# RMO Practice Board!

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 $f$ Note by Harsh Shrivastava
5 years, 6 months ago

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## Comments

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Since 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$\dfrac{4}{8}=\sqrt{f}\\ \Longrightarrow \dfrac{1}{256}=f$,hence there is only one value of $f=\dfrac{1}{256}$.Is this correct Harsh?

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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.

- 5 years, 5 months ago

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Yes , its one of the short Newton identities.

- 5 years, 5 months ago

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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 .

- 5 years, 6 months ago

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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

- 5 years, 6 months ago

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A very elegant solution

- 1 year, 8 months ago

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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.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

$\QED$

- 5 years, 6 months ago

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Find all positive integer solutions $a$, $b$ and $c$ such that

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

- 5 years, 6 months ago

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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 $a,b$

- 5 years, 6 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.

- 5 years, 6 months ago

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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 $\pm 1 \pmod{6}$.

- 5 years, 6 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.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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I didn't get it.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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 $\dfrac{41}{42}$

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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 $(2,7)$.

- 5 years, 6 months ago

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would you please explain how x = -19 mody

- 5 years, 6 months ago

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Putting $x^y-y^x=xy^2-19$ in $\pmod{y}$, we get $x=-19 \pmod{y}$.

- 5 years, 6 months ago

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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).

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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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. 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.

- 5 years, 6 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.

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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Isn't this from INMO 2013 ??

- 5 years, 5 months ago

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Yes it is.

- 5 years, 5 months ago

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

- 5 years, 5 months ago

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let n be an integral solution (if possible). then n [n^3-an^2-bn-c] = d. Now for the equation to hold n divides d. so n is less than or equal to d. If n is negative -ax^3 is positive and is greater than bx^2 as a is greater than b.similarly for the next term and the total value becomes positive but known the value is 0.hence our assumption that n is negative is wrong .so n is positive and less than d. Now as n<d <a, n^4<an^3 and -bn^2 and -cx and -d all are negative,so the sum is also negative which was again 0. hence contradiction. So no integral roots exists.

- 1 year, 1 month ago

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

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

- 5 years, 6 months ago

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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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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,$\dfrac{3}{4}>ab+bc+ca$?

- 5 years, 6 months ago

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using cauchy again, like this

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

sum is cyclic

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Prove that $2 < (\dfrac{n+2}{n+1}) ^ {(n+1)}$ for all integers $n > 0$.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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 $p^{2}$.

- 5 years, 6 months ago

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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 ?

- 5 years, 6 months ago

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Please post the proof, thanks!

- 5 years, 6 months ago

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My proof is incomplete I could not prove that a is divisible by p^2

- 5 years, 6 months ago

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Post the proof that a is divisible by p, Thanks!

- 5 years, 6 months ago

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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 out

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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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 $\angle BAC$.

- 5 years, 6 months ago

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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 :)

- 5 years, 6 months ago

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Shrihari,we get that $CF=2DF$

- 5 years, 6 months ago

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I don't think so check again.

- 5 years, 6 months ago

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Applying Menelaus' theorem, ( (BE/EC)(CF/FD)(DA/AB)=1

- 5 years, 6 months ago

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BE/EC is 2 and DA/AB is 1/2. So CF/FD is 1

- 5 years, 6 months ago

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Yup I also got 90 degrees.

- 5 years, 6 months ago

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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?

- 5 years, 6 months ago

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FD=FC=FA . Great !

- 5 years, 6 months ago

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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 it

- 5 years, 6 months ago

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I also need some NT probs...

- 5 years, 6 months ago

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Even I need NT problems...

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Pune region.

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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).

- 5 years, 6 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 $(AP×AB)+(AR×AD)=AQ×AC$.

- 5 years, 6 months ago

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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]

$|AB|*|AP|+|AR|*|AD|=|AQ|*|AC|$

- 5 years, 6 months ago

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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 $axy+byz+czx=abc$.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Please post the solution , Thanks!

- 5 years, 6 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.

- 5 years, 6 months ago

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Harsh,could you please add a figure?

- 5 years, 6 months ago

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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 $\Delta CRP \sim \Delta CAS$

- 5 years, 6 months ago

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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 $BC$ .

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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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 $2a^{2}<5b$.

- 5 years, 6 months ago

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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 $5b > 2a^{2}$.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Prove that

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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First use an image uploading website (I prefer postimg.org ), upload your picture, and copy the image link. The format for uploading images in a solution is

![Alternate text](image URL)

. Type the text which you wish to be displayed if the image doesn't load in the third brackets, and paste the image URL in the second brackets. In the following example, the image URL is http://s12.postimg.org/5qhlgca71/untitled.png, and I wrote the code [broken link: http://s12.postimg.org/5qhlgca71/untitled.png]. Here's how it appears:

this is a copy paste from this note on moderators

- 5 years, 6 months ago

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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 $m$.

- 5 years, 6 months ago

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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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Its easy.Only one solution (2,2)

- 5 years, 6 months ago

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Wrong. Check again.

- 5 years, 6 months ago

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I think he means (4,2).

- 5 years, 6 months ago

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Ya I meant (4,2). Sorry,typing error. :P

- 5 years, 6 months ago

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Its okay , please post your method. :)

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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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

- 5 years, 6 months 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.

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Oh nice !

- 5 years, 6 months ago

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Try this set guys.

- 5 years, 6 months ago

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

$max(k)=?$

Details:

$b|a\rightarrow$ $a$ is divisible by $b$.

- 5 years, 6 months ago

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1991

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Shrihari,I have posted the general result of that combinatorics problem in its 'discuss solution' part. Now try to prove the general result .

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

- 5 years, 6 months ago

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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 advance

- 5 years, 6 months ago

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Yes..... Can you post solution ?

- 5 years, 6 months ago

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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 :)

- 5 years, 6 months ago

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Done :)

- 5 years, 6 months ago

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Thanks

- 5 years, 6 months ago

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Let $x,y$ be real numbers such that $x,y > 1$

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

- 5 years, 6 months ago

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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

- 5 years, 6 months ago

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

Then it will be easy manipulate the expression.

- 5 years, 6 months ago

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Nice idea!

- 5 years, 6 months ago

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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, $\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 8$.

- 5 years, 6 months ago

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

- 5 years, 6 months ago

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Do you mean that I haven't proven that minimum value of $\frac{y^{2}}{y-1}$ is $4$?

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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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 $a^{2}b^{2}c^{2}≤\frac{1}{54}$.

- 5 years, 6 months ago

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See my note

- 5 years, 6 months ago

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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?

- 5 years, 6 months ago

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1/2

- 5 years, 6 months ago

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Correct. Can you show your steps?

- 5 years, 6 months ago

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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.

- 5 years, 6 months ago

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