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# This note has been used to help create the RMO Math Contest Preparation wiki

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!

Note by Satvik Golechha
3 years ago

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

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

- 3 years ago

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

- 2 years ago

@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.

- 3 years 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 it

- 3 years 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.

- 3 years ago

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

- 2 years ago

awesome method!!

- 3 years ago

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).

- 3 years ago

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 $$\pi$$.)

- 3 years ago

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?

- 1 year, 10 months ago

Anybody got the sixth onr

- 2 years ago

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.

- 2 years, 12 months ago

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

- 3 years 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.

- 3 years ago

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.

- 3 years ago

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

- 3 years 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.

- 3 years ago

I have uploaded the Mumbai region Paper here

- 3 years ago

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....

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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

- 3 years ago

Answer to some of the questions

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

4)$$x$$=$$y$$=$$z$$=2014

6) $$100$$

- 3 years ago

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

- 3 years ago

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

- 3 years ago

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

- 3 years 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.

- 3 years ago

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

- 3 years ago

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

- 3 years ago

The question paper of Delhi was different.

- 3 years ago

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

- 3 years ago

The paper was very easy compared to previous years'.

- 3 years ago

Jharkhand's paper is similar to Rajasthan 's.

- 3 years ago

I'm posting all of them

- 3 years ago

There are many sets of question papers.

- 3 years ago

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.

- 2 years ago

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

- 3 years ago

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

- 3 years ago

@Satvik Golechha This paper is of which state?

- 3 years ago

Here is the paper for Rajasthan region.

- 3 years ago

What is the answer for Q.6?

- 3 years ago

Comment deleted Dec 08, 2014

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.

- 3 years 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.

- 3 years ago

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

- 3 years ago

Problem 6 is a simplified version of IMO 1975 Problem 4

- 3 years 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.

- 3 years ago

Did it by the same method

- 2 years, 2 months ago

I too added wrong & got 103 @Satvik Golechha !! Btw what would be cutoff??

- 3 years ago

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

- 3 years ago

In Q6) I showed every steps but counted wrong so what will i get for that question?@Satvik Golechha

- 3 years 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?

- 3 years ago

Almost three :( What about you??

- 3 years ago

4 completely right, one 75% correct, one 20% correct.

- 3 years ago

That's Good!!!!

- 3 years ago

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

- 3 years ago

Comment deleted May 06, 2015

Did u give RMO??

- 3 years ago

Comment deleted May 06, 2015

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

- 3 years ago

Actually I am in $$10^{th}$$

- 3 years ago

Then too youhave a good grip over maths, as its difficult for a 10th grade student for doing all this

- 3 years ago

Ans to Q6) 173

- 2 years, 10 months ago

Sorry, but the answer is $$100$$.

- 2 years, 10 months ago