# RMO 2016 Delhi Region

Here is the paper of RMO 2016 Delhi.Pls post answers and solutions to all questions. Also tell ur marks and estimated cut off Thanks!

Note by Kaustubh Miglani
3 years, 1 month ago

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Problem 1

We can see that as we move the point $P$ on the circumference of the circle$[$excluding $X$ and $Y],$the $\angle XPY=\angle XP_1Y$ remains constant.So this shows that $AB=A_1B_1.$Now we use extended sin rule to complete the problem.
Let the circum-radius of $\triangle PAB$ be $R$ and $\triangle P_1A_1B_1$ be $R_1.$
In $\triangle PAB, \frac{AB}{sin\angle P}=2R$ and in $\triangle P_1A_1B_1,\frac{A_1B_1}{sin\angle P_1}=\frac{AB}{sin\angle P}=2R_1.$ Therefore $2R=2R_1\Rightarrow\boxed {R=R_1}.$Hence Proved.

- 3 years, 1 month ago

nice solution

- 3 years, 1 month ago

Hey can u explain in detail why A1B1=AB

- 2 years, 10 months ago

Because angle subtended by both of these chords at centre are equal. angle A1 X A =P X P1=P Y P1=B Y B1

- 2 years, 10 months ago

Which class are u in and which school?

- 2 years, 10 months ago

I am graduating from IIT Bombay. What is ur favourite college bro?

- 2 years, 10 months ago

IIT Is the one Bombay ,Perhaps

- 2 years, 10 months ago

Prefered Jee rank?mine was 17

- 2 years, 10 months ago

Hi Where Can I find some good material for RMO this year? Last time I failed miserably :'( could solve just one... Thanks :)

- 2 years, 5 months ago

would u like to join my RMO/INMO group. if yes then give ur email.

- 2 years, 5 months ago

Yeah he is good.Invite him

- 2 years, 5 months ago

- 2 years, 5 months ago

- 2 years, 5 months ago

- 2 years, 5 months ago

he is already in there @ayush rai

- 2 years, 4 months ago

is dis question addressed to me?

- 2 years, 5 months ago

My email yashmehan@gmail.com Sorry for the late reply I do not often open up Brilliant Thanks!!!

- 2 years, 3 months ago

My email is adityak1135@gmail.com I know that this is a really late reply but I didn't know about it earlier ;)

- 2 years, 3 months ago

first solve all previous year question papers, then use brilliant if ur aspiration is limited to rmo not inmo or selection camp P.S. i did 4 at my time

- 2 years, 5 months ago

Do u remember books you used for iit preparation.I am an iit aspirant

- 2 years, 5 months ago

hcv,Bmsharma,resnick,irodov,krotov,kolenkow -Phy atkins,rc mukerje,bahadur,jd lee-Chm sl loney,hallknight,mostly tatamcgraw buks and tons of practise question papers here at allen.u can download them online also.Ncert is must .Hope it helps.

- 2 years, 5 months ago

I m in class 10 And from bbps dw. But ur profile says u live in noida

- 2 years, 10 months ago

Problem 5 1)its the easiest- use the identity ${ (x+y+z) }^{ 2 }+{ (x+y-z) }^{ 2 }+{ (x-y+z) }^{ 2 }+{ (-x+y+z) }^{ 2 }={ (2x) }^{ 2 }+{ (2y) }^{ 2 }+{ (2z) }^{ 2 }$ which gives infinite solutions in integers x,y,z.u can check it satisfis all other cond.s 2)solve this one other way round, assume 2016 = ${ (2x) }^{ 2 }+{ (2y) }^{ 2 }+{ (2z) }^{ 2 }$ then it implies 504 = ${ (x) }^{ 2 }+{ (y) }^{ 2 }+{ (z) }^{ 2 }$ by inspection x=22 y=4 z=6, then use the identity Problem 2 -we obtain nth term in terms of n,a,b then use strong induction.

- 2 years, 5 months ago

the question says pairwise distinct positive integers with no common factor, doesn't that mean they're pairwise coprime?

- 2 years, 4 months ago

The question mentions that they all don't have a factor in common, and that doesn't mean they are pairwise relatively prime. If x,y, and z are relatively prime, then we are done because then the gcd of these 7 terms would be 1 :)

- 2 years, 3 months ago

oh, thank you i didn't understand the question properly then

- 2 years, 2 months ago

Can anyone post complete solution to 2nd and last problem?

- 2 years, 4 months ago

Problem 4

By the Cauchy-Schwarz inequality, $\left(\dfrac{1}{a}+1+c\right)(a^{3}+b^{2}+c) \geq (a+b+c)^{2}=9$

$\implies (ac+a+1)(a^{3}+b^{2}+c) \geq 9a$

Thus, we get

$f(a,b,c)=\quad \sum {\dfrac{(ac+a+1)a}{(ac+a+1)({a}^{3}+{b}^{2}+{c})}} \leq \quad \sum \dfrac{a(ac+a+1)}{9a}$

$\implies f(a,b,c) \leq \dfrac{1}{9}[(ac+a+1)+(ba+b+1)+(cb+c+1)]$

$=\dfrac{1}{9}[(a+b+c)+(1+1+1)+(ab+bc+ca)] \leq \dfrac{1}{9}[3+3+3]=\boxed{1}$

Here we used the identity $(a+b+c)^{2} \geq 3(ab+bc+ca)$ to get $ab+bc+ca \leq 3$

- 2 years, 3 months ago

Solution to Problem 1. Let $P_{1}$ and $P_{2}$ be the two points on $\odot\omega_{1}$ as $P$ varies across the circumference of $\odot \omega_{1}.$ Similarly, let $P_{1}X$ and $P_{1}Y$ meet $\odot\omega_{2}$ at $A_{1}$ , $B_{1}$ and $P_{2}X$ and $P_{2}Y$ meet $\odot\omega2$ at $A_{2}$ , $B_{2}$ respectively.

Clearly , $\angle XP_{1}Y = \angle XP_{2}Y$ and $\angle XB_{1}Y = \angle XB_{2}Y$

Adding these two $eq^{n}$ we get:: $\angle A_{1}XB_{1} = \angle A_{2}XB_{2}$ => $A_{1}B_{1} = A_{2}B_{2}.$

Also, $\Delta XP_{1}Y \sim \Delta A_{1}P_{1}B_{1}$ => $\dfrac{Radius XP_{1}Y}{Radius A_{1}P_{1}B_{1}}$ = $\dfrac{XY}{A_{1}B_{1}}$ ........[ $1$ ]

Similarly, $\Delta XP_{2}Y \sim \Delta A_{2}P_{2}B_{2}$

=> $\dfrac{Radius XP_{2}Y}{Radius A_{2}P_{2}B_{2}}$ = $\dfrac{XY}{A_{2}B_{2}}$

=> $\dfrac{Radius XP_{1}Y}{Radius A_{2}P_{2}B_{2}}$ = $\dfrac{XY}{A_{1}B_{1}}$ ........[ $2$ ]

[Because $\Delta XP_{1}Y$ and $\triangle XP_{2}Y$ have the same circumcircle. & $A_{1}B_{1} = A_{2}B_{2}$]

From $eq^{n}$ $[1]$ and $[2]$ we get that :: Radius $A_{1}P_{1}B_{1}$ = Radius $A_{2}P_{2}B_{2}.$

$K.I.P.K.I.G$

- 2 years, 7 months ago

Diagram to problem 1.........

- 2 years, 7 months ago

Can anyone give complete solution of 2 and 6

- 2 years, 4 months ago

@Azimuddin Sheikh At least two cards have same suit let P1, P2 has same suit then the assistant will turn down one of P1, P2 (say P1) and arrange P2, P3, P4, P5 in such a way that P2 takes first position so the magician know that the card turned down and the card having first position have same suit. This will help him to identify suit of P1 now there are 6 ways to arrange P3 P4 P5 as LMH, LHM, MLH, MHL, HLM, HML (low number, middle number, high number) If difference of numbers obtained on P1 and P2 is 6 then turn down the higher one and give first position to lower one. Let us assume P1 has number k and P2 has no k + 1, k + 2, k + 3, k + 4, k + 5 or k + 6. If P2 has k + 1 then show LMH. If P2 has k + 2 then show LHM and soon. If difference is > 6 then turn down lower one and give first position to higher one and add 1,2,3,4,5,6 for arrangements LMH, LHM, MLH, MHL, HLM, HML in the higher number for example adding 4 to 11 indicates 2.

- 2 years, 4 months ago

The last combinatorics question is more or less the same as the first question in the chapter combinatorics from the book 'The art and craft of problem-solving'

- 2 years, 3 months ago

@Azimuddin Sheikh Idk latex so cant post soln to 2

- 2 years, 4 months ago

Pls do provide the solution of the last problem

- 2 years, 3 months ago

Diagram to problem $3$

- 2 years, 7 months ago

Instead,use A-Humpty point lemma, which proves it without anything.

- 2 years, 5 months ago

Yup, and then repeat the same procedure to prove the lemma in examination.

- 2 years, 5 months ago

So in exam, to do any inequality, we can use only upto am-gm? not even cauchy? or can we not use even weighted am-gm?

- 2 years, 5 months ago

Hm... All those words go above me. Humpty-Dumpty theorem (Both different) are well-known. Although not that standard to be used directly in exams. I have seen guys proving each and every lemma, all for the sake of marks only/.

- 2 years, 5 months ago

so by standard textbooks, do they mean ncert?

- 2 years, 4 months ago

What? No.......

- 2 years, 4 months ago

You can use AM-GM and weighted AM-GM because at times some problems are not easy to prove without weighted AM-GM{RMO 2012 paper1 p3}. Use Cauchy-Schwarz too because it is of the basic inequalities. If you use a lemma like the Titu's lemma, then better prove it for RMO, I guess no need to prove that in INMO, not IMO definitely!

- 2 years, 2 months ago

Solution to $3$

Const:-- Join $BP$ , $BQ$ , $CP$ & $CQ.$

Clearly, Using Alternate Segment Theorem , we get:$\color{#D61F06}{\angle BAQ = \angle BQP}$ and $\color{#3D99F6}{\angle BAP = \angle BPQ}$.--- $[1]$

Also Using the isosceles triangle property, we get :: $\color{#D61F06}{\angle PQC = \angle BQP}$ & $\color{#3D99F6}{\angle QPC = \angle BPQ}$-----$[2]$

Thus from $eq^{n}$ $[1]$ and $[2]$ we get: $\color{#D61F06}{\angle BAQ = \angle PQC}$ & $\color{#3D99F6}{\angle BAP = \angle QPC}$.

Adding these two results we get: $\angle BAQ + \angle BAP$ = $\angle PQC + \angle QPC$

Thus $\angle PCQ$ = $180^{\circ}-[\angle PQC + \angle QPC]$ = $180^{\circ}-[\angle BAQ + \angle BAP]$

Therefore, $\angle PCQ + PAQ$ = $180^{\circ}-[\angle BAQ + \angle BAP] + [\angle BAP + \angle BAQ]$ = $180^{\circ}$

=>$\quad APCQ$ is cyclic

=> $\angle PAC= \color{#D61F06}{\angle PQC = \angle BAQ}$

=> $\angle PAC = \angle BAQ$

- 2 years, 7 months ago

Need a solution for q 2 and 6

- 2 years, 4 months ago

Please explain the statement above the concluding statement

- 1 year, 9 months ago