# 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
6 years, 7 months 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

- 6 years, 7 months ago

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

- 5 years, 7 months 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.

- 6 years, 7 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 it

- 6 years, 7 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.

- 6 years, 7 months ago

awesome method!!

- 6 years, 7 months ago

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

- 5 years, 7 months ago

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.

- 2 years, 10 months 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).

- 6 years, 7 months 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 $\pi$.)

- 6 years, 7 months ago

The question paper of Delhi was different.

- 6 years, 7 months ago

There are many sets of question papers.

- 6 years, 7 months ago

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

- 6 years, 7 months ago

The paper was very easy compared to previous years'.

- 6 years, 7 months ago

I'm posting all of them

- 6 years, 7 months ago

Jharkhand's paper is similar to Rajasthan 's.

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 years, 7 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.

- 6 years, 7 months ago

Answer to some of the questions

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

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

6) $100$

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 years, 7 months 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....

- 6 years, 7 months ago

I have uploaded the Mumbai region Paper here

- 6 years, 7 months ago

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

- 6 years, 7 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.

- 6 years, 7 months 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.

- 6 years, 7 months ago

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

- 6 years, 7 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.

- 6 years, 7 months 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.

- 6 years, 7 months ago

Anybody got the sixth onr

- 5 years, 8 months 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?

- 5 years, 5 months ago

What is the answer for Q.6?

- 6 years, 7 months ago

@Satvik Golechha This paper is of which state?

- 6 years, 7 months ago

Here is the paper for Rajasthan region.

- 6 years, 7 months ago

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

- 6 years, 7 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.

- 6 years, 7 months 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.

- 5 years, 7 months ago

Ans to Q6) 173

- 6 years, 5 months ago

Sorry, but the answer is $100$.

- 6 years, 5 months ago