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 Problem 1

We can see that as we move the point PP on the circumference of the circle[[excluding XX and Y],Y],the XPY=XP1Y\angle XPY=\angle XP_1Y remains constant.So this shows that AB=A1B1.AB=A_1B_1.Now we use extended sin rule to complete the problem.
Let the circum-radius of PAB\triangle PAB be RR and P1A1B1\triangle P_1A_1B_1 be R1.R_1.
In PAB,ABsinP=2R\triangle PAB, \frac{AB}{sin\angle P}=2R and in P1A1B1,A1B1sinP1=ABsinP=2R1.\triangle P_1A_1B_1,\frac{A_1B_1}{sin\angle P_1}=\frac{AB}{sin\angle P}=2R_1. Therefore 2R=2R1R=R1.2R=2R_1\Rightarrow\boxed {R=R_1}.Hence Proved.

Ayush G Rai - 3 years, 1 month ago

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nice solution

abhishek alva - 3 years, 1 month ago

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Hey can u explain in detail why A1B1=AB

Kaustubh Miglani - 2 years, 10 months ago

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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

Yuvraj Singh - 2 years, 10 months ago

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@Yuvraj Singh Which class are u in and which school?

Kaustubh Miglani - 2 years, 10 months ago

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

Yuvraj Singh - 2 years, 10 months ago

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@Yuvraj Singh IIT Is the one Bombay ,Perhaps

Kaustubh Miglani - 2 years, 10 months ago

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@Kaustubh Miglani Prefered Jee rank?mine was 17

Yuvraj Singh - 2 years, 10 months ago

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

Yash Mehan - 2 years, 5 months ago

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

Ayush G Rai - 2 years, 5 months ago

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@Ayush G Rai Yeah he is good.Invite him

Kaustubh Miglani - 2 years, 5 months ago

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@Kaustubh Miglani what about u?

Ayush G Rai - 2 years, 5 months ago

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@Ayush G Rai what about me?

Kaustubh Miglani - 2 years, 5 months ago

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@Kaustubh Miglani wont u join our group?

Ayush G Rai - 2 years, 5 months ago

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@Ayush G Rai he is already in there @ayush rai

abhishek alva - 2 years, 4 months ago

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@Ayush G Rai is dis question addressed to me?

Yuvraj Singh - 2 years, 5 months ago

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

Yash Mehan - 2 years, 3 months ago

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

Aditya Khurmi - 2 years, 3 months ago

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@Yash Mehan 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

Yuvraj Singh - 2 years, 5 months ago

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

Kaustubh Miglani - 2 years, 5 months ago

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@Kaustubh Miglani 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.

Yuvraj Singh - 2 years, 5 months ago

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

Kaustubh Miglani - 2 years, 10 months ago

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Problem 5 1)its the easiest- use the identity (x+y+z)2+(x+yz)2+(xy+z)2+(x+y+z)2=(2x)2+(2y)2+(2z)2{ (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{ (2x) }^{ 2 }+{ (2y) }^{ 2 }+{ (2z) }^{ 2 } then it implies 504 = (x)2+(y)2+(z)2{ (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.

Yuvraj Singh - 2 years, 5 months ago

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

Govind 771 - 2 years, 4 months ago

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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 :)

Aditya Khurmi - 2 years, 3 months ago

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

Govind 771 - 2 years, 2 months ago

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

Azimuddin Sheikh - 2 years, 4 months ago

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Problem 4

By the Cauchy-Schwarz inequality, (1a+1+c)(a3+b2+c)(a+b+c)2=9\left(\dfrac{1}{a}+1+c\right)(a^{3}+b^{2}+c) \geq (a+b+c)^{2}=9

    (ac+a+1)(a3+b2+c)9a\implies (ac+a+1)(a^{3}+b^{2}+c) \geq 9a

Thus, we get

f(a,b,c)=(ac+a+1)a(ac+a+1)(a3+b2+c)a(ac+a+1)9af(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}

    f(a,b,c)19[(ac+a+1)+(ba+b+1)+(cb+c+1)]\implies f(a,b,c) \leq \dfrac{1}{9}[(ac+a+1)+(ba+b+1)+(cb+c+1)]

=19[(a+b+c)+(1+1+1)+(ab+bc+ca)]19[3+3+3]=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)23(ab+bc+ca)(a+b+c)^{2} \geq 3(ab+bc+ca) to get ab+bc+ca3ab+bc+ca \leq 3

Aditya Khurmi - 2 years, 3 months ago

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Solution to Problem 1. Let P1P_{1} and P2P_{2} be the two points on ω1\odot\omega_{1} as PP varies across the circumference of ω1.\odot \omega_{1}. Similarly, let P1XP_{1}X and P1YP_{1}Y meet ω2\odot\omega_{2} at A1A_{1} , B1B_{1} and P2XP_{2}X and P2YP_{2}Y meet ω2\odot\omega2 at A2A_{2} , B2B_{2} respectively.

Clearly , XP1Y=XP2Y\angle XP_{1}Y = \angle XP_{2}Y and XB1Y=XB2Y\angle XB_{1}Y = \angle XB_{2}Y

Adding these two eqneq^{n} we get:: A1XB1=A2XB2\angle A_{1}XB_{1} = \angle A_{2}XB_{2} => A1B1=A2B2.A_{1}B_{1} = A_{2}B_{2}.

Also, ΔXP1YΔA1P1B1\Delta XP_{1}Y \sim \Delta A_{1}P_{1}B_{1} => RadiusXP1YRadiusA1P1B1\dfrac{Radius XP_{1}Y}{Radius A_{1}P_{1}B_{1}} = XYA1B1\dfrac{XY}{A_{1}B_{1}} ........[ 11 ]

Similarly, ΔXP2YΔA2P2B2\Delta XP_{2}Y \sim \Delta A_{2}P_{2}B_{2}

=> RadiusXP2YRadiusA2P2B2\dfrac{Radius XP_{2}Y}{Radius A_{2}P_{2}B_{2}} = XYA2B2\dfrac{XY}{A_{2}B_{2}}

=> RadiusXP1YRadiusA2P2B2\dfrac{Radius XP_{1}Y}{Radius A_{2}P_{2}B_{2}} = XYA1B1\dfrac{XY}{A_{1}B_{1}} ........[ 22 ]

[Because ΔXP1Y\Delta XP_{1}Y and XP2Y\triangle XP_{2}Y have the same circumcircle. & A1B1=A2B2A_{1}B_{1} = A_{2}B_{2}]

From eqneq^{n} [1][1] and [2][2] we get that :: Radius A1P1B1A_{1}P_{1}B_{1} = Radius A2P2B2.A_{2}P_{2}B_{2}.

K.I.P.K.I.GK.I.P.K.I.G

The almighty knows it all. - 2 years, 7 months ago

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

The almighty knows it all. - 2 years, 7 months ago

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Can anyone give complete solution of 2 and 6

Azimuddin Sheikh - 2 years, 4 months ago

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@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.

Kaustubh Miglani - 2 years, 4 months ago

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@Kaustubh Miglani 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'

Aditya Khurmi - 2 years, 3 months ago

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@Azimuddin Sheikh Idk latex so cant post soln to 2

Kaustubh Miglani - 2 years, 4 months ago

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Pls do provide the solution of the last problem

Pranav Pundeer - 2 years, 3 months ago

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Diagram to problem 33

The almighty knows it all. - 2 years, 7 months ago

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

Yuvraj Singh - 2 years, 5 months ago

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

The almighty knows it all. - 2 years, 5 months ago

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@The almighty Knows It All. 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?

Govind 771 - 2 years, 5 months ago

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@Govind 771 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/.

The almighty knows it all. - 2 years, 5 months ago

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@The almighty Knows It All. so by standard textbooks, do they mean ncert?

Govind 771 - 2 years, 4 months ago

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@Govind 771 What? No.......

The almighty knows it all. - 2 years, 4 months ago

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@Govind 771 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!

Aditya Khurmi - 2 years, 2 months ago

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Solution to 33

Const:-- Join BPBP , BQBQ , CPCP & CQ.CQ.

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

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

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

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

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

Therefore, PCQ+PAQ\angle PCQ + PAQ = 180[BAQ+BAP]+[BAP+BAQ]180^{\circ}-[\angle BAQ + \angle BAP] + [\angle BAP + \angle BAQ] = 180180^{\circ}

=>APCQ\quad APCQ is cyclic

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

=> PAC=BAQ\angle PAC = \angle BAQ

The almighty knows it all. - 2 years, 7 months ago

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Need a solution for q 2 and 6

Azimuddin Sheikh - 2 years, 4 months ago

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Please explain the statement above the concluding statement

Rishabh Mahajan - 1 year, 9 months ago

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