At last, nervousness comes down. Exam got over. Thanks to Brilliant, it suited my learning style. I like to learn new stuff through discussion and solving new problems,

Now, here is the paper. I think I have done a decent job

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## Comments

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TopNewestLet \(\boxed{(x+\frac{1}{x})=m}.......(1)\) . Therefore \[(x^{3}+\frac{1}{x^{3}})+3(x+\frac{1}{x})=m^{3}\] \[\boxed{(x^{3}+\frac{1}{x^{3}})=m^{3}-3m}.......(2)\] Squaring both sides , we get \[\boxed{(x^{6}+\frac{1}{x^{6}})=m^{6}+9m^{2}-6m^{4}-2}........(3)\] Now rewriting the given expression in terms of \(m\) we get \[=\frac{m^{6}-(m^{6}+9m^{2}-6m^{4}-2)-2}{m^{3}+m^{3}-3m}\] \[=\frac{6m^{4}-9m^{2}}{2m^{3}-3m}\] \[=3m\] By \(A.M-G.M\) \[x+\frac{1}{x}\geq2\] Therefore \(3m\geq6\) Hence the minimum value of the given expression is \(6\)

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Did the same

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Question 5 is trivial. Add 1 to the LHS and then it becomes a perfect cube. Then try proving that the RHS is a cube lying between x^3 and (x+2)^3 forcing it to be (x+1)^3. thus u obtain that x=y. Substitute that and get solutions as (1,1),(10,10)

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Perfect solution!

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\(\boxed{CP=CD,EP=PF}.....(1)\) (Since \(C_{1}P,C_{2}P\) are perpendicular to \(CD,EF\) respectively) Angle\(CPF\) =Angle\(EPF\) Angle\(CPE\) =Angle\(FPD.....(2)\) Using \((1),(2)\) Triangle CPF is congruent to triangle EPD (BY SAS Test)

Triangle CPE is congruent to triangle FPD (BY SAS Test)

Therefore CF=ED,CE=FD...(I). And Angle\(FCD\) =Angle\(EDC\) Which implies CF||ED....(3) Similarly we prove CE||FD...(4) Hence using (I),(3),(4) we prove that CEDF is a rectangle.

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What is the cutoff for qualifying to the next round?

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I too want to know the same .... i am nervous about that .. If anyone gets to know it ... PLEASE post it in this discussion ... let the Tamil Nadu people in Brilliant know it ...

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is trichy a centre for RMO

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LOL you were damn lucky to get selected.

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@ Ganesh Ayyappan Hi I am Anurag from Mumbai and currently, I am in 8th standard and selected for RMO.I am able to solve only three or four problems from each RMO paper or even 2 if it is difficult. For sums I am not able to solve, I am able to get through halfway and then I can't solve because I don't the formula or else, I am missing something. Is 3 or 4 problems from 6 alright to get through RMO and go to INMO? Please let me know. Do follow me. I will be readily answering any sum on Brilliant.

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I have got the proof for the 4 th one

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Friends ... The RMO results are out ... Fortunately I hav cleared it and now can write INMO ... but i feel nervous coz i am hardly able to do 1 sum frm previous year papers ... so i wish to get some tips regareding INMO

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Congratulations!

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thanks ... wat about you brother? .. i tink ur region results came before ours ...

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I got #1 and #3 right. For #4 I showed that CDEF is a parallelogram since P is the midpoint of CD and EF. Finally for #6a) I got 722 by some arithmetic an error. How much marks would I be approximately getting? Any guesses for the cutoff? Thanks!

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Q3) Clearly \( N \equiv 0 \equiv 80 \quad (\mod 4) \).

Now we construct a table for \(2^{5^{n}} \quad (\mod 25) \).

\[ \Large{\underline { \begin{matrix} n & & n\quad \left( \mod 25 \right) \end{matrix} } \\ \begin{matrix} 1 & & 7 \\ 2 & & 7 \\ 3 & & 7 \\ & . & \\ & . & \\ & . & \\ & & \end{matrix}}\]

This gives \(N \equiv 5 \equiv 80 \quad (\mod 25) \). Hence the answer is \( \boxed{80}\)

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Both your and Svatejas's solutions are good.

What about question 5? How you did?

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I did it like Shrihari B.

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Also, Do you appeared in RMO? From Delhi region?

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People .. Any idea abt cutoff?? ..

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Is 6(a) 721 and 6(b) 5004?

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Yeah ... I too got the same

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Yeah .I too got the same.

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hey mohan i got 5005 dah

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3) 80

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What method did u use ? mod 100 ?

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That is probably the easiest method.

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Yeah .. I too got the Same

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i hope pic is clear ... else jus reply to me ... i will type out the full question

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Someone please post solution for q6, it was the easiest but sill want to know how others did it

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Use Principle of Inclusion and Exclusion.

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Got through with part a, finding part b difficult

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I think this region's paper was the easiest one.

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In fact i found this the toughest paper !

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@Shrihari B I feel that Mumbai region was the hardest. Definitely my region was the easiest but still I couldn't quality that :( Why aren't you participating in the INMO board anyone?

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besides being a 14 yr old ... u are a level 5 solver in all topics u hav took ... tats really appreciable ... so dont worry abt this yr RMO .... GOOD LUCK

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The answers i got

1) 6

3) 80

5) x=y=1 & x=y=10

6) A - 721 & B - 5004

i verified once that all are right .. and i tink some of my friends havent noticed that this discussion is put up ....

@Harsh Shrivastava @Adarsh Kumar @Mehul Chaturvedi @Kartik Sharma and some more of my BRILLIANT friends ... awaiting ur valuable replies

Also ... can sum1 post soln of Q2 & 4??

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\(Sol^{n}\):

\(1st\) case: When \(\angle BDP = \angle CDQ\) then \(MP = MQ.\)

Constructions: Produce \(QM\) to \(Q'\) such that \(MQ = MQ'.\) Join \( BQ' , PQ\) and \(PQ'.\)

Here clearly, quadrilateral \(BQCQ'\) is a \(|| gm\). \(=>\) \(\angle BQ' = CQ\) --- \([1]\) and also, \(\angle Q'BM = \angle C\) \(=>\) \(\angle Q'BP = \angle B +\angle C = 180^{\circ}- \angle A = \angle PDQ\) --- \([2]\)

In \(\triangle BDP\) and \(\Delta CDQ,\) \(\angle BDP = \angle CDQ\) and \(\angle BPD = \angle CQD\) \(=90^{\circ}.\) \(=>\) \(\Delta BDP \sim \Delta CDQ\) \(=>\) \(\dfrac{BP}{CQ}= \dfrac{DP}{DQ}\) \(=>\) \(\dfrac{BP}{BQ'} = \dfrac{DP}{DQ}\) (Using - \([1]\) ) \(=>\) \(\dfrac{BP}{DP} = \dfrac{BQ'}{DQ}\)

Now, in \(\Delta BQ'P\) & \(\Delta PDQ\) , \(\dfrac{BP}{DP} = \dfrac{BQ'}{DQ}\) and \(\angle Q'BP = \angle PDQ\) [From \(eq^{n}\)] - \(2\) \(=>\) \(\Delta BQ'P \sim \Delta PDQ\) \(=>\) \(\angle BPQ' = \angle DPQ\) \(=>\) \(\angle BPQ' + \angle Q'PQ = \angle Q'PQ + \angle DPQ\) \(=>\) \(90^{\circ} = \angle Q'PQ + \angle DPQ = \angle QPQ'\) \(=>\) \(\Delta QPQ'\) is right angled at \(P\).

And we know that the midpoint of the hypotenuse of a right angled triangle is circumcenter of that \(\Delta\). \(=>\) \(MQ = MP\) \( = MQ'\)

Now, just reverse this process to prove that: If \(MP = MQ\) then \(\angle BDP = \angle CDQ\).

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What do you expect the cutoff to be?

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Since i do not hav any idea .. Im asking here ...

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Problem 4 is easy. One can easily notice that \(EP = FP\) and \(DP = CP\) [Because, the \(\perp\) from the center to the chord bisects the chord]. => \(FDEC\) is a \(|| gm.\) Now, since \(DACB\) is cyclic, \(DP\times CP = AP\times BP = DP^{2}\) -- \(1\). Similarly, since \(FAEB\) is cyclic, \(EP\times FP = AP\times BP = FP^{2}\)-- \(2\) The two results \(1 and 2\) tells that, \(DP = FP => 2DP = 2FP => DC = FE\). So, quad \(DFCE\) is a \(||gm\) with the diagonals equal. => \(CEDF\) is a rectangle

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