# 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

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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 \(6\) – Shivam Jadhav · 1 year, 6 months ago

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– Ganesh Ayyappan · 1 year, 6 months ago

Did the sameLog in to reply

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, 6 months ago

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– Svatejas Shivakumar · 1 year, 6 months ago

Perfect solution!Log in to reply

\(\boxed{CP=CD,EP=PF}.....(1)\) (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, 6 months ago

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What is the cutoff for qualifying to the next round? – Mohamed Shuaib Hasan · 1 year, 6 months ago

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– Ganesh Ayyappan · 1 year, 6 months ago

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 ...Log in to reply

– Sriram Venkatesan · 1 year, 5 months ago

is trichy a centre for RMOLog in to reply

– Ganesh Ayyappan · 1 year, 5 months ago

Bro ... I dont know abt trichy ...Log in to reply

I have got the proof for the 4 th one – Snehan Jayakumar · 1 year, 4 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, 5 months ago

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– Svatejas Shivakumar · 1 year, 5 months ago

Congratulations!Log in to reply

– Ganesh Ayyappan · 1 year, 5 months ago

thanks ... wat about you brother? .. i tink ur region results came before ours ...Log in to reply

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, 6 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, 6 months ago

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What about question 5? How you did? – Priyanshu Mishra · 1 year, 6 months ago

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– Lakshya Sinha · 1 year, 6 months ago

I did it like Shrihari B.Log in to reply

– Priyanshu Mishra · 1 year, 6 months ago

Ok. Which book do you have regarding Olympiad geometry?Log in to reply

– Lakshya Sinha · 1 year, 6 months ago

In mathematics for me, Brilliant is enough and some other sites, which i believe you follow (Ao...)Log in to reply

Also, Do you appeared in RMO? From Delhi region? – Priyanshu Mishra · 1 year, 6 months ago

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– Jack Frost · 1 year, 6 months ago

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

– Lakshya Sinha · 1 year, 6 months ago

Mine too, I think the cut off would be 51Log in to reply

– Jack Frost · 1 year, 6 months ago

Ha that makes me kind of more nervous since I'm expcecting about 45-55. :)Log in to reply

– Lakshya Sinha · 1 year, 6 months ago

Me I am getting only 17-22.Log in to reply

People .. Any idea abt cutoff?? .. – Ganesh Ayyappan · 1 year, 6 months ago

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Is 6(a) 721 and 6(b) 5004? – Svatejas Shivakumar · 1 year, 6 months ago

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– Ganesh Ayyappan · 1 year, 6 months ago

Yeah ... I too got the sameLog in to reply

– Bala Vidyadharan · 1 year, 6 months ago

Yeah .I too got the same.Log in to reply

– Madhav Srirangan · 1 year, 6 months ago

hey mohan i got 5005 dahLog in to reply

3) 80 – Akshat Sharda · 1 year, 6 months ago

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– Shrihari B · 1 year, 6 months ago

What method did u use ? mod 100 ?Log in to reply

– Svatejas Shivakumar · 1 year, 6 months ago

That is probably the easiest method.Log in to reply

– Shrihari B · 1 year, 6 months ago

But can u explain how did u solve that question by mod 100 ? I could not solve it that wayLog in to reply

– Svatejas Shivakumar · 1 year, 6 months ago

\(32^{5}=2^{25} \equiv 2^{10} \times 2^{10} \times 2^{5} \equiv 24^{2} \times 32 \equiv 32 \pmod{100}\). 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 \(80\).Log in to reply

– Ganesh Ayyappan · 1 year, 6 months ago

Yeah .. I too got the SameLog in to reply

i hope pic is clear ... else jus reply to me ... i will type out the full question – Ganesh Ayyappan · 1 year, 6 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, 5 months ago

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– Svatejas Shivakumar · 1 year, 5 months ago

Use Principle of Inclusion and Exclusion.Log in to reply

– Silver Vice · 1 year, 5 months ago

Got through with part a, finding part b difficultLog in to reply

I think this region's paper was the easiest one. – Priyanshu Mishra · 1 year, 6 months ago

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– Shrihari B · 1 year, 5 months ago

In fact i found this the toughest paper !Log in to reply

@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? – Svatejas Shivakumar · 1 year, 5 months ago

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– Shrihari B · 1 year, 5 months ago

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 participatingLog in to reply

– Svatejas Shivakumar · 1 year, 5 months ago

No problem. I was just asking.Log in to reply

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, 5 months ago

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– Svatejas Shivakumar · 1 year, 5 months ago

I actually became level 5 by mostly doing level 1-3 problems, especially the unrated problems worth 100 points.Log in to reply

– Ganesh Ayyappan · 1 year, 5 months ago

its okay ... do level 4 frm now onwards ... do level 5 ocassionally ... make sure u do gud progress in algebra geometry and inequalitiesLog in to reply

– Svatejas Shivakumar · 1 year, 5 months ago

I am actually in 8thLog in to reply

– Ganesh Ayyappan · 1 year, 5 months ago

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 takeLog in to reply

– Svatejas Shivakumar · 1 year, 5 months ago

Thank you very much for your encouragement. Wish all the very best to everyone appearing for INMO.Log in to reply

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, 6 months ago

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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 \(\angle BDP = \angle CDQ\). – Rohit Camfar · 1 month, 3 weeks ago

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– Jack Frost · 1 year, 6 months ago

What do you expect the cutoff to be?Log in to reply

– Ganesh Ayyappan · 1 year, 6 months ago

Since i do not hav any idea .. Im asking here ...Log in to reply

– Rohit Camfar · 1 month, 3 weeks ago

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 rectangleLog in to reply