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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 \(\angle ABC\) 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 \(x\) and \(y\) such that \[x^2+2y^2+\dfrac{1}{2} \leq x(2y+1)\]

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

**4** Find all positive real number triplets \((x,y,z)\) which satisfy \[2x-2y+\dfrac{1}{z}=\dfrac{1}{2014}; 2y-2z+\dfrac{1}{x}=\dfrac{1}{2014}; 2z-2x+\dfrac{1}{y}=\dfrac{1}{2014}\]

**5** Let \(ABC\) be a triangle. Let \(X\) be on the segment \(\overline{BC}\), such that \(AB=AX\). Let \(AX\) meet the circumcircle \(\Gamma\) of triangle \(ABC\) again at \(D\). Show that the circumcentre of \(\Delta BDX\) lies on \(\Gamma\).

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

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!

## Comments

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TopNewest(4) Add all three equations.

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

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

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

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 · 2 years, 9 months ago

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– Rohan Rajpal · 1 year, 9 months ago

Is it wrong if we open 3/2014 and compare values?Log in to reply

@Satvik Golechha Proof to the \(3\)rd question:For \[n(2310-n)\] to be a multiple of \(2310,\)both of them have to contribute factors of \(2310.\)Let us take \[2310=n*x\] thus, \(n\) and \(x\) are both factors of \(2310.\)Now,let us replace \(n\) by \[\dfrac{2310}{x}\] simplifying gives \[\dfrac{2310^{2}*(x-1)}{x^{2}}.\]But,\[\dfrac{2310^{2}}{x^{2}}=n^{2}.\]Putting this into the expression gives\[:n^{2}*(x-1)=n^{2}*x-n^{2}.\]But,\[n*x=2310.\]Substituting gives,\[2310*n-n^{2}.\]But,\(n^{2}\) can't be divisible by \(2310\) as in the prime factorisation of \(2310\) all the numbers are distinct,then how can a square exist.Hence proved. – Adarsh Kumar · 2 years, 9 months ago

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– Subrata Saha · 2 years, 9 months ago

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 itLog in to reply

– Salmaan Shahid · 2 years, 9 months ago

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.Log in to reply

– Utkarsh Chaturvedi · 1 year, 9 months ago

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

– Adarsh Kumar · 2 years, 9 months ago

awesome method!!Log in to reply

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 · 2 years, 9 months ago

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Problem 5.Easy one. Any point \(V\) on \(\bigcirc ABC\) satisfies\[\measuredangle BVD=\measuredangle BAD=\measuredangle BAX=2\measuredangle BXA=2\measuredangle BXD\]

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

(All the angles are directed mod \(\pi\).) – Jubayer Nirjhor · 2 years, 9 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 · 1 year, 7 months ago

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Anybody got the sixth onr – Devang Patil · 1 year, 10 months ago

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Short solution to #1: Let \(D\) be the foot of the perpendicular from \(B\) to \(AC\). It is well-known that \(BR\) and \(BD\) are isogonal. Then since \(AXRB\) is cyclic and we have \(\angle RXC=\angle RBA=\angle CBD=90^\circ-\angle BCA\). Now let \(XR\) intersect \(BC\) at a point \(P\); then since \(\angle RXC+\angle PCX=90^\circ\) we easily get \(\angle XPC=90^\circ\). Done. – David Altizio · 2 years, 9 months ago

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I gave RMO from UP and got 4 questions correct , will i reach stage II (INMO)? – Devang Agarwal · 2 years, 9 months ago

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– Satvik Golechha · 2 years, 9 months ago

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.Log in to reply

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. – One Top · 2 years, 9 months agoLog in to reply

@Satvik, If you've solved Q1, will you please post the solution here? – One Top · 2 years, 9 months ago

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– Satvik Golechha · 2 years, 9 months ago

I did it by extending perpendicular from \(R\) to \(\overline{BC}\) and cutting \(\overline{AC}\) in \(X\), and then proved that \(ARBX\) is a cyclic quadrilateral.Log in to reply

I have uploaded the Mumbai region Paper here – Pranshu Gaba · 2 years, 9 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 · 2 years, 9 months ago

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What could be the expected cutoff for this set of RMO paper ? – Abhigyan Shekhar · 2 years, 9 months ago

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– Abhigyan Shekhar · 2 years, 9 months ago

Well, I think in MP region the cutoff would not exceed 50 marks anywayLog in to reply

– Satvik Golechha · 2 years, 9 months ago

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

Answer to some of the questions

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

4)\(x\)=\(y\)=\(z\)=2014

6) \(100\) – Parth Lohomi · 2 years, 9 months ago

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Why 17? Earlier, it used to be 30 right? – Krishna Ar · 2 years, 9 months ago

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– Satvik Golechha · 2 years, 9 months ago

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

– Subrata Saha · 2 years, 9 months ago

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

– Satvik Golechha · 2 years, 9 months ago

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.Log in to reply

– Ashu Dablo · 2 years, 9 months ago

Dude, you got any idea how many from Delhi region get selected??Log in to reply

– Krishna Ar · 2 years, 9 months ago

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

The question paper of Delhi was different. – Aneesh Kundu · 2 years, 9 months ago

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– Satvik Golechha · 2 years, 9 months ago

This is of Rajasthan. May I please get the Delhi paper? Thanks.Log in to reply

– Aneesh Kundu · 2 years, 9 months ago

The paper was very easy compared to previous years'.Log in to reply

– Subrata Saha · 2 years, 9 months ago

Jharkhand's paper is similar to Rajasthan 's.Log in to reply

– Aneesh Kundu · 2 years, 9 months ago

I'm posting all of themLog in to reply

– Mahimn Bhatt · 2 years, 9 months ago

There are many sets of question papers.Log in to reply

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 · 1 year, 9 months ago

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I attempted 5 and all of them are 100 percent correct? what are my chances? – Siddharth Kumar · 2 years, 9 months ago

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– Siddharth Kumar · 2 years, 9 months ago

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

@Satvik Golechha This paper is of which state? – Pranjal Jain · 2 years, 9 months ago

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Here is the paper for Rajasthan region. – Pranjal Jain · 2 years, 9 months ago

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What is the answer for Q.6? – Ankush Tiwari · 2 years, 9 months ago

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– Joel Tan · 2 years, 9 months ago

S (n) is congruent to n mod 9, so n is congruent to 2 mod 9. However, the sum of digits of any such n is less than or equal to 9+9+2=20. The sum of digits of a number less than or equal to 20 cannot be greater than 1+9=10, so for all n <1000 congruent to 2 mod 9, S (S (n))=2. Thus answer is (1000-100)/9=100.Log in to reply

– Abhigyan Shekhar · 2 years, 9 months ago

Well, many have have counted to get the answer.Only few got the answer correct but I don't think the approach is right.Log in to reply

– Satvik Golechha · 2 years, 9 months ago

That's really cool. All of us were making cases and counting. Extra marks for extra Elegance!Log in to reply

IMO 1975 Problem 4 – Ishan Singh · 2 years, 9 months ago

Problem 6 is a simplified version ofLog in to reply

– Gautam Sharma · 2 years, 9 months ago

that could be done by multinomial theorem we have to find coeff of \(x^{20},x^{11}\) and we will get 100. And Multinomial gives ordered and all distinct pairs so no need to worry for duplicacy. And also 3 pairs when S(n)=2. For 11------> 61 numbers. For 20------>36 numbers . For 2-------->3 numbers.Log in to reply

– Prakher Gaushal · 1 year, 11 months ago

Did it by the same methodLog in to reply

@Satvik Golechha !! Btw what would be cutoff?? – Harsh Shrivastava · 2 years, 9 months ago

I too added wrong & got 103Log in to reply

– Satvik Golechha · 2 years, 9 months ago

I expect cutoff to be around 4.685954 questions out of 6. Many are getting 4 correct.Log in to reply

@Satvik Golechha – Harsh Shrivastava · 2 years, 9 months ago

In Q6) I showed every steps but counted wrong so what will i get for that question?Log in to reply

– Satvik Golechha · 2 years, 9 months ago

Exactly the case with me! I made three cases, and made some error(s) while counting. I expect them to give around 1/3 of the total marks for the question. How many questions did you get right in RMO?Log in to reply

– Harsh Shrivastava · 2 years, 9 months ago

Almost three :( What about you??Log in to reply

– Satvik Golechha · 2 years, 9 months ago

4 completely right, one 75% correct, one 20% correct.Log in to reply

– Harsh Shrivastava · 2 years, 9 months ago

That's Good!!!!Log in to reply

– Satvik Golechha · 2 years, 9 months ago

Marks are OK, but since we have only 15 selections from Rajasthan, there's quite a lesser chance of selections.Log in to reply

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– Subrata Saha · 2 years, 9 months ago

Did u give RMO??Log in to reply

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– Subrata Saha · 2 years, 9 months ago

U are surely going to qualify, I hope u are in 8 and at this stage 4.5 in RMO, its outstanding.Log in to reply

– Parth Lohomi · 2 years, 9 months ago

Actually I am in \(10^{th}\)Log in to reply

– Mahimn Bhatt · 2 years, 9 months ago

Then too youhave a good grip over maths, as its difficult for a 10th grade student for doing all thisLog in to reply

Ans to Q6) 173 – Kunal Jain · 2 years, 7 months ago

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– Harsh Shrivastava · 2 years, 7 months ago

Sorry, but the answer is \(100\).Log in to reply