RMO 2015 - Tamilnadu & Pondicherry region

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

#Algebra #Geometry #NumberTheory #Polynomials #RMO #Math

Note by Ganesh Ayyappan
5 years, 5 months ago

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Comments

Let (x+1x)=m.......(1)\boxed{(x+\frac{1}{x})=m}.......(1) . Therefore (x3+1x3)+3(x+1x)=m3(x^{3}+\frac{1}{x^{3}})+3(x+\frac{1}{x})=m^{3} (x3+1x3)=m3−3m.......(2)\boxed{(x^{3}+\frac{1}{x^{3}})=m^{3}-3m}.......(2) Squaring both sides , we get (x6+1x6)=m6+9m2−6m4−2........(3)\boxed{(x^{6}+\frac{1}{x^{6}})=m^{6}+9m^{2}-6m^{4}-2}........(3) Now rewriting the given expression in terms of mm we get =m6−(m6+9m2−6m4−2)−2m3+m3−3m=\frac{m^{6}-(m^{6}+9m^{2}-6m^{4}-2)-2}{m^{3}+m^{3}-3m} =6m4−9m22m3−3m=\frac{6m^{4}-9m^{2}}{2m^{3}-3m} =3m=3m By A.M−G.MA.M-G.M x+1x≥2x+\frac{1}{x}\geq2 Therefore 3m≥63m\geq6 Hence the minimum value of the given expression is 66

Shivam Jadhav - 5 years, 5 months ago

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Did the same

Ganesh Ayyappan - 5 years, 5 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)

Shrihari B - 5 years, 5 months ago

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Perfect solution!

A Former Brilliant Member - 5 years, 5 months ago

CP=CD,EP=PF.....(1)\boxed{CP=CD,EP=PF}.....(1) (Since C1P,C2PC_{1}P,C_{2}P are perpendicular to CD,EFCD,EF respectively) AngleCPFCPF =AngleEPFEPF AngleCPECPE =AngleFPD.....(2)FPD.....(2) Using (1),(2)(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 AngleFCDFCD =AngleEDCEDC 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 - 5 years, 5 months ago

What is the cutoff for qualifying to the next round?

Mohamed Shuaib Hasan - 5 years, 5 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 - 5 years, 5 months ago

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is trichy a centre for RMO

Sriram Venkatesan - 5 years, 5 months ago

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

Ganesh Ayyappan - 5 years, 5 months ago

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

Ganesh Ayyappan - 5 years, 5 months ago

3) 80

Akshat Sharda - 5 years, 5 months ago

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Yeah .. I too got the Same

Ganesh Ayyappan - 5 years, 5 months ago

What method did u use ? mod 100 ?

Shrihari B - 5 years, 5 months ago

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That is probably the easiest method.

A Former Brilliant Member - 5 years, 5 months ago

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

Shrihari B - 5 years, 5 months ago

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@Shrihari B – 325=225≡210×210×25≡242×32≡32(mod100)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 3232 is 3232. Hence last two digits of NN is last two digits of 2015×32=644802015 \times 32=64480 which is 8080.

A Former Brilliant Member - 5 years, 5 months ago

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

A Former Brilliant Member - 5 years, 5 months ago

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Yeah ... I too got the same

Ganesh Ayyappan - 5 years, 5 months ago

Yeah .I too got the same.

Bala vidyadharan - 5 years, 5 months ago

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hey mohan i got 5005 dah

madhav srirangan - 5 years, 5 months ago

People .. Any idea abt cutoff?? ..

Ganesh Ayyappan - 5 years, 5 months ago

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

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

nn(mod  25)‾172737... \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≡5≡80(mod  25)N \equiv 5 \equiv 80 \quad (\mod 25) . Hence the answer is 80 \boxed{80}

Department 8 - 5 years, 5 months ago

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Both your and Svatejas's solutions are good.

What about question 5? How you did?

Priyanshu Mishra - 5 years, 5 months ago

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I did it like Shrihari B.

Department 8 - 5 years, 5 months ago

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@Department 8 – Ok. Which book do you have regarding Olympiad geometry?

Priyanshu Mishra - 5 years, 5 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...)

Department 8 - 5 years, 5 months ago

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@Department 8 – 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 - 5 years, 5 months ago

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

Department 8 - 5 years, 5 months ago

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@Department 8 – Ha that makes me kind of more nervous since I'm expcecting about 45-55. :)

Jack Frost - 5 years, 5 months ago

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

Department 8 - 5 years, 5 months ago

@Department 8 – Ya, AOPS is full of resources regarding any topic, contests.

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

Priyanshu Mishra - 5 years, 5 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!

Jack Frost - 5 years, 5 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

Ganesh Ayyappan - 5 years, 5 months ago

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Congratulations!

A Former Brilliant Member - 5 years, 5 months ago

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

Ganesh Ayyappan - 5 years, 5 months ago

I have got the proof for the 4 th one

Snehan Jayakumar - 5 years, 4 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.

Anurag Ramachandran - 3 years, 9 months ago

LOL you were damn lucky to get selected.

Anurag CHS - 2 years, 8 months ago

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 - 5 years, 5 months ago

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What do you expect the cutoff to be?

Jack Frost - 5 years, 5 months ago

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

Ganesh Ayyappan - 5 years, 5 months ago

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

SolnSol^{n}:

1st1st case: When ∠BDP=∠CDQ\angle BDP = \angle CDQ then MP=MQ.MP = MQ.

Constructions: Produce QMQM to Q′Q' such that MQ=MQ′.MQ = MQ'. Join BQ′,PQ BQ' , PQ and PQ′.PQ'.

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

In △BDP\triangle BDP and ΔCDQ,\Delta CDQ, ∠BDP=∠CDQ\angle BDP = \angle CDQ and ∠BPD=∠CQD\angle BPD = \angle CQD =90∘.=90^{\circ}. =>=> ΔBDP∼ΔCDQ\Delta BDP \sim \Delta CDQ =>=> BPCQ=DPDQ\dfrac{BP}{CQ}= \dfrac{DP}{DQ} =>=> BPBQ′=DPDQ\dfrac{BP}{BQ'} = \dfrac{DP}{DQ} (Using - [1][1] ) =>=> BPDP=BQ′DQ\dfrac{BP}{DP} = \dfrac{BQ'}{DQ}

Now, in ΔBQ′P\Delta BQ'P & ΔPDQ\Delta PDQ , BPDP=BQ′DQ\dfrac{BP}{DP} = \dfrac{BQ'}{DQ} and ∠Q′BP=∠PDQ\angle Q'BP = \angle PDQ [From eqneq^{n}] - 22 =>=> ΔBQ′P∼ΔPDQ\Delta BQ'P \sim \Delta PDQ =>=> ∠BPQ′=∠DPQ\angle BPQ' = \angle DPQ =>=> ∠BPQ′+∠Q′PQ=∠Q′PQ+∠DPQ\angle BPQ' + \angle Q'PQ = \angle Q'PQ + \angle DPQ =>=> 90∘=∠Q′PQ+∠DPQ=∠QPQ′90^{\circ} = \angle Q'PQ + \angle DPQ = \angle QPQ' =>=> ΔQPQ′\Delta QPQ' is right angled at PP.

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

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

Vishwash Kumar ΓΞΩ - 4 years, 1 month ago

Problem 4 is easy. One can easily notice that EP=FPEP = FP and DP=CPDP = CP [Because, the ⊥\perp from the center to the chord bisects the chord]. => FDECFDEC is a ∣∣gm.|| gm. Now, since DACBDACB is cyclic, DP×CP=AP×BP=DP2DP\times CP = AP\times BP = DP^{2} -- 11. Similarly, since FAEBFAEB is cyclic, EP×FP=AP×BP=FP2EP\times FP = AP\times BP = FP^{2}-- 22 The two results 1and21 and 2 tells that, DP=FP=>2DP=2FP=>DC=FEDP = FP => 2DP = 2FP => DC = FE. So, quad DFCEDFCE is a ∣∣gm||gm with the diagonals equal. => CEDFCEDF is a rectangle

Vishwash Kumar ΓΞΩ - 4 years, 1 month ago

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

Priyanshu Mishra - 5 years, 5 months ago

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In fact i found this the toughest paper !

Shrihari B - 5 years, 5 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?

A Former Brilliant Member - 5 years, 5 months ago

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@A Former 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 - 5 years, 5 months ago

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@Ganesh Ayyappan – I am actually in 8th

A Former Brilliant Member - 5 years, 5 months ago

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@A Former 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 - 5 years, 5 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.

A Former Brilliant Member - 5 years, 5 months ago

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

A Former Brilliant Member - 5 years, 5 months ago

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@A Former 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 - 5 years, 5 months ago

@A Former 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 - 5 years, 5 months ago

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@Shrihari B – No problem. I was just asking.

A Former Brilliant Member - 5 years, 5 months ago

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

Silver Vice - 5 years, 5 months ago

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Use Principle of Inclusion and Exclusion.

A Former Brilliant Member - 5 years, 5 months ago

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Got through with part a, finding part b difficult

Silver Vice - 5 years, 5 months ago
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