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.

@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

@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.

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.

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 :)

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.

@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
–
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'

@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/.

@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!

Clearly, Using Alternate Segment Theorem , we get:\( \color{red}{\angle BAQ = \angle BQP}\) and \( \color{Blue}{\angle BAP = \angle BPQ}\).--- \([1]\)

Also Using the isosceles triangle property, we get :: \( \color{red}{\angle PQC = \angle BQP}\) & \( \color{blue}{\angle QPC = \angle BPQ}\)-----\([2]\)

Thus from \(eq^{n}\) \([1]\) and \([2]\) we get: \( \color{red}{\angle BAQ = \angle PQC}\) & \( \color{blue}{\angle BAP = \angle QPC}\).

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

Easy Math Editor

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boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

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## Comments

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TopNewestProblem 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.

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nice solution

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Hey can u explain in detail why A1B1=AB

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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

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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.

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

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The question mentions that they

alldon'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 :)Log in to reply

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

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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 \(ab+bc+ca \leq 3\)

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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 \(A_{2}P_{2}B_{2}.\)

\(K.I.P.K.I.G\)

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

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Can anyone give complete solution of 2 and 6

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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.

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@Azimuddin Sheikh Idk latex so cant post soln to 2

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Pls do provide the solution of the last problem

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Diagram to problem \(3\)

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

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

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Solution to \(3\)

Const:-- Join \(BP\) , \(BQ\) , \(CP\) & \(CQ.\)

Clearly, Using Alternate Segment Theorem , we get:\( \color{red}{\angle BAQ = \angle BQP}\) and \( \color{Blue}{\angle BAP = \angle BPQ}\).--- \([1]\)

Also Using the isosceles triangle property, we get :: \( \color{red}{\angle PQC = \angle BQP}\) & \( \color{blue}{\angle QPC = \angle BPQ}\)-----\([2]\)

Thus from \(eq^{n}\) \([1]\) and \([2]\) we get: \( \color{red}{\angle BAQ = \angle PQC}\) & \( \color{blue}{\angle BAP = \angle QPC}\).

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

=> \(\angle PAC= \color{red}{\angle PQC = \angle BAQ}\)

=> \(\angle PAC = \angle BAQ\)

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Need a solution for q 2 and 6

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Please explain the statement above the concluding statement

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