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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
9 months, 2 weeks ago

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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 · 9 months, 1 week ago

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@Shivam Jadhav Did the same Ganesh Ayyappan · 9 months, 1 week 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 · 9 months, 1 week ago

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@Shrihari B Perfect solution! Svatejas Shivakumar · 9 months, 1 week 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 · 9 months, 1 week ago

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What is the cutoff for qualifying to the next round? Mohamed Shuaib Hasan · 9 months, 1 week ago

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@Mohamed Shuaib Hasan 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 · 9 months, 1 week ago

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@Ganesh Ayyappan is trichy a centre for RMO Sriram Venkatesan · 9 months ago

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@Sriram Venkatesan Bro ... I dont know abt trichy ... Ganesh Ayyappan · 8 months, 4 weeks ago

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I have got the proof for the 4 th one Snehan Jayakumar · 7 months, 2 weeks 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 · 8 months, 3 weeks ago

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@Ganesh Ayyappan Congratulations! Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar thanks ... wat about you brother? .. i tink ur region results came before ours ... Ganesh Ayyappan · 8 months, 3 weeks 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 · 9 months, 1 week 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 · 9 months, 1 week ago

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@Lakshya Sinha Both your and Svatejas's solutions are good.

What about question 5? How you did? Priyanshu Mishra · 9 months, 1 week ago

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@Priyanshu Mishra I did it like Shrihari B. Lakshya Sinha · 9 months, 1 week ago

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@Lakshya Sinha Ok. Which book do you have regarding Olympiad geometry? Priyanshu Mishra · 9 months, 1 week 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 · 9 months, 1 week 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 · 9 months, 1 week 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 · 9 months, 1 week ago

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@Jack Frost Mine too, I think the cut off would be 51 Lakshya Sinha · 9 months, 1 week ago

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@Lakshya Sinha Ha that makes me kind of more nervous since I'm expcecting about 45-55. :) Jack Frost · 9 months, 1 week ago

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@Jack Frost Me I am getting only 17-22. Lakshya Sinha · 9 months, 1 week ago

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People .. Any idea abt cutoff?? .. Ganesh Ayyappan · 9 months, 1 week ago

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Is 6(a) 721 and 6(b) 5004? Svatejas Shivakumar · 9 months, 1 week ago

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@Svatejas Shivakumar Yeah ... I too got the same Ganesh Ayyappan · 9 months, 1 week ago

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@Svatejas Shivakumar Yeah .I too got the same. Bala Vidyadharan · 9 months, 1 week ago

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@Bala Vidyadharan hey mohan i got 5005 dah Madhav Srirangan · 9 months, 1 week ago

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3) 80 Akshat Sharda · 9 months, 2 weeks ago

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@Akshat Sharda What method did u use ? mod 100 ? Shrihari B · 9 months, 1 week ago

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@Shrihari B That is probably the easiest method. Svatejas Shivakumar · 9 months, 1 week ago

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@Svatejas Shivakumar But can u explain how did u solve that question by mod 100 ? I could not solve it that way Shrihari B · 9 months, 1 week 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 \(80\). Svatejas Shivakumar · 9 months, 1 week ago

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@Akshat Sharda Yeah .. I too got the Same Ganesh Ayyappan · 9 months, 1 week ago

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i hope pic is clear ... else jus reply to me ... i will type out the full question Ganesh Ayyappan · 9 months, 2 weeks 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 · 8 months, 3 weeks ago

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@Silver Vice Use Principle of Inclusion and Exclusion. Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar Got through with part a, finding part b difficult Silver Vice · 8 months, 3 weeks ago

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I think this region's paper was the easiest one. Priyanshu Mishra · 9 months, 1 week ago

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@Priyanshu Mishra In fact i found this the toughest paper ! Shrihari B · 8 months, 3 weeks ago

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@Shrihari B @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 · 8 months, 3 weeks ago

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@Svatejas Shivakumar 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 · 8 months, 3 weeks ago

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@Shrihari B No problem. I was just asking. Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar 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 · 8 months, 3 weeks 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. Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar 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 · 8 months, 3 weeks ago

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@Ganesh Ayyappan I am actually in 8th Svatejas Shivakumar · 8 months, 3 weeks ago

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@Svatejas Shivakumar 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 · 8 months, 3 weeks ago

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@Ganesh Ayyappan Thank you very much for your encouragement. Wish all the very best to everyone appearing for INMO. Svatejas Shivakumar · 8 months, 3 weeks 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 · 9 months, 1 week ago

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@Ganesh Ayyappan What do you expect the cutoff to be? Jack Frost · 9 months, 1 week ago

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@Jack Frost Since i do not hav any idea .. Im asking here ... Ganesh Ayyappan · 9 months, 1 week ago

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