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RMO 2015 - Tamilnadu & Pondicherry region

This note has been used to help create the RMO Math Contest Preparation wiki

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

Note by Ganesh Ayyappan
1 year, 11 months ago

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  Easy Math Editor

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Let \(\boxed{(x+\frac{1}{x})=m}.......(1)\) . 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 \(6\)

Shivam Jadhav - 1 year, 11 months ago

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Did the same

Ganesh Ayyappan - 1 year, 10 months ago

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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)

Shrihari B - 1 year, 11 months ago

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Perfect solution!

Brilliant Member - 1 year, 11 months ago

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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.

Shivam Jadhav - 1 year, 11 months ago

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What is the cutoff for qualifying to the next round?

Mohamed Shuaib Hasan - 1 year, 11 months ago

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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 ...

Ganesh Ayyappan - 1 year, 11 months ago

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is trichy a centre for RMO

Sriram Venkatesan - 1 year, 10 months ago

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@Sriram Venkatesan Bro ... I dont know abt trichy ...

Ganesh Ayyappan - 1 year, 10 months ago

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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.

Anurag Ramachandran - 2 months ago

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I have got the proof for the 4 th one

Snehan Jayakumar - 1 year, 9 months ago

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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

Ganesh Ayyappan - 1 year, 10 months ago

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Congratulations!

Brilliant Member - 1 year, 10 months ago

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

Ganesh Ayyappan - 1 year, 10 months ago

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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!

Jack Frost - 1 year, 10 months ago

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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 \( \boxed{80}\)

Lakshya Sinha - 1 year, 10 months ago

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Both your and Svatejas's solutions are good.

What about question 5? How you did?

Priyanshu Mishra - 1 year, 10 months ago

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I did it like Shrihari B.

Lakshya Sinha - 1 year, 10 months ago

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@Lakshya Sinha Ok. Which book do you have regarding Olympiad geometry?

Priyanshu Mishra - 1 year, 10 months ago

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@Priyanshu Mishra In mathematics for me, Brilliant is enough and some other sites, which i believe you follow (Ao...)

Lakshya Sinha - 1 year, 10 months ago

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@Lakshya Sinha Ya, AOPS is full of resources regarding any topic, contests.

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

Priyanshu Mishra - 1 year, 10 months ago

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@Lakshya Sinha What do you guys expect the cutoff to be? I'm really nervous about the result as this is my first time writing RMO.

Jack Frost - 1 year, 10 months ago

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@Jack Frost Mine too, I think the cut off would be 51

Lakshya Sinha - 1 year, 10 months ago

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@Lakshya Sinha Ha that makes me kind of more nervous since I'm expcecting about 45-55. :)

Jack Frost - 1 year, 10 months ago

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@Jack Frost Me I am getting only 17-22.

Lakshya Sinha - 1 year, 10 months ago

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People .. Any idea abt cutoff?? ..

Ganesh Ayyappan - 1 year, 10 months ago

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

Brilliant Member - 1 year, 11 months ago

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Yeah ... I too got the same

Ganesh Ayyappan - 1 year, 11 months ago

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Yeah .I too got the same.

Bala Vidyadharan - 1 year, 10 months ago

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hey mohan i got 5005 dah

Madhav Srirangan - 1 year, 10 months ago

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3) 80

Akshat Sharda - 1 year, 11 months ago

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What method did u use ? mod 100 ?

Shrihari B - 1 year, 11 months ago

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That is probably the easiest method.

Brilliant Member - 1 year, 11 months ago

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@Brilliant Member But can u explain how did u solve that question by mod 100 ? I could not solve it that way

Shrihari B - 1 year, 11 months ago

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@Shrihari B \(32^{5}=2^{25} \equiv 2^{10} \times 2^{10} \times 2^{5} \equiv 24^{2} \times 32 \equiv 32 \pmod{100}\). Hence last two digits of every fifth power of \(32\) is \(32\). Hence last two digits of \(N\) is last two digits of \(2015 \times 32=64480\) which is \(80\).

Brilliant Member - 1 year, 11 months ago

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Yeah .. I too got the Same

Ganesh Ayyappan - 1 year, 11 months ago

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

Ganesh Ayyappan - 1 year, 11 months ago

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

Silver Vice - 1 year, 10 months ago

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Use Principle of Inclusion and Exclusion.

Brilliant Member - 1 year, 10 months ago

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Got through with part a, finding part b difficult

Silver Vice - 1 year, 10 months ago

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

Priyanshu Mishra - 1 year, 10 months ago

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In fact i found this the toughest paper !

Shrihari B - 1 year, 10 months ago

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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?

Brilliant Member - 1 year, 10 months ago

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@Brilliant Member Hi Svatejas sorry for not being able to participate in the INMO practise board.Actually I had my INMO training camp for four days and so hardly got any time to log in to brilliant. But it gets over tomorrow so i will be participating

Shrihari B - 1 year, 10 months ago

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@Shrihari B No problem. I was just asking.

Brilliant Member - 1 year, 10 months ago

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@Brilliant Member dont worry bro ... u r a 9th grader i guess .. u still hav 2 yrs ... for me .. this is the first and last time i cud hav written RMO (I was eligible to write RMO during 9th and 10th but wasnt aware about what was it) ... so only i am nervous about INMO ...

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

Ganesh Ayyappan - 1 year, 10 months ago

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@Ganesh Ayyappan I actually became level 5 by mostly doing level 1-3 problems, especially the unrated problems worth 100 points.

Brilliant Member - 1 year, 10 months ago

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@Brilliant Member its okay ... do level 4 frm now onwards ... do level 5 ocassionally ... make sure u do gud progress in algebra geometry and inequalities

Ganesh Ayyappan - 1 year, 10 months ago

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@Ganesh Ayyappan I am actually in 8th

Brilliant Member - 1 year, 10 months ago

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@Brilliant Member oh ... well and good .... u hav 3 more yrs .. GOOD LUCK ... dont at all regret or tink abt wat u did this RMO ... i am telling this coz i am able to see ur potential through the novel methods u suggest and a LEVEL 5 guy in Algebra Geometry and Combinatorics .... in 8th ... i wasnt this well-equipped bro ... u hav a long way to go ... Wish u a BRIGHT COLORFUL FUTURE in the field u want to take

Ganesh Ayyappan - 1 year, 10 months ago

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@Ganesh Ayyappan Thank you very much for your encouragement. Wish all the very best to everyone appearing for INMO.

Brilliant Member - 1 year, 10 months ago

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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??

Ganesh Ayyappan - 1 year, 11 months ago

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Diagram : Solution to problem 2. @Ganesh Ayyappan Here is a solution for the problem (2).

\(Sol^{n}\):

\(1st\) 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 \(\angle BDP = \angle CDQ\).

Rohit Camfar - 6 months, 1 week ago

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What do you expect the cutoff to be?

Jack Frost - 1 year, 10 months ago

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

Ganesh Ayyappan - 1 year, 10 months ago

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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

Rohit Camfar - 6 months, 1 week ago

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