RMO Practice Board!

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

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

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Here is a problem to start with :

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

x84x7+7x6+ax5+bx4+cx3+dx2+ex+f=0x^{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 ff

Note by Harsh Shrivastava
3 years, 10 months ago

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Since we were given that the roots were positive reals,we get motivated and tempted to use,R.M.SA.MG.MH.MR.M.S\geq A.M\geq G.M\geq H.M,also since we have been given,α\sum \alpha and αβ\sum \alpha\beta we try to find the value of α2=422×7=2\sum \alpha^2=4^2-2\times 7=2.We apply R.M.SA.MR.M.S \geq A.M 28481212\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.MA.M=G.M,hence48=f81256=f\dfrac{4}{8}=\sqrt[8]{f}\\ \Longrightarrow \dfrac{1}{256}=f,hence there is only one value of f=1256f=\dfrac{1}{256}.Is this correct Harsh?

Adarsh Kumar - 3 years, 10 months ago

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In your solution you need prove that (cyclicω)2cyclicω1ω2=cyclicω2(\sum_{cyclic}ω)^{2}-\sum_{cyclic}ω_{1}ω_{2}=\sum_{cyclic}ω^{2}

Shivam Jadhav - 3 years, 10 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.

Trevor Arashiro - 3 years, 9 months ago

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@Trevor Arashiro Yes , its one of the short Newton identities.

Nihar Mahajan - 3 years, 9 months ago

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If x,y,zx,y,z are positive real numbers, prove that,

(x+y+z)2(yz+zx+xy)23(y2+yz+z2)(z2+zx+x2)(x2+xy+y2)\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 - 3 years, 10 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 43T\frac{4}{\sqrt{3}}T where T is the area of the triangle....then use T=abc4RT=\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

Souryajit Roy - 3 years, 10 months ago

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A very elegant solution

Popular Power - 2 days, 15 hours ago

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You have a set of consecutive positive integers,{1,2,3,...,199,200}\{1,2,3,...,199,200\},if you choose 101101 numbers from these at random,then prove that there at-least two numbers of which one divides the other.

Adarsh Kumar - 3 years, 10 months ago

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If xx and yy are integers such that y2+3(xy)2=30x2+517y^2+3(xy)^2=30x^2+517, Prove that 3(xy)2=5883(xy)^2=588

Department 8 - 3 years, 10 months ago

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Let y2=by^{2} = b and x2=ax^{2} = a.

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

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

    (b10)(3a+1)=507×1=13×39=3×169\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)=(±2,±7)(x ,y) = ( \pm 2, \pm 7).

This forces 3(xy)2=5883(xy)^{2}= 588.

\QED\QED

Harsh Shrivastava - 3 years, 10 months ago

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Find all positive integer solutions aa, bb and cc such that

7a2+11b2=c77.\dfrac {7}{a^2} + \dfrac {11}{b^2} = \dfrac {c}{77}.

Sharky Kesa - 3 years, 10 months ago

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

7a2+11b2=c77.\dfrac {7 }{a^2} + \dfrac {11 }{b^2} = \dfrac{c}{77}.

Case 1: c>1c>1

This forces a2<7 and b2<11a^{2} < 7 \text{ and } b^{2} < 11

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

Case 2: c77c \leq 77

Since c77c \leq 77 this implies c771\dfrac{c}{77} \leq 1.

Let c77=1k\dfrac{c}{77} = \dfrac{1}{k} for some integer k1k \geq 1.

    c=77k\implies c = \dfrac{77}{k}

Now since cc is a positive integer , this forces k | 77k \text{ | } 77.

Possible values for k=1,7,11,77k = 1,7,11,77.

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

Harsh Shrivastava - 3 years, 10 months ago

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Find all primes pp such that for all prime qq, the remainder of pp upon divison by qq is squarefree.

Dev Sharma - 3 years, 10 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 ±1(mod6)\pm 1 \pmod{6}.

A Former Brilliant Member - 3 years, 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 - 3 years, 10 months ago

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

Shrihari B - 3 years, 10 months ago

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I didn't get it.

Saarthak Marathe - 3 years, 10 months ago

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Let x,yx,y and zz be positive real numbers such that x+y+z=1x+y+z=1. Prove that yz1+x+zx1+y+xy1+z14\frac{yz}{1+x}+\frac{zx}{1+y}+\frac{xy}{1+z} \le \frac{1}{4}.

A Former Brilliant Member - 3 years, 10 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.

Saarthak Marathe - 3 years, 10 months ago

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

x+y+z+3x1+3y1+3z1=2(x+2+y+2+z+2)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 - 3 years, 10 months ago

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

Surya Prakash - 3 years, 10 months ago

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If a,b,ca,b,c are positive integers such that 1a+1b+1c<1\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}<1. Prove that the maximum value of 1a+1b+1c\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} is 4142\dfrac{41}{42}

A Former Brilliant Member - 3 years, 10 months ago

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If none of a,b,c=2a,b,c=2, the maximum value of the expression will be 13+14+15=4760<4142\frac{1}{3}+\frac{1}{4}+\frac{1}{5}=\frac{47}{60}<\frac{41}{42}. So, one of them has to 22 say aa. So we have to prove that the maximum value of 1b+1c\frac{1}{b}+\frac{1}{c} is 1021\frac{10}{21}. If none of b,c=3b,c=3, the maximum of the expression will be 14+15=920<1021\frac{1}{4}+\frac{1}{5}=\frac{9}{20}<\frac{10}{21}. So, one them has to be 33 say bb. Therefore, 1c17\frac{1}{c} \le \frac{1}{7}.Therefore the maximum value of cc is 77 and the maximum value of 1a+1b+1c\frac{1}{a}+\frac{1}{b}+\frac{1}{c} is 4142\frac{41}{42}.

Not sure if this is the best method.

A Former Brilliant Member - 3 years, 10 months ago

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Find all primes x,yx,y such that xyyx=xy219x^{y}-y^{x}=xy^{2}-19.Enjoy solving this.

Souryajit Roy - 3 years, 10 months ago

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Using FLT, we have

xyx(mody)x19(mody)x+190(mody)x^y \equiv x \pmod{y} \Rightarrow x \equiv -19 \pmod {y} \Rightarrow x+19 \equiv 0 \pmod{y}

yxy(modx)y19(modx)19y0(modx)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=2x=2 or y=2y=2. We now separate this into 2 cases:

Case 1: y=2y=2

When y=2y=2, we have 192=170(modx)19 - 2 = 17 \equiv 0 \pmod{x}, which implies x=17x=17. The solution in this case is (17,2)(17, 2). Checking, however, we find that this clearly doesn't work. So, there are no solutions in this case.

Case 2: x=2x=2

When x=2x=2, we have 19+2=210(mody)19+2 = 21 \equiv 0 \pmod {y}, which implies y=3y=3 or y=7y=7. Checking, we find both these solutions work. Thus, 2 solutions exist in this case: (2,3)(2, 3) and (2,7)(2,7).

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

Sharky Kesa - 3 years, 10 months ago

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would you please explain how x = -19 mody

Dev Sharma - 3 years, 10 months ago

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@Dev Sharma Putting xyyx=xy219x^y-y^x=xy^2-19 in (mody)\pmod{y}, we get x=19(mody)x=-19 \pmod{y}.

Sharky Kesa - 3 years, 10 months ago

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Using FLT and modulo xx and yy we get that,x+190(mody),y190(modx)x+19 \equiv 0 \pmod{y},y-19 \equiv 0 \pmod{x},now just use parity(as suggested by @Sharky Kesa).

Adarsh Kumar - 3 years, 10 months ago

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Let pp and qq be distinct primes. Show that

j=1(p1)/2[qjp]+j=1(q1)/2[pjq]=(p1)(q1)4\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.

A Former Brilliant Member - 3 years, 10 months ago

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Let S=(x,y)x,yN,1x(p1)/2,1y(q1)/2S={(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 SS can be partitioned into two sets S1S_1 and S2S_2 according as qx>pyqx>py or qx<pyqx<py. Note that there are no pairs in such SS such that qx=pyqx=py.

The set S1S_1 can be described as the set of all pairs (x,y)(x,y) satisfying 1x(p1)/2,1y<qx/p1 \le x \le (p-1)/2, \quad 1 \le y < qx/p. Then S1S_1 has j=1(p1)/2qjp\displaystyle \sum_{j=1}^{(p-1)/2}{\frac{qj}{p}} elements.

Similarly, S2S_2 has i=1(q1)/2piq\displaystyle \sum_{i=1}^{(q-1)/2}{\frac{pi}{q}}. Hence, we get the required result.

A Former Brilliant Member - 3 years, 10 months ago

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Let a,b,c,dNa,b,c,d \in \mathbb{N} such that abcda \ge b \ge c \ge d and d0d \neq 0. Show that the equation x4ax3bx2cxd=0x^4 - ax^3 - bx^2 - cx -d = 0 has no integer solution in xx.

Please post a "complete solution" like you would do in RMO.

Nihar Mahajan - 3 years, 10 months ago

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I'll have a crack at it. As d0d \neq 0, x0x \neq 0.

Denote y=1xy = \frac{1}{x}.

We have

x4=ax3+bx2+cx+dx^4 = ax^3 + bx^2 + cx + d \quad \quad

1=ay+by2+cy3+dy4(1)1 = ay + by^2 + cy^3 + dy^4 \quad \quad (1)

Now, we separate this into 2 cases:

Case 1:

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

an+bn2+cn3+dn4=1\dfrac {-a}{n} + \dfrac {b}{n^2} + \dfrac {-c}{n^3} + \dfrac {d}{n^4} = 1

an+bn2+cn+dn4=1\dfrac {-an+b}{n^2}+\dfrac{-cn+d}{n^4}=1

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

Case 2:

x=nx = n where nNn \in \mathbb{N}, then y=1ny=\dfrac{1}{n}. Substituting this into (1)(1), we get

an+bn2+cn3+dn4=1\dfrac {a}{n} + \dfrac {b}{n^2} + \dfrac {c}{n^3} + \dfrac {d}{n^4} = 1

This implies that n1n \neq 1, and

an<1\dfrac {a}{n} < 1

1an+an2+an3+an4<i=1ani=an11 \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

na>1\dfrac {n}{a} > 1

n1a<1\dfrac {n-1}{a} < 1

There are no such nn that satisfy the above two inequalities. Thus, no solutions exist.

Sharky Kesa - 3 years, 10 months ago

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Isn't this from INMO 2013 ??

Shrihari B - 3 years, 9 months ago

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Yes it is.

Nihar Mahajan - 3 years, 9 months ago

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

Shrihari B - 3 years, 9 months ago

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

Prove that ab+bc+ca34ab+bc+ca \leq \frac{3}{4}

Harsh Shrivastava - 3 years, 10 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(a2+b2+c2)>(a+b+c)23(a^2 + b^2 + c^2) > (a + b + c)^2

so

a2+b2+c2>3/4a^2 + b^2 + c^2 > 3/4

using cauchy again,

(a2+b2+c2)2>(ab+bc+ca)2(a^2 + b^2 + c^2)^2 > (ab + bc + ca)^2

3/4 > ab + bc + ca

Dev Sharma - 3 years, 10 months ago

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Nice solution. You could also have used the fact that (a+b+c)23(ab+bc+ca)(a+b+c)^{2} \ge 3(ab+bc+ca)

A Former Brilliant Member - 3 years, 10 months ago

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

Saarthak Marathe - 3 years, 10 months ago

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You got the following two inequalities,a2+b2+c2>ab+bc+caanda2+b2+c2>33a^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,34>ab+bc+ca\dfrac{3}{4}>ab+bc+ca?

Adarsh Kumar - 3 years, 10 months ago

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@Adarsh Kumar using cauchy again, like this

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

sum is cyclic

Dev Sharma - 3 years, 10 months ago

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If a,b0a,b \ge 0. Prove that 2(a(a+b)3+ba2+b2)3(a2+b2)\sqrt{2}(\sqrt{a(a+b)^{3}}+b \sqrt{a^{2}+b^{2}}) \le 3(a^{2}+b^{2}).

A Former Brilliant Member - 3 years, 10 months ago

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

Harsh Shrivastava - 3 years, 10 months ago

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

Shrihari B - 3 years, 10 months ago

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Let p>3p>3 be a prime number. Suppose k=1p11k=ab\displaystyle \sum_{k=1}^{p-1}{\frac{1}{k}}=\frac{a}{b} where gcd(a,b)=1\gcd(a,b)=1. Prove that aa is divisible by p2p^{2}.

A Former Brilliant Member - 3 years, 10 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 ?

Shrihari B - 3 years, 10 months ago

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Please post the proof, thanks!

Harsh Shrivastava - 3 years, 10 months ago

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

Shrihari B - 3 years, 10 months ago

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

Harsh Shrivastava - 3 years, 10 months ago

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@Harsh Shrivastava 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

Shrihari B - 3 years, 10 months ago

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I have the solution only till aa is divisible by pp. To prove that aa is divisible by p2p^{2} is given below the solution as a challenge.

A Former Brilliant Member - 3 years, 10 months ago

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

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

Harsh Shrivastava - 3 years, 10 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 :)

Shrihari B - 3 years, 10 months ago

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Shrihari,we get that CF=2DFCF=2DF

Saarthak Marathe - 3 years, 10 months ago

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@Saarthak Marathe I don't think so check again.

Shrihari B - 3 years, 10 months ago

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

Saarthak Marathe - 3 years, 10 months ago

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

Shrihari B - 3 years, 10 months ago

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Yup I also got 90 degrees.

Harsh Shrivastava - 3 years, 10 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 BAC=90\angle BAC = 90^\circ , do you see why?

Nihar Mahajan - 3 years, 10 months ago

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@Nihar Mahajan FD=FC=FA . Great !

Shrihari B - 3 years, 10 months ago

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@Shrihari B 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

Shrihari B - 3 years, 10 months ago

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@Shrihari B I also need some NT probs...

Harsh Shrivastava - 3 years, 10 months ago

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@Harsh Shrivastava Even I need NT problems...

Nihar Mahajan - 3 years, 10 months ago

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

Shrihari B - 3 years, 10 months ago

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@Shrihari B Pune region.

Nihar Mahajan - 3 years, 10 months ago

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There are given 25002^{500} points on a circle labeled 1,2,...,25001, 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.

Department 8 - 3 years, 10 months ago

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Given seven arbitrary distinct real numbers, show that there exist two numbers xx and yy such thst 0<xy1+xy<130<\dfrac{x-y}{1+xy}<\dfrac{1}{\sqrt{3}}.

A Former Brilliant Member - 3 years, 10 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).

Shrihari B - 3 years, 10 months ago

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If a circle goes through point AA of a parallelogram ABCDABCD, cuts the two sides AB,ADAB,AD and the diagonal ACAC at points P,RP,R and QQ respectively. Prove that (AP×AB)+(AR×AD)=AQ×AC(AP×AB)+(AR×AD)=AQ×AC.

A Former Brilliant Member - 3 years, 10 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 ASAS be the diameter of given circle.

AB+AD=ACAB+AD=AC

Therefore taking dot product withASAS

AB.AS+AD.AS=AC.ASAB.AS+AD.AS=AC.AS

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

ABAP+ARAD=AQAC|AB|*|AP|+|AR|*|AD|=|AQ|*|AC|

Saarthak Marathe - 3 years, 10 months ago

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Let ABCABC be a triangle and PP be a point inside it.Let x,y,zx,y,z denote the lengths of PAPA,PBPB,PCPC.Let a,b,ca,b,c be the lengths of sides BCBC,CACA,ABAB.Find all points PP such that axy+byz+czx=abcaxy+byz+czx=abc.

Souryajit Roy - 3 years, 10 months ago

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Find the least positive integer n such that 2549 divides n25452541n^{2545} - 2541

Dev Sharma - 3 years, 10 months ago

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Please post the solution , Thanks!

Harsh Shrivastava - 3 years, 10 months ago

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

Harsh Shrivastava - 3 years, 10 months ago

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Harsh,could you please add a figure?

Adarsh Kumar - 3 years, 10 months ago

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Hint: Let QPCO={R}\overrightarrow{QP} \cap \overleftrightarrow{CO} = \{R\} and COΠ={S}\overleftrightarrow{CO} \cap \Pi = \{S\} and then it will suffice to prove that ΔCRPΔCAS\Delta CRP \sim \Delta CAS

Nihar Mahajan - 3 years, 10 months ago

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An easy INMO geometry problem:

Let ABCABC be a triangle with circumcircle Γ\Gamma. Let MM be a point in the interior of triangle ABC ABC which is also on the bisector of A\angle A. Let AM,BM,CMAM, BM, CM meet Γ \Gamma in A1,B1,C1 A_{1}, B_{1}, C_{1} respectively. Suppose PP is the point of intersection of A1C1 A_{1}C_{1} with ABAB and QQ is the point of intersection of A1B1A_{1}B_{1} with ACAC. Prove that PQPQ is parallel to BCBC .

Nihar Mahajan - 3 years, 10 months ago

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Let a,bR,a0a,b \in R, a \neq 0. Show that a2+b2+1a2+ba3a^{2}+b^{2}+\frac{1}{a^{2}}+\frac{b}{a} \ge \sqrt{3}.

A Former Brilliant Member - 3 years, 10 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.

Shrihari B - 3 years, 10 months ago

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If a,b,c,d,ea,b,c,d,e are real numbers, prove that the roots of x5+ax4+bx3+cx2+dx+e=0x^{5}+ax^{4}+bx^{3}+cx^{2}+dx+e=0 cannot all be real if 2a2<5b2a^{2}<5b.

A Former Brilliant Member - 3 years, 10 months ago

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Let roots be p,q,r,s,tp,q,r,s,t.

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

Therefore Sum of squares of roots = a22ba^{2} - 2b

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

a22ba25a^{2} -2b \geq \frac{a^{2}}{5}

    2a25b\implies 2a^{2} \geq 5b

Therefore the polynomial cannot have all roots real if 5b>2a2 5b > 2a^{2}.

Harsh Shrivastava - 3 years, 10 months ago

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Show that n5+n4+1n^{5}+n^{4}+1 is not prime for n>1n>1.

A Former Brilliant Member - 3 years, 10 months ago

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n2+n+1n^{2} + n + 1 is factor of the given expression.Hence the result is obvious.

Harsh Shrivastava - 3 years, 10 months ago

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Prove that

1<11001+11002+.........+13001<431<\frac { 1 }{ 1001 } +\frac { 1 }{ 1002 } +.........+\frac { 1 }{ 3001 } <\frac { 4 }{ 3 }

Satyajit Ghosh - 3 years, 10 months ago

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

Saarthak Marathe - 3 years, 10 months ago

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

Satyajit Ghosh - 3 years, 10 months ago

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@Satyajit Ghosh 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.

Saarthak Marathe - 3 years, 10 months ago

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

Saarthak Marathe - 3 years, 10 months ago

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@Saarthak Marathe 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 Alternate text. 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 This is an image file :D This is an image file :D. Here's how it appears:

this is a copy paste from this note on moderators

Satyajit Ghosh - 3 years, 10 months ago

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If mm is a positive real number and satisfies the equation 2m=1+1m+m2142m=1+\frac{1}{m}+\sqrt{m^{2}-\frac{1}{4}} Find mm.

Shivam Jadhav - 3 years, 10 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

Shrihari B - 3 years, 10 months ago

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Find all pairs (m,n)(m,n) such that 2m+3n2^m+3^n is a perfect square. (A popular NT problem)

Nihar Mahajan - 3 years, 10 months ago

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Its easy.Only one solution (2,2)

Saarthak Marathe - 3 years, 10 months ago

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Wrong. Check again.

Nihar Mahajan - 3 years, 10 months ago

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@Nihar Mahajan I think he means (4,2).

A Former Brilliant Member - 3 years, 10 months ago

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@A Former Brilliant Member Ya I meant (4,2). Sorry,typing error. :P

Saarthak Marathe - 3 years, 10 months ago

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@Saarthak Marathe Its okay , please post your method. :)

Nihar Mahajan - 3 years, 10 months ago

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@Nihar Mahajan Taking mod 4 we get that m,n should be even. Then using gcd(2m,3n)=1 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 m=4,n=2 are the solutions.

Saarthak Marathe - 3 years, 10 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.

Rakhi Bhattacharyya - 3 years, 10 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

Shrihari B - 3 years, 10 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.

Shrihari B - 3 years, 10 months ago

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Its easy , note that BP=QC=sbBP=QC=s-b where ss is the semiperimeter of the triangle and bb is length of side opposite to B \angle B respectively. Since MM is mid point of BCBC , the result follows.

Nihar Mahajan - 3 years, 10 months ago

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

A Former Brilliant Member - 3 years, 10 months ago

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@A Former Brilliant Member In the instructions given in question paper , it is said that protractors are not allowed.

Nihar Mahajan - 3 years, 10 months ago

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Oh nice !

Shrihari B - 3 years, 10 months ago

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Try this set guys.

Akshat Sharda - 3 years, 10 months ago

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1991k(199019911992+199219911990)1991^{k}|(1990^{1991^{1992}}+1992^{1991^{1990}})

max(k)=?max(k)=?


Details:

bab|a\rightarrow aa is divisible by bb.

Akshat Sharda - 3 years, 10 months ago

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1991

Saarthak Marathe - 3 years, 10 months ago

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

Shrihari B - 3 years, 10 months ago

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@Shrihari B 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

Saarthak Marathe - 3 years, 10 months ago

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@Saarthak Marathe 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

Shrihari B - 3 years, 10 months ago

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Yes..... Can you post solution ?

Akshat Sharda - 3 years, 10 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 :)

Shrihari B - 3 years, 10 months ago

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Done :)

A Former Brilliant Member - 3 years, 10 months ago

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Thanks

Shrihari B - 3 years, 10 months ago

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

Prove that x2y1+y2x18\dfrac{x^{2}}{y-1} + \dfrac{y^{2}}{ x-1} \geq 8

Harsh Shrivastava - 3 years, 10 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

Shrihari B - 3 years, 10 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.

Harsh Shrivastava - 3 years, 10 months ago

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Nice idea!

Nihar Mahajan - 3 years, 10 months ago

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

x2y1+y2x1y2y1+y2x14+y2x1    x2y14\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 x2y1\frac{x^{2}}{y-1} is 44 and it occurs if and only if x=y=2x=y=2.

Therefore, x2y1+y2x18\frac{x^{2}}{y-1}+\frac{y^{2}}{x-1} \ge 8.

A Former Brilliant Member - 3 years, 10 months ago

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Hmm you have not proven that min value of x2y1\frac{x^{2}}{y-1} is 4.

Harsh Shrivastava - 3 years, 10 months ago

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@Harsh Shrivastava Do you mean that I haven't proven that minimum value of y2y1\frac{y^{2}}{y-1} is 44?

A Former Brilliant Member - 3 years, 10 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.

Satyajit Ghosh - 3 years, 10 months ago

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a,b,ca,b,c are real numbers such that a+b+c=0a+b+c=0 and a2+b2+c2=1a^2+b^2+c^2=1. Prove that a2b2c2154a^{2}b^{2}c^{2}≤\frac{1}{54}.

Akshat Sharda - 3 years, 10 months ago

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See my note

Satyajit Ghosh - 3 years, 10 months ago

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A barrel contains 2n2n balls, numbered 11 to 2n2n. 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?

A Former Brilliant Member - 3 years, 10 months ago

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1/2

Saarthak Marathe - 3 years, 10 months ago

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Correct. Can you show your steps?

A Former Brilliant Member - 3 years, 10 months ago

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Write the sum Sn=k=0n(1)k(nk)k3+9k2+26k+24S_n= \displaystyle \sum_{k=0}^n{\frac{(-1)^{k}{n \choose{k}}}{k^{3}+ 9k^{2}+26k+24}} in the form p(n)q(n)\frac{p(n)}{q(n)} where p(n),q(n)p(n),q(n) are polynomials with integer coefficients.

A Former Brilliant Member - 3 years, 10 months ago

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We first note that k3+9k2+26k+24=(k+2)(k+3)(k+4)k^{3}+9k^{2}+26k+24=(k+2)(k+3)(k+4).Hence, Sn=k=0n(1)k(nk)(k+2)(k+3)(k+4)S_n=\displaystyle \sum_{k=0}^n{\frac{(-1)^{k}{n \choose{k}}}{(k+2)(k+3)(k+4)}}