RMO 2014 Full Paper Discussion

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Regional Mathematical Olympiad-2014

Time: 3 hours

Instructions:

\bullet Calculators (in any form) and protractors are not allowed.

\bullet Rulers and compasses are allowed.

\bullet Answer all the questions.

\bullet All questions carry equal marks. Maximum marks: 102.


Questions:-

1 Let ABC be an acute-angled triangle and suppose ABC\angle ABC is the largest angle of the triangle. Let R be it's circumcentre. Let the circumcircle of triangle ARB cut AC again in X. Prove that RX is perpendicular to BC.


2 Find all real numbers xx and yy such that x2+2y2+12x(2y+1)x^2+2y^2+\dfrac{1}{2} \leq x(2y+1)


3 Prove that there does not exist any positive integer n<2310n<2310, such that n(2310n)n(2310-n) is a multiple of 23102310.


4 Find all positive real number triplets (x,y,z)(x,y,z) which satisfy 2x2y+1z=12014;2y2z+1x=12014;2z2x+1y=120142x-2y+\dfrac{1}{z}=\dfrac{1}{2014}; 2y-2z+\dfrac{1}{x}=\dfrac{1}{2014}; 2z-2x+\dfrac{1}{y}=\dfrac{1}{2014}


5 Let ABCABC be a triangle. Let XX be on the segment BC\overline{BC}, such that AB=AXAB=AX. Let AXAX meet the circumcircle Γ\Gamma of triangle ABCABC again at DD. Show that the circumcentre of ΔBDX\Delta BDX lies on Γ\Gamma.


6 For any natural number nn, let S(n)S(n) denote the sum of digits of nn. Find the number of all 3-digit numbers nn such that S(S(n))=2S(S(n))=2.


RMO was held in various centres in India on 7th December. You can take it as a test. Please let me know your score. 17 students will be selected from my state, Rajasthan. Pls Like and Reshare and Enjoy!

Note by Satvik Golechha
4 years, 6 months ago

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(4) Add all three equations.

1x+1y+1z=32014\frac {1}{x}+\frac {1}{y}+\frac {1}{z}=\frac {3}{2014}

Multiply the first equation by z, second by x, third by y and add all the three new equations to get x+y+z2014=3    x+y+z=3×2014\frac {x+y+z}{2014}=3 \implies x+y+z=3×2014.

Thus (x+y+z)(1x+1y+1z)=9(x+y+z)(\frac {1}{x}+\frac {1}{y}+\frac {1}{z})=9.

But Cauchy-Schwarz inequality tells us that the expression is greater than or equal to 9 with equality iff corresponding terms have the same ratio ie. x=y=z. Hence x=y=z=2014 is the only solution

Joel Tan - 4 years, 6 months ago

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Is it wrong if we open 3/2014 and compare values?

Rohan Rajpal - 3 years, 6 months ago

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@Satvik Golechha Proof to the 33rd question:For n(2310n)n(2310-n) to be a multiple of 2310,2310,both of them have to contribute factors of 2310.2310.Let us take 2310=nx2310=n*x thus, nn and xx are both factors of 2310.2310.Now,let us replace nn by 2310x\dfrac{2310}{x} simplifying gives 23102(x1)x2.\dfrac{2310^{2}*(x-1)}{x^{2}}.But,23102x2=n2.\dfrac{2310^{2}}{x^{2}}=n^{2}.Putting this into the expression gives:n2(x1)=n2xn2.:n^{2}*(x-1)=n^{2}*x-n^{2}.But,nx=2310.n*x=2310.Substituting gives,2310nn2.2310*n-n^{2}.But,n2n^{2} can't be divisible by 23102310 as in the prime factorisation of 23102310 all the numbers are distinct,then how can a square exist.Hence proved.

Adarsh Kumar - 4 years, 6 months ago

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This I did by using the fact that 2310 is square free and that NCERT procedure that if p divides n^2, then p divindes n. Thus showing 2310 is a factor of n. And hence n can never be less than it

Brilliant Member - 4 years, 6 months ago

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I did it like shubham saha (that's his name :P ) . I was really doubtful about this one as I couldn't imagine an RMO question to be of such(easy) level.

salmaan shahid - 4 years, 6 months ago

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awesome method!!

Adarsh Kumar - 4 years, 6 months ago

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Precisely! But do the RMO examiners give the full 17 marks for such a short method?

Utkarsh Chaturvedi - 3 years, 6 months ago

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Simpler fact is that if n (2310-n) is divisible by 2310, n^2 is divisible by 2310 and since no square except 1 divides 2310, n=2310k for k>0 and k being integer, implying there does not exist solution.

Anurag Ramachandran - 8 months, 3 weeks ago

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Proof for 2: The inequality is equivalent to (x - 2y)^2 + (x - 1)^2 <= 0. But a squared number is always greater than or equal to 0, which implies that there is a real number in the equality case, which implies that we solve the equation (x - 2y)^2 + (x - 1)^2 = 0 and this is possible when x = 2y = 1, implying the ordered pair (x, y) = (1, 1/2).

John Ashley Capellan - 4 years, 6 months ago

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Problem 5. Easy one. Any point VV on ABC\bigcirc ABC satisfies

BVD=BAD=BAX=2BXA=2BXD\measuredangle BVD=\measuredangle BAD=\measuredangle BAX=2\measuredangle BXA=2\measuredangle BXD

so ABC\bigcirc ABC is the locus of all points VV such that BVD=2BXD\measuredangle BVD=2\measuredangle BXD. The conclusion is now evident.

(All the angles are directed mod π\pi.)

Jubayer Nirjhor - 4 years, 6 months ago

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The question paper of Delhi was different.

Aneesh Kundu - 4 years, 6 months ago

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There are many sets of question papers.

Mahimn Bhatt - 4 years, 6 months ago

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This is of Rajasthan. May I please get the Delhi paper? Thanks.

Satvik Golechha - 4 years, 6 months ago

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The paper was very easy compared to previous years'.

Aneesh Kundu - 4 years, 6 months ago

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I'm posting all of them

Aneesh Kundu - 4 years, 6 months ago

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Jharkhand's paper is similar to Rajasthan 's.

Brilliant Member - 4 years, 6 months ago

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Why 17? Earlier, it used to be 30 right?

Krishna Ar - 4 years, 6 months ago

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Dude, you got any idea how many from Delhi region get selected??

Ashu Dablo - 4 years, 6 months ago

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30(Not too sure though- You can check last yr's result +cutoff)

Krishna Ar - 4 years, 6 months ago

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Exactly, it used to be 30 till last year. And I can't possible answer "Why?".

Satvik Golechha - 4 years, 6 months ago

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Well, perhaps government has understood that people in RAJASTHAN don't want to be mathematician but IITian. Just joking...

Brilliant Member - 4 years, 6 months ago

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@Brilliant Member Well, actually, it's partly true. People don't even know 'mathematician' is someone; everyone's running after IIT. And it'sn't their fault; in their opinion IIT will get them jobs, and money.

Satvik Golechha - 4 years, 6 months ago

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Answer to some of the questions

2) 1,121, \dfrac{1}{2}

4)xx=yy=zz=2014

6) 100100

Parth Lohomi - 4 years, 6 months ago

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What could be the expected cutoff for this set of RMO paper ?

Abhigyan Shekhar - 4 years, 6 months ago

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Well, I think in MP region the cutoff would not exceed 50 marks anyway

Abhigyan Shekhar - 4 years, 6 months ago

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I guess around 4 questions right outta 6. This year paper was a bit easier than last year.

Satvik Golechha - 4 years, 6 months ago

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Proof for 3: We consider the equation 2310n - n^2 = 2310k which implies that n^2 is congruent to 0 (mod 2310) with the condition that n^2 < 2310^2. Consider the prime factors of 2310 = 5 . 2 . 7 . 11 . 3 in which it has no factor that is a perfect square which implies that there is no n that satisfies the divisibility and n^2 < 2310^2....

John Ashley Capellan - 4 years, 6 months ago

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I have uploaded the Mumbai region Paper here

Pranshu Gaba - 4 years, 6 months ago

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@Satvik, If you've solved Q1, will you please post the solution here?

Ajit Athle - 4 years, 6 months ago

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I did it by extending perpendicular from RR to BC\overline{BC} and cutting AC\overline{AC} in XX, and then proved that ARBXARBX is a cyclic quadrilateral.

Satvik Golechha - 4 years, 6 months ago

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Q1: Draw RM perpendicular to AB. RM bisects /ARB since RA = RB = the circumradius of Tr. ABC. Extend XR to meet BC in N. Now /ARM=/BRM=C and /MAR=/MBR=90 - C. Hence /_ XBR= /XAR= 90-B since quad XABR is concyclic. Now /ABX = (90 - C)- (90-B)=B-C) which is a +ve quantity since B is the largest /_ in the triangle. Therefore /ARX=B - C. This makes /MRX= C+B-C = B. In other words, quad. BMRN is concyclic and thus RN or RX is perpendicular to BC since /_RMB=90 by construction.

Ajit Athle - 4 years, 6 months ago

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I gave RMO from UP and got 4 questions correct , will i reach stage II (INMO)?

Devang Agarwal - 4 years, 6 months ago

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Was it the same paper as this one? If yes, then you may have a chance, but many got 4 correct. If it was any other paper, you have a good chance. All the best.

Satvik Golechha - 4 years, 6 months ago

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Short solution to #1: Let DD be the foot of the perpendicular from BB to ACAC. It is well-known that BRBR and BDBD are isogonal. Then since AXRBAXRB is cyclic and we have RXC=RBA=CBD=90BCA\angle RXC=\angle RBA=\angle CBD=90^\circ-\angle BCA. Now let XRXR intersect BCBC at a point PP; then since RXC+PCX=90\angle RXC+\angle PCX=90^\circ we easily get XPC=90\angle XPC=90^\circ. Done.

David Altizio - 4 years, 5 months ago

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Anybody got the sixth onr

Devang Patil - 3 years, 6 months ago

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Do the books prescribed by hbcse suffice for rmo preparation?. I am in class 9 now, and looking forward to write rmo next year. Can anyone suggest good preparation material for the exam?

Sidharth Nair - 3 years, 3 months ago

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What is the answer for Q.6?

Ankush Tiwari - 4 years, 6 months ago

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@Satvik Golechha This paper is of which state?

Pranjal Jain - 4 years, 6 months ago

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Here is the paper for Rajasthan region.

Pranjal Jain - 4 years, 6 months ago

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I attempted 5 and all of them are 100 percent correct? what are my chances?

Siddharth Kumar - 4 years, 6 months ago

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Was not able to solve the first question... and also had a different method for question 2, did it by discriminant and functions.

Siddharth Kumar - 4 years, 6 months ago

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The sixth one is easy. Here's how to solve it: S(S(n)) = 2, therefore the values of S(n) can be 2, 11, or 20. (you can check, these are the only possible values.

Case I: S(n) = 2, possible values = 101, 110, 200

Case II: S(n) = 11 -Subcase I: when H = 1 (H is the hundreds digit), for T = 1 to 9 and the corresponding values of O (tens and ones), we get 9 values (119, 128, 137, 146, 155, 164, 173, 182, 191) -Subcase II: when H = 2, here, the order of T and O goes from T = 0 to 9 and the corresponding O values, this produces 10 values (i.e. 209, 218, 227, 236, 245, 254, 263, 272, 281, 290) If you go for further values of H, you will see that the number of integers satisfying S(n) = 11 reduces by 1 each time (from H = 2 to H = 9) the summation of values in this case therefore is 9 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 = 61

Case III: S(n) = 20, similar to Case II, instead that the value of H is from 2 to 9 and that the no. of permissible integers increases by 1. You can go over the subcases yourself. The summation comes out to be 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

Grand total of three cases = 3 + 61 + 36 = 100.

Utkarsh Chaturvedi - 3 years, 6 months ago

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Ans to Q6) 173

Kunal Jain - 4 years, 4 months ago

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Sorry, but the answer is 100100.

Harsh Shrivastava - 4 years, 3 months ago

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