10 months ago

Well its easy just use Bernoulli's inequality that (1+x)^n > 1+nxLog in to reply

Prove that

\(1<\frac { 1 }{ 1001 } +\frac { 1 }{ 1002 } +.........+\frac { 1 }{ 3001 } <\frac { 4 }{ 3 } \) – Satyajit Ghosh · 10 months ago

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– Saarthak Marathe · 10 months ago

It is easy. I got a calculus and a non calculus answerLog in to reply

– Satyajit Ghosh · 10 months ago

Since Rmo is a precalculus olympiad, answer without calculus will be better:PLog in to reply

– Saarthak Marathe · 10 months ago

Can u tell me how to add an image in the answer?Log in to reply

Alternate text

This is an image file :D

this is a copy paste from this note on moderators – Satyajit Ghosh · 10 months ago

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– Saarthak Marathe · 10 months ago

I think we can use calculus in rmo .But if we use it we need to be very careful in our steps and not miss any points.Log in to reply

Show that \(n^{5}+n^{4}+1\) is not prime for \(n>1\). – Svatejas Shivakumar · 10 months ago

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– Harsh Shrivastava · 10 months ago

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

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 \(2a^{2}<5b\). – Svatejas Shivakumar · 10 months ago

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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 \( 5b > 2a^{2}\). – Harsh Shrivastava · 10 months ago

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Let \(a,b \in R, a \neq 0\). Show that \(a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{b}{a} \ge \sqrt{3}\). – Svatejas Shivakumar · 10 months ago

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– Shrihari B · 10 months ago

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.Log in to reply

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 \(BC\) . – Nihar Mahajan · 10 months ago

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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. – Harsh Shrivastava · 10 months ago

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– Adarsh Kumar · 10 months ago

Harsh,could you please add a figure?Log in to reply

Hint:Let \(\overrightarrow{QP} \cap \overleftrightarrow{CO} = \{R\}\) and \(\overleftrightarrow{CO} \cap \Pi = \{S\}\) and then it will suffice to prove that \(\Delta CRP \sim \Delta CAS\) – Nihar Mahajan · 10 months agoLog in to reply

Find the least positive integer n such that 2549 divides \(n^{2545} - 2541\) – Dev Sharma · 10 months ago

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– Harsh Shrivastava · 9 months, 4 weeks ago

Please post the solution , Thanks!Log in to reply

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 \(axy+byz+czx=abc\). – Souryajit Roy · 10 months ago

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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 \((AP×AB)+(AR×AD)=AQ×AC\). – Svatejas Shivakumar · 10 months ago

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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]

\(|AB|*|AP|+|AR|*|AD|=|AQ|*|AC| \) – Saarthak Marathe · 9 months, 4 weeks ago

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similartriangle \(CBA\).( \(\Delta PQR \sim \Delta CBA\) by \(AAA\) similarity.) – Svatejas Shivakumar · 9 months, 4 weeks agoLog in to reply

Given seven arbitrary distinct real numbers, show that there exist two numbers \(x\) and \(y\) such thst \(0<\dfrac{x-y}{1+xy}<\dfrac{1}{\sqrt{3}}\). – Svatejas Shivakumar · 10 months ago

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– Shrihari B · 10 months ago

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).Log in to reply

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. – Lakshya Sinha · 10 months, 1 week ago

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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 ..... – Shrihari B · 9 months ago

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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. – Shrihari B · 9 months, 2 weeks ago

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this set. Can you suggest me some good books for Geometry? My geometry is horrible. – Svatejas Shivakumar · 9 months ago

See the problems ofLog in to reply

– Shrihari B · 9 months ago

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

@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 – Shrihari B · 9 months, 2 weeks ago

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– Adarsh Kumar · 9 months, 2 weeks ago

BTW how many did you solve?Log in to reply

– Svatejas Shivakumar · 9 months ago

I didn't qualify :(Log in to reply

– Shrihari B · 9 months ago

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 ...Log in to reply

– Svatejas Shivakumar · 9 months ago

Thanks for your encouragement. I will definitely participate in the INMO board and help you reach IMO this year!Log in to reply

– Shrihari B · 9 months ago

Frankly speaking even INMO is impossible for me .... but I am just enjoying math. I have got an excuse of not studying for IIT-JEE till Jan 17(INMO).Log in to reply

– Svatejas Shivakumar · 9 months ago

You're not an IIT-JEE aspirant?Log in to reply

– Svatejas Shivakumar · 9 months, 2 weeks ago

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 :(Log in to reply

– Adarsh Kumar · 9 months, 2 weeks ago

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

Btw my region results are not out yet. – Harsh Shrivastava · 9 months, 2 weeks ago

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– Shrihari B · 9 months, 2 weeks ago

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

– Harsh Shrivastava · 9 months, 2 weeks ago

Chhattisgrah State.Log in to reply

All the best to everyone writing RMO! – Svatejas Shivakumar · 9 months, 3 weeks ago

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– Shrihari B · 9 months, 3 weeks ago

All the best everyone ! :)Log in to reply

– Saarthak Marathe · 9 months, 3 weeks ago

All the best everyone.!!Log in to reply

– Nihar Mahajan · 9 months, 3 weeks ago

All the best everyone! :)Log in to reply

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. – Svatejas Shivakumar · 9 months, 3 weeks ago

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– Shrihari B · 9 months, 2 weeks ago

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

– Shrihari B · 9 months, 3 weeks ago

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.Log in to reply

– Svatejas Shivakumar · 9 months, 3 weeks ago

yes n is a natural number. Sorry forgot to mention itLog in to reply

– Saarthak Marathe · 9 months, 3 weeks ago

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

– Svatejas Shivakumar · 9 months, 3 weeks ago

Range.Log in to reply

For any integer \(a\), prove that \(a^{37} \equiv a \pmod{1729}\). – Svatejas Shivakumar · 9 months, 3 weeks ago

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– Shrihari B · 9 months, 3 weeks ago

Its easy just apply keep applying FLT for various factors of 1729Log in to reply

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. – Svatejas Shivakumar · 10 months ago

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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!! – Svatejas Shivakumar · 9 months, 4 weeks ago

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– Saarthak Marathe · 9 months, 3 weeks ago

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

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? – Svatejas Shivakumar · 10 months ago

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– Saarthak Marathe · 9 months, 4 weeks ago

1/2Log in to reply

– Svatejas Shivakumar · 9 months, 4 weeks ago

Correct. Can you show your steps?Log in to reply

\(a,b,c\) are real numbers such that \(a+b+c=0\) and \(a^2+b^2+c^2=1\). Prove that \(a^{2}b^{2}c^{2}≤\frac{1}{54}\). – Akshat Sharda · 10 months ago

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note – Satyajit Ghosh · 10 months ago

See myLog in to reply

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. – Satyajit Ghosh · 10 months ago

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