RMO Practice Board!

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

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

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

$\QED$

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

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

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 $a,b$

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

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

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.

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

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.

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 $(2,7)$.

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).

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.

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.

Isn't this from INMO 2013 ??

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.

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

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

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 ?

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 :)

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).

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]

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

Please post the solution , Thanks!

Harsh,could you please add a figure?

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.

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

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

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

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

Its easy.Only one solution (2,2)

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

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.

1991

Done :)

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

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

See my note

1/2

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!!

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

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

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.

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

All the best everyone! :)

All the best everyone.!!

All the best everyone ! :)

BTW how many did you solve?

Congo!

Btw my region results are not out yet.

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

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 :(

I didn't qualify :(

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