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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, 10 months 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 $$6$$

- 1 year, 10 months ago

Did the same

- 1 year, 10 months ago

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)

- 1 year, 10 months ago

Perfect solution!

- 1 year, 10 months ago

$$\boxed{CP=CD,EP=PF}.....(1)$$ (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.

- 1 year, 10 months ago

What is the cutoff for qualifying to the next round?

- 1 year, 10 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 ...

- 1 year, 10 months ago

is trichy a centre for RMO

- 1 year, 9 months ago

Bro ... I dont know abt trichy ...

- 1 year, 9 months ago

@ 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.

- 1 month ago

I have got the proof for the 4 th one

- 1 year, 8 months ago

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

- 1 year, 9 months ago

Congratulations!

- 1 year, 9 months ago

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

- 1 year, 9 months ago

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!

- 1 year, 10 months ago

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 $$\boxed{80}$$

- 1 year, 10 months ago

Both your and Svatejas's solutions are good.

What about question 5? How you did?

- 1 year, 10 months ago

I did it like Shrihari B.

- 1 year, 10 months ago

Ok. Which book do you have regarding Olympiad geometry?

- 1 year, 9 months ago

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

- 1 year, 9 months ago

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

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

- 1 year, 9 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.

- 1 year, 9 months ago

Mine too, I think the cut off would be 51

- 1 year, 9 months ago

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

- 1 year, 9 months ago

Me I am getting only 17-22.

- 1 year, 9 months ago

People .. Any idea abt cutoff?? ..

- 1 year, 10 months ago

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

- 1 year, 10 months ago

Yeah ... I too got the same

- 1 year, 10 months ago

Yeah .I too got the same.

- 1 year, 10 months ago

hey mohan i got 5005 dah

- 1 year, 10 months ago

3) 80

- 1 year, 10 months ago

What method did u use ? mod 100 ?

- 1 year, 10 months ago

That is probably the easiest method.

- 1 year, 10 months ago

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

- 1 year, 10 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 $$80$$.

- 1 year, 10 months ago

Yeah .. I too got the Same

- 1 year, 10 months ago

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

- 1 year, 10 months ago

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

- 1 year, 9 months ago

Use Principle of Inclusion and Exclusion.

- 1 year, 9 months ago

Got through with part a, finding part b difficult

- 1 year, 9 months ago

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

- 1 year, 10 months ago

In fact i found this the toughest paper !

- 1 year, 9 months ago

@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?

- 1 year, 9 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 participating

- 1 year, 9 months ago

No problem. I was just asking.

- 1 year, 9 months ago

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

- 1 year, 9 months ago

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

- 1 year, 9 months ago

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

- 1 year, 9 months ago

I am actually in 8th

- 1 year, 9 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 take

- 1 year, 9 months ago

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

- 1 year, 9 months ago

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??

- 1 year, 10 months ago

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 $$\angle BDP = \angle CDQ$$.

- 5 months, 2 weeks ago

What do you expect the cutoff to be?

- 1 year, 10 months ago

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

- 1 year, 10 months 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 rectangle

- 5 months, 2 weeks ago