Synthetic solution to number 1: Since the triangle is isosceles, \(A,I,H\) are collinear, where \(I,H\) are the incenter and orthocenter respectively. Let \(AH\cap BC, \odot (ABC)=D,E\). It is well known that \(DE=DH\), thus \(H\) lies on the the incircle implies \(DE=DH=2ID\). Using an incenter property from triangles - incenter, we know

So why aren't we considering negative values of \(tan\) because \(tan\) ranges from -infinity to +infinity.

I think that we should also include in the proof that the triangle is an acute angled triangle. (because in acute angled triangle, all angles will be less than \(90\) degrees and since we know that \(tan\) is positive in first quadrant \((0\) to \(90)\) degrees, we can directly eliminate the negative values of \(tan\) in the question.)

So, therefore I think you are not considering the negative values of \(\boxed{tan(B)}\).

I think we have to mention in the solution that since the orhtocentre of the triangle lies inside the triangle, it's an acute angled triangle, and therefore we are taking only the positive values of \(tan(B)\).

Shourya Pandey how many questions did you solve in INMO?Which books did you study for RMO and INMO??

Solution to number 6: The sequence is arithemtic so we can let \(a_n=a_1+(n-1)d\). Note that \(m=a_i\) for some \(i\) \(\iff \) \(\displaystyle \frac {m-a_1}{d}\in \mathbb{Z}\). Since \(a_{p+1}=a_1+pd, a_{q+1}=a_1+qd\), the following is true:
\[\frac {a_1^2-a_1}{d}\in \mathbb{Z}\\ \frac {a_1^2-a_1}{d}+2a_1p+p^2d\in \mathbb{Z}\\ \frac {a_1^2-a_1}{d}+2a_1p+p^2d\in \mathbb{Z}\]

It is not difficult to show that \(a_1,d\in \mathbb{Q}\), thus we can assume \(\displaystyle a_1=\frac {x_1}{y_1}, d=\frac {x_2}{y_2}\) with \(x_1,y_1,x_2,y_2\in \mathbb{Z}\) and \((x_1,y_1)=1, (x_2,y_2)=1\). It suffices to show that \(y_1=y_2=\pm 1\).

Plugging these into the first expression gives \(\frac {(x_1^2-x_1y_1)y_2}{y_1^2x_2}\in \mathbb{Z}\iff y_1^2x_2|(x_1^2-x_1y_1)y_2\). We can simplify this further by considering the relatively prime condition. Note that \((y_1,x_1)=1\implies (y_1,x_1-y_1)=1\), thus \((y_1^2, x_1^2-x_1y_1)=1\). Hence \(\displaystyle y_1^2|y_2 (\star)\).

Applying the same substitution for the second equation gives \(y_1y_2|2px_1y_2+p^2x_2y_1\implies y_2|p^2x_2y_1\implies\) \(\displaystyle y_2|p^2y_1\). From \((\star\) we can let \(y_2=y_1^2k\) for some integer \(k\), thus \(y_2|p^2y_1\implies y_1^2k|p^2y_1\implies y_1k|p^2\). Similarly, the analogous operation on the third equation gives \(y_1k|q^2\). This means \(y_1k\) is a common divisor of \(p^2,q^2\). Since \((p^2,q^2)=1\), \(|y_1k|=1\implies |y_1|=1, |y_2|=1\), which is what we want to prove.

I am Anay Karnik from Mumbai, just received my scorecard, got 51.
Let's make a list to determine the cut-off.

Anay Karnik (Mumbai) - 51

By the way, the question paper uploaded is actually mine (I know from the wrinkles on the bottom of the paper) I was the first one to reach home, and scan my paper and upload it on AOPS. The person who uploaded it here has copied mine from there!

@Raushan Sharma
–
@Shrihari B, @Raushan Sharma: The cutoff really doesn't matter much to me... but even if the paper is easy, the probability that the cutoff will go above 60 is really low... well anyway I hope you guys get selected whatever be the cutoff... :-D

Yes I copied it from Aops. Don't remember who had posted it but I guess it must be you :P I had earlier mentioned in the note that I copied it from Aops as I didn't write INMO. Probably a moderator edited it. Anyway thanks for uploading the paper, had great fun solving or at least trying to solve the problems :)

(i)By the well ordering principle the set of positive integral values of \(T^k(n)\) must have a least value(positive integral) say m. Let's assume that \(m \neq 1\).

If m is even:\[T(m)=\dfrac{m}{2}<m\]

If m is odd:\[T(T(m))=\dfrac{m+1}{2}<m\]

Both cases lead to a contradiction. Hence our assumption is wrong.

A short proof to problem 4 is to consider 2017 very close points, say, the angle made by the minor arc formed by the farthest two points is less than \(\frac {2\pi}{n} \) . Complete the 2017 regular n-gons formed by these points. Clearly(one shouod explain this in the examination) no two n-gons have any points in common. Therefore, by the pigeonhole principle, at least one of these n-gons has no red point as its vertex.

Hey, Shourya, tell me what's wrong in this proof. Let's label 2 adjacent red points to be \(A_1\) and \(A_2\). Now we divide the arc between the 2 point into an arbitrarily large number of equal parts, say, \(10^9\), and we mark points labelled as \(R_1\, R_2\) and so on to denote the boundary of the sectors. Now we draw a regular n-gon with \(A_1\) as one of the vertex. It may or may not contain any other red point. Now we rotate the n-gon so that the vertex shifts to \(R_1\). This n-gon also may have a red point on it. If it doesn't then this is our required polygon. If it does then we rotate it again to \(R_2\) and so on. We have only 2016 red points hence a maximum number of 2015 of the regular n-gons out of the \(10^9\) polygons so created will have red points as one of the vertices. Hence the others are our required polygons. Hence proved.

I am not sure about it. But I feel that you should have proved that the the n-gons are regular. See the arrangement of the points is not regular in the case you have mentioned. So getting a regular n-gon will be a very special case. You need to prove that the special case exists.

@Aditya Kumar
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Huh????? what are you talking about... you want an n-gon with blue vertices which are infinite in number in contrast to the red ones which are only 2016 in number. A regular n-gon with blue vertices is not a special case..

@Abhishek Bakshi
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The question is concerned about obtaining a regular n-gon. For that you have to prove that the sides are equal and all angles are equal. Now according to your explanation, the \(R_{n}\) distribution is irregular. Hence obtaining a regular n-gon becomes a special case.

@Aditya Kumar
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I think you have misunderstood my solution..I am not saying that a regular n-gon will be formed using the points of \(R_n\). I am saying that just take one point of \(R_n\)and draw a regular n-gon with it as one of the vertices. Such a polygon always exists and we have to just prove that out of the \(10^9\) polygons such formed, the 2016 red points don't lie on all of them which is quite obvious. I believe that you understand the sarcasm too...haha..

A lengthy proof to problem 4:
Let the circle given in the question be the unit circle centred at the origin, WLOG.
Each point has a unique polar coordinate \( (1, \theta) \), where \( 0 \le \theta <2\pi \). We will express all points (on this circle) only by its polar angle.
Let \(\theta_{i} , i= 1,2,3,...,2016 \) denote the 2016 red points on the circle. For convinience, denote \( \frac {\theta}{\pi} \) as \(b_i\)

LEMMA 1 :- There exists a number \(\alpha \), such that \( \alpha - b_i\) is irrational, for all \(i\).
(Try proving this. I'll post the proof later).

Now, we choose the points of the regular n-gon to be \( \alpha\pi , \alpha\pi + \frac {2\pi}{n} , ..., \alpha\pi + \frac {2(n-1)\pi}{n} \). Here \(\alpha \) may not lie in \([0, 2\pi) \), but it does not matter.

Suppose some vertex of this n-gon coincides with some red point, say \( \alpha\pi + \frac {2l\pi}{n} = \theta_i+ 2k\pi\), where \(l\) is one of\(0,1,2,...,n-1\) and \(k\) is an integer. Then rearranging it gives
\(\alpha - b_i= 2k - \frac {2l}{n} \), a contradiction to the lemma.

Thanks guys! A lot of people have been asking me the same question lately, thus I have written an article on it on my site w/ my friends: http://mathometer.weebly.com/preparing-for-inmo.html
The site is a mini success, garnering 12k views in the 3 weeks of its existence :)

@Rohit Kumar
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No. They do award full marks. If you are able to prove something crucial using coordinate geometry and you mention the geometrical interpretation of the coordinate bash you have done, then marks are awarded according to how crucial it is in the pure geometric solution. However, if you are just blindly doing algebra and reach some weird coordinates ( even though you are really close), you get a zero. Also, you lose all marks if you say " clearly .... simplifies to ... ", where it isn't very clear how the previous step would mean the next.

This is why it is often recommended to use coordinate only as a tool to prove small steps in the problems. Of course, it is different if you're very good at bashing.

Hey guys I would Like if everyone posted their expected marks here. That would help us to get a rough idea of the cutoff. Starting with me ... I am expecting 34 :( After u post your marks I will edit it in here and then delete your comments to avoid a lot of junk.

(I)Let \(a=b=1\). Therefore \(c^3-2c+1=0\). Solving we get a solution as \(c=\dfrac{-1+\sqrt{5}}{2}\). Hence it is not necessary for \(a=b=c\)

(II)We may assume \(a=max(a,b,c)\) since \(a(a^3+b^3)=b(b^3+c^3)\) and \(a \ge b\), \(a^3+b^3 \le b^3+c^3\) or \(c \ge a\). This forces \(a=c\). Similarly, \(a(a^3+b^3)=c(c^3+a^3)\) or \(b=c\).

@Brilliant Member
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I solved 1,2,3 please see no. 3 which i posted above.I assumed a>=b>=c but not a>=c>=b in the second one and justified the first case of no. 2.how much should I get in no. 2.

Yes, absolutely correct. Just I assumed a=b=2, and then c=\(\sqrt{5} - 1\) for the 1st one, and 2nd one same assumption WLOG, and then some manipulation.

I wanna ask is this correct for no. 3
T(2)=1
T(3)=4,=>T^(3)=1
T(4)=2,=>T^2(4)=1
1)We prove by induction.Assume for some 4 \leq m \geq 2n and some k,we can get T^k(m)=1.
Case 1) for m=2n+1
T(m)=2n+2;=>T^2(m)=n+1 < 2n
So we can get for some k, T^k(2n+1)=1
Similarly for case 2) .So this proves our assumption.
2) T^{k+2} (n)=1;for c{k+2} values.
Case 1)n=2m
T^{k+2} (2m)=T^{k+1} (m)=1;for c{k+1} values.
Case 2)n=2m-1
T^{k+2} (2m-1)=T^{k+1} (2m)=T^k (m)=1;for c{k} values.
This shows c{k+2}=c{k+1}+c{k}.

Isn't this one very simple? There are infinitely many points which are blue. There are also infinitely many points which are blue between two red points. There are only 2016 points, so definitely there are infinitely many regular n-gons which means that there are infinitely many regular n-gons with blue vertices.

Yes, this question I couldn't understand what it meant, or what I needed to show, coz there are infinitely many points on the circle, so... what does this mean!!

@Akash Deep
–
I think the papers were comparable. Although the paper this year was slightly difficult on an average, but the 5th question this year wasn't impossible in the time-limit ( unlike last year's).

@Samarth Agarwal
–
Construct a regular polygon with one of its vertices fixed at a red point.
Now measure the positive anticlockwise angles between all vertice and all red point.
Rotate the polygon by half of the smallest of these angles.
Now every vertex initially on a blue point must land on a blue point, as the rotation was by half of the smallest angle.
Every vertex on a red point must land on a blue point, as the polygon was rotated.

If a vertex is already on a red point, measure the angle from adjacent red points.

@Shourya Pandey
–
i did 1,
2 (in 2nd i just showed for the first case that all 3 of them cannot be distinct , means any 2 are equal , after taking this i showed that if a=b=x and x and c satisfy x^3 +c^3 - 2c^2x = 0 then there is no need for x = c)whereas the 2nd part of 2nd question i did correctly.
in 3rd i did first part correctly and if s(k) denotes the set containing elements for which t^k(n) in ,2nd part i showed that s(k) is a subset of s(k+2) and showed that from every element of s(k+1) we can generate a corresponding element of s(k+2), then i showed that if there is any element in s(k+2) it is either of s(k) or generated from s(k+1)
4- did not attempt
5- wrote that R*S stuff and left it.
6. - just wrote that for an a.p to have all elements integal , c.d and first term must be integral with a useless proof of this.
(please let me know if i have any chances and how much marks can i get)

@Harsh Shrivastava
–
I thought u are in class 11....but if u have done 2 then u may get inmo merit certificate and direct eligibility for inmo next year

@Harsh Shrivastava
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Do u know the solution top fifth one? ......using rs=area I reached somewhere and I needed only to prove that bd^2=area of abc.....not getting after that

@Brilliant Member
–
I also got the first part fully, but in the second part I tried many things, but didn't land up with the result. What do u think is the partition of marks in Q.3??

@Raushan Sharma
–
Since i am using brilliant on phone its difficult to use latex once my pc is repaired i would write a sol. To 3rd one ....its very interesting

@Brilliant Member
–
I would soon write a complete sol. Of ques. 3.... the second part involves cases taking a no. In the set to be once even and then odd... (I would post complete solution soon)

@Samarth Agarwal
–
Yes, I also got to the same to prove BD^2 = Area, then I was trying with Stewart's theorem, and also dropped perpendiculars from A and C on BD, and tried to find the value of BD^2 in two different ways, and equate, but the time was up, and it was incomplete LOL. I think this problem was quite manipulative.

@Samarth Agarwal
–
No, no, actually one of my fellow mates Vishal Raj solved it fully using Stewart's theorem only, and his strategy was also the same as writing BD^2 in two different ways, and then manipulating. So, we can expect 10 marks I hope

@Deekshith Kanagala
–
@deekshith kanagala: how much are you expecting... I think this year's cutoff will be higher than that of last year... it should be around 55.. those who get selected would have solved the 1st, 2nd and 4th problems fully and 1st part of the 3rd problem...

@Deekshith Kanagala
–
@deekshith kanagala: nice score... you are quite accurate as you got the marks for 2 questions.... but i don't think the cutoff will go so low... btw I am in 12th standard and i gave inmo last 2 years... unfortunately i missed the cutoff last year by 6 marks, so i have a little knowledge about what the cutoff could be...

## Comments

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TopNewestSynthetic solution to number 1: Since the triangle is isosceles, \(A,I,H\) are collinear, where \(I,H\) are the incenter and orthocenter respectively. Let \(AH\cap BC, \odot (ABC)=D,E\). It is well known that \(DE=DH\), thus \(H\) lies on the the incircle implies \(DE=DH=2ID\). Using an incenter property from triangles - incenter, we know\[\frac {AI}{ID}=\frac {EI}{DE}=\frac {ID+DE}{DE}=\frac {3}{2}\].

Hence by the angle bisector theorem, \[\frac {AB}{BC}=\frac {AB}{2BD}=\frac {AI}{2ID}=\boxed {\frac {3}{4}}\]

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This is the shortest and easiest way to solve the question and get 17/17......+1 :)

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Answer to question number 1:Let the foot of the perpendicular from \(A\) to \(BC\) be \(M\).

Let \(H,I\) be the orthocentre and incentre respectively.

(I)Since in an isosceles triangle orthocentre and incentre are collinear .

Proof of

statement I:\(\triangle(ABM)\) is congruent to \(\triangle(ACM)\) (By \(RHS\) test ) \(AB=AC,AM=AM,\angle(AMB)=\angle(AMC)\)

Therefore \(M\) is a midpoint of \(BC\). Let \(J\) be the intersection of incircle with \(BC\).

Therefore \(\angle(IJC)=90°\). \(2BJ=AB+BC-AC,2CJ=BC+AC-AB\)

But, \(AB=AC\)

Therefore \(BJ=CJ,BJ+CJ=BC\) .

This implies \(J\) is the midpoint of \(BC\) , but we proved that \(M\) is the midpoint of \(BC\) .

Therefore \(J,M\) must coincide.

This proves that \(I\) lies on \(AM\).

Therefore \(A,M,H,I\) are colinear.

Since \(AB=AC\) ,\(\angle(B)=\angle(C)\) Now \(\boxed{tan(\frac{B}{2})=\frac{IM}{BM}}.......(1)\) ) \(\angle(HBC=(90-C)=(90-B)\)

Therefore \(tan(90-B)=\frac{HM}{BM}\)

But \(2IM=HM\) since \(H,I,M\) are collinear and also lie on a circle with \(I\) as the centre .

hence \(\boxed{Cot(B)=\frac{2IM}{BM}}.......(2)\)

Dividing \((1)\) by \((2)\) We get \(\frac{1}{2}=tan(B)tan(\frac{B}{2})\)

After solving this we get, \(tan(B)=\frac{\sqrt{5}}{2}\)

But \((90-\frac{A}{2})=B\) Therefore \(tan(\frac{A}{2})=\frac{2}{\sqrt{5}}\)

This implies \(sin(\frac{A}{2})=\frac{2}{3}......(3)\)

In triangle \(ABM\) \(sin(A/2)=BM/AB=BC/2AB....(4)\)

Therefore using \((3),(4)\) we get

\[\frac{AB}{BC}=\frac{3}{4}\]

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Shivam Jadhav I have a doubt.After solving \(tan(B)tan(\frac{B}{2})=\frac{1}{2}\), I'm getting the following values:-

\(tan(\frac{B}{2})=\frac{1}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\)

\(tan(B)=\frac{\sqrt{5}}{2}, -\frac{\sqrt{5}}{2}\)

So why aren't we considering negative values of \(tan\) because \(tan\) ranges from

-infinity to +infinity.I think that we should also include in the proof that the triangle is an acute angled triangle. (because in acute angled triangle, all angles will be less than \(90\) degrees and since we know that \(tan\) is positive in first quadrant \((0\) to \(90)\) degrees, we can directly eliminate the negative values of \(tan\) in the question.)So, therefore I think you are not considering the negative values of \(\boxed{tan(B)}\).Log in to reply

Ratio of two sides can't be negative .

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And we know that value of \(tan\) is negative in the second quadrant (between \(90\) to \(180\) degrees).

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Indeed. It should be mentioned in the solution, at least.

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I think we have to mention in the solution that since the orhtocentre of the triangle lies inside the triangle, it's an acute angled triangle, and therefore we are taking only the positive values of \(tan(B)\).Shourya Pandey how many questions did you solve in INMO?Which books did you study for RMO and INMO??Log in to reply

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5TH QUESTION

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5TH QUESTION

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5TH QUESTION

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Solution to number 6: The sequence is arithemtic so we can let \(a_n=a_1+(n-1)d\). Note that \(m=a_i\) for some \(i\) \(\iff \) \(\displaystyle \frac {m-a_1}{d}\in \mathbb{Z}\). Since \(a_{p+1}=a_1+pd, a_{q+1}=a_1+qd\), the following is true: \[\frac {a_1^2-a_1}{d}\in \mathbb{Z}\\ \frac {a_1^2-a_1}{d}+2a_1p+p^2d\in \mathbb{Z}\\ \frac {a_1^2-a_1}{d}+2a_1p+p^2d\in \mathbb{Z}\]It is not difficult to show that \(a_1,d\in \mathbb{Q}\), thus we can assume \(\displaystyle a_1=\frac {x_1}{y_1}, d=\frac {x_2}{y_2}\) with \(x_1,y_1,x_2,y_2\in \mathbb{Z}\) and \((x_1,y_1)=1, (x_2,y_2)=1\). It suffices to show that \(y_1=y_2=\pm 1\).

Plugging these into the first expression gives \(\frac {(x_1^2-x_1y_1)y_2}{y_1^2x_2}\in \mathbb{Z}\iff y_1^2x_2|(x_1^2-x_1y_1)y_2\). We can simplify this further by considering the relatively prime condition. Note that \((y_1,x_1)=1\implies (y_1,x_1-y_1)=1\), thus \((y_1^2, x_1^2-x_1y_1)=1\). Hence \(\displaystyle y_1^2|y_2 (\star)\).

Applying the same substitution for the second equation gives \(y_1y_2|2px_1y_2+p^2x_2y_1\implies y_2|p^2x_2y_1\implies\) \(\displaystyle y_2|p^2y_1\). From \((\star\) we can let \(y_2=y_1^2k\) for some integer \(k\), thus \(y_2|p^2y_1\implies y_1^2k|p^2y_1\implies y_1k|p^2\). Similarly, the analogous operation on the third equation gives \(y_1k|q^2\). This means \(y_1k\) is a common divisor of \(p^2,q^2\). Since \((p^2,q^2)=1\), \(|y_1k|=1\implies |y_1|=1, |y_2|=1\), which is what we want to prove.

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I could reach to the equation (*) but then beated around the Bush after that how much will i get on this question then?

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You can get 3/4 of the total marks of the question for your solution. If a question is of 16.6 marks then i think you should get around 12 marks.

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Which books did you for both RMO and INMO?Log in to reply

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I could only prove a and d are rational. How much should I expect ?( I did mention that those 3 expressions were integers.)

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I did the same. I really don't think there is any other solution to it. Nice write up, by the way.

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Nice solution. Sequence problems are very fond for the problem posers as it requires many topics.

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I am Anay Karnik from Mumbai, just received my scorecard, got 51. Let's make a list to determine the cut-off.

Anay Karnik (Mumbai) - 51

By the way, the question paper uploaded is actually mine (I know from the wrinkles on the bottom of the paper) I was the first one to reach home, and scan my paper and upload it on AOPS. The person who uploaded it here has copied mine from there!

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How many questions have u attempted?

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4, no chance for me, cutoff will be 65 or above.

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@Anay Karnik: 65 is too high.... it would be around 55..

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@Shrihari B, @Raushan Sharma: The cutoff really doesn't matter much to me... but even if the paper is easy, the probability that the cutoff will go above 60 is really low... well anyway I hope you guys get selected whatever be the cutoff... :-D

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Yes I copied it from Aops. Don't remember who had posted it but I guess it must be you :P I had earlier mentioned in the note that I copied it from Aops as I didn't write INMO. Probably a moderator edited it. Anyway thanks for uploading the paper, had great fun solving or at least trying to solve the problems :)

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no problem

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3.

(i)By the well ordering principle the set of positive integral values of \(T^k(n)\) must have a least value(positive integral) say m. Let's assume that \(m \neq 1\).

If m is even:\[T(m)=\dfrac{m}{2}<m\]

If m is odd:\[T(T(m))=\dfrac{m+1}{2}<m\]

Both cases lead to a contradiction. Hence our assumption is wrong.

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i too did exactly the same thing

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A short proof to problem 4 is to consider 2017 very close points, say, the angle made by the minor arc formed by the farthest two points is less than \(\frac {2\pi}{n} \) . Complete the 2017 regular n-gons formed by these points. Clearly(one shouod explain this in the examination) no two n-gons have any points in common. Therefore, by the pigeonhole principle, at least one of these n-gons has no red point as its vertex.

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Hey, Shourya, tell me what's wrong in this proof. Let's label 2 adjacent red points to be \(A_1\) and \(A_2\). Now we divide the arc between the 2 point into an arbitrarily large number of equal parts, say, \(10^9\), and we mark points labelled as \(R_1\, R_2\) and so on to denote the boundary of the sectors. Now we draw a regular n-gon with \(A_1\) as one of the vertex. It may or may not contain any other red point. Now we rotate the n-gon so that the vertex shifts to \(R_1\). This n-gon also may have a red point on it. If it doesn't then this is our required polygon. If it does then we rotate it again to \(R_2\) and so on. We have only 2016 red points hence a maximum number of 2015 of the regular n-gons out of the \(10^9\) polygons so created will have red points as one of the vertices. Hence the others are our required polygons. Hence proved.

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That's pretty much same to what Shourya did. I think it's correct.

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I am not sure about it. But I feel that you should have proved that the the n-gons are regular. See the arrangement of the points is not regular in the case you have mentioned. So getting a regular n-gon will be a very special case. You need to prove that the special case exists.

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regular n-gon. For that you have to prove that the sides are equal and all angles are equal. Now according to your explanation, the \(R_{n}\) distribution is irregular. Hence obtaining a regular n-gon becomes a special case.Log in to reply

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this.

Ooh sorry I actually misunderstood the solution. Meanwhile have a look atLog in to reply

@Aditya Kumar: what do u think about the above solution??

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Brilliant ! 17/17.

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You too have a good solution to the problem.

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A lengthy proof to problem 4: Let the circle given in the question be the unit circle centred at the origin, WLOG. Each point has a unique polar coordinate \( (1, \theta) \), where \( 0 \le \theta <2\pi \). We will express all points (on this circle) only by its polar angle. Let \(\theta_{i} , i= 1,2,3,...,2016 \) denote the 2016 red points on the circle. For convinience, denote \( \frac {\theta}{\pi} \) as \(b_i\)

LEMMA 1 :- There exists a number \(\alpha \), such that \( \alpha - b_i\) is irrational, for all \(i\). (Try proving this. I'll post the proof later).

Now, we choose the points of the regular n-gon to be \( \alpha\pi , \alpha\pi + \frac {2\pi}{n} , ..., \alpha\pi + \frac {2(n-1)\pi}{n} \). Here \(\alpha \) may not lie in \([0, 2\pi) \), but it does not matter.

Suppose some vertex of this n-gon coincides with some red point, say \( \alpha\pi + \frac {2l\pi}{n} = \theta_i+ 2k\pi\), where \(l\) is one of\(0,1,2,...,n-1\) and \(k\) is an integer. Then rearranging it gives \(\alpha - b_i= 2k - \frac {2l}{n} \), a contradiction to the lemma.

This solves the problem.

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Anyone who cleared INMO?

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I did (got 73). You should join the discussion on AoPS, my username is TheOneYouWant there... looks like someone below was impersonating me :P

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Thanks guys! A lot of people have been asking me the same question lately, thus I have written an article on it on my site w/ my friends: http://mathometer.weebly.com/preparing-for-inmo.html The site is a mini success, garnering 12k views in the 3 weeks of its existence :)

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Congrats! Shubham on getting selected for INMO. Which standard you are currently in?? Can you tell/guide me on how did you prepare for INMO please??Log in to reply

Pls can you advise me on how to prepare effectively for INMO. I am not in AOPS.

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Official solutions are out! Click here

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I used simple co-ordinate geometry to solve question 1. Will that be considered?

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Yes. If your solution is entirely correct, full marks will be awarded.

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Do they award full marks ?(I heard from someone that they don't award you full marks even if your solution is correct )

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This is why it is often recommended to use coordinate only as a tool to prove small steps in the problems. Of course, it is different if you're very good at bashing.

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Hey guys I would Like if everyone posted their expected marks here. That would help us to get a rough idea of the cutoff. Starting with me ... I am expecting 34 :( After u post your marks I will edit it in here and then delete your comments to avoid a lot of junk.

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A friend of mine got five and a half questions correct.

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I am also expecting about 35 to 40.

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2.

(I)Let \(a=b=1\). Therefore \(c^3-2c+1=0\). Solving we get a solution as \(c=\dfrac{-1+\sqrt{5}}{2}\). Hence it is not necessary for \(a=b=c\)

(II)We may assume \(a=max(a,b,c)\) since \(a(a^3+b^3)=b(b^3+c^3)\) and \(a \ge b\), \(a^3+b^3 \le b^3+c^3\) or \(c \ge a\). This forces \(a=c\). Similarly, \(a(a^3+b^3)=c(c^3+a^3)\) or \(b=c\).

Hence it is necessary for \(a=b=c\).

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The second part is NOT SYMMETRIC, but only CYCLIC. Therefore \(a \ge b \ge c\) and \(a \ge c \ge b\) are two different cases.

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Yes you are right. Is my solution correct now?

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Hey Shourya I didn't take the second case i.e. a>=c>=b. How many marks will be deducted ???

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I assumed symmetric and the first inequality you wrote .how much should I get?

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Am I correct?

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Yes, absolutely correct. Just I assumed a=b=2, and then c=\(\sqrt{5} - 1\) for the 1st one, and 2nd one same assumption WLOG, and then some manipulation.

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You know anyone from your region and how much he/she did?

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Sorry I don't

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Ooohk

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what are your marks

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what is the cuttoff inmo 2016

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Got the scorecard....could only make 35.... :(

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I wanna ask is this correct for no. 3 T(2)=1 T(3)=4,=>T^(3)=1 T(4)=2,=>T^2(4)=1 1)We prove by induction.Assume for some 4 \leq m \geq 2n and some k,we can get T^k(m)=1. Case 1) for m=2n+1 T(m)=2n+2;=>T^2(m)=n+1 < 2n So we can get for some k, T^k(2n+1)=1 Similarly for case 2) .So this proves our assumption. 2) T^{k+2} (n)=1;for c

{k+2} values. Case 1)n=2m T^{k+2} (2m)=T^{k+1} (m)=1;for c{k+1} values. Case 2)n=2m-1 T^{k+2} (2m-1)=T^{k+1} (2m)=T^k (m)=1;for c{k} values. This shows c{k+2}=c{k+1}+c{k}.Log in to reply

4.

Isn't this one very simple? There are infinitely many points which are blue. There are also infinitely many points which are blue between two red points. There are only 2016 points, so definitely there are infinitely many regular n-gons which means that there are infinitely many regular n-gons with blue vertices.

Am I missing something obvious?

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Yes, this question I couldn't understand what it meant, or what I needed to show, coz there are infinitely many points on the circle, so... what does this mean!!

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Sort of. Just put it on paper vividly.

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what do u think is this years paper easier or last years

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If a vertex is already on a red point, measure the angle from adjacent red points.

Hence proved.

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Can anybody please explain me question no. \(3\) properly?Rohit you belong to which region? Which books did you study for RMO and INMO??Log in to reply

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Does anyone knows what will be the cutoff this year

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It will be three and half questions surely.

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I feel the same.

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How many did you solved?

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Only 2... :( btw how many did u solved?

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btw which two?

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I tried to use coordinate bash in 5th but calculations were tooo lengthy so left it.

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r'(AB+BD+AD)/2+r'(BD+CD+BC)/2=r(AB+BC+AC)/2

On solving (1/r')=(1/r)+(BD/area of ABC)

I was struck after that now we only need to prove BD^2=ac/2 but how?.....

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Comment deleted Jan 18, 2016

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Try zooming in.

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I am able to see it clearly. If you are still not able to see the problems, I will type the problems here.

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Yes now it is visible. Thanks.

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@deekshith kanagala: how much are you expecting... I think this year's cutoff will be higher than that of last year... it should be around 55.. those who get selected would have solved the 1st, 2nd and 4th problems fully and 1st part of the 3rd problem...

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@deekshith kanagala: nice score... you are quite accurate as you got the marks for 2 questions.... but i don't think the cutoff will go so low... btw I am in 12th standard and i gave inmo last 2 years... unfortunately i missed the cutoff last year by 6 marks, so i have a little knowledge about what the cutoff could be...

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I am getting 65

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Hi. It is great to see this. Which region are you from, and what is your name?

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You got your scorecard??

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