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New year with a new beginning! And a recap of 2024..and all the best for INMO 2025!

Hi everyone! Happy New Year :)  Thank you so much for 95k+ views!!! How was everyone's 2024? What are everyone's resolutions? ( Do write down in the comment section! And you can come back 1 year later to see if you made them possible!). And.. What about me?  A Better human being Well, I want to become a better human being this year compared to last year. From a very young age, my father has been saying to me, "It does not matter if you are a good mathematician, but you should be a nice human being." As a teenager, I never took the statement seriously. Well, all that mattered to me was to do good mathematically. Why should I care about other people's feelings? These were all my thoughts in high school.  So I ended up saying a few hurtful statements without realising that they were hurtful.  I never actually cared throughout my high school. You know, the world is too big, if I hurt person A, no worries, I will move on to person B and start a new friendship! As a res...
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How to prepare for INMO

Since INMO is coming up, it would be nice to write a post about it! A lot of people have been asking me for tips. To people who are visiting this site for the first time, hi! I am Sunaina Pati, an undergrad student at Chennai Mathematical Institute. I was an INMO awardee in 2021,2022,2023. I am also very grateful to be part of the India EGMO 2023 delegation. Thanks to them I got a silver medal!  I think the title of the post might be clickbait for some. What I want to convey is how I would have prepared for INMO if I were to give it again. Anyway, so here are a few tips for people! Practice, practice, practice!! I can not emphasize how important this is. This is the only way you can realise which areas ( that is combinatorics, geometry, number theory, algebra) are your strength and where you need to work on. Try the problems as much as you want, and make sure you use all the ideas you can possibly think of before looking at a hint. So rather than fixing time as a measure to dec...

Some problems in Olympiad Graph theory!

Hello there! It has been a long time since I uploaded a post here. I recently took a class at the European Girls' Mathematical Olympiad Training Camp 2024, held at CMI. Here are a few problems that I discussed! My main references were Po-Shen Loh's Graph theory Problem set (2008), Adrian tang's Graph theory problem set (2012) and Warut Suksompong's Graph Cycles and Olympiad Problems Handout and AoPS. I also referred to Evan Chen's Graph theory Otis Problem set for nice problems! Text Book Problems which are decent A connected graph G is said to be k-vertex-connected (or k-connected) if it has more than k vertices and remains connected whenever fewer than k vertices are removed. Show that every k-connected graph of order atleast 2k contains a cycle of length at least 2k. We begin with a lemma. Prove that a graph G of order n \geq 2k is k connected then every 2 disjoint set V_1 and V_2 of k distinct vertices each, there exist k...

Calkin-Wilf Tree

I gave this talk at CMI STEMS final camp 2024. Definitions Before proceeding, we must be clear about what our title means. What do you mean by Counting? What do we mean by the term counting? We are going to prove that Rational numbers are countable . That is, there is a bijection between natural numbers and rational numbers. A bijective function f:X\rightarrow Y is a one-to-one (injective) and onto (surjective) mapping of a set X to a set Y. Note that every bijection from set X to a set Y also has an inverse function from set Y to set X. But how are we going to create the bijection? We will first create a bijection between the Natural numbers and Positive rationals. Let f(1),f(2),\dots be the mapping from natural numbers from \Bbb{N}\rightarrow +\Bbb{Q}. Then, note that there is also a bijection from \Bbb{N}\rightarrow -\Bbb{Q} by simply mapping i\in \Bbb{N}\rightarrow -f(i). And to create the bijection from g:\Bbb{N} \rightarrow \Bbb{Q}, c...

Orders and Primitive roots

 Theory  We know what Fermat's little theorem states. If p is a prime number, then for any integer a, the number a^p − a is an integer multiple of p. In the notation of modular arithmetic, this is expressed as a^{p}\equiv a{\pmod {p}}.
So, essentially, for every (a,m)=1, {a}^{\phi (m)}\equiv 1 \pmod {m}. But \phi (m) isn't necessarily the smallest exponent. For example, we know 4^{12}\equiv 1\mod 13 but so is 4^6. So, we care about the "smallest" exponent d such that a^d\equiv 1\mod m given (a,m)=1.  Orders Given a prime p, the order of an integer a modulo p, p\nmid a, is the smallest positive integer d, such that a^d \equiv 1 \pmod p. This is denoted \text{ord}_p(a) = d. If p is a primes and p\nmid a, let d be order of a mod p. Then a^n\equiv 1\pmod p\implies d|n. Let n=pd+r, r\ll d. Which implies a^r\equiv 1\pmod p. But d is the smallest natural number. So r=0. So d|n. Show that n divid...

EGMO 2024 P2

  This problem is the same level as last year's P2 or a bit harder, I feel.  No hand diagram because I didn't use any diagram~ (I head solved it) Problem: Let ABC be a triangle with AC>AB , and denote its circumcircle by \Omega and incentre by I. Let its incircle meet sides BC,CA,AB at D,E,F respectively. Let X and Y be two points on minor arcs \widehat{DF} and \widehat{DE} of the incircle, respectively, such that \angle BXD = \angle DYC. Let line XY meet line BC at K. Let T be the point on \Omega such that KT is tangent to \Omega and T is on the same side of line BC as A. Prove that lines TD and AI meet on \Omega. We begin with the following claim! Points B,X,Y,C-are concyclic Because CD-is tangent to the incircle, we get that \angle CYD=\angle BXD and \angle CDY=\angle DXY. So \angle BXD+\angle DXY+YCB=180 \implies \angle BXY+\angle YCB=180.
  Also note that K-B-C is radical axis of the inci...

IMO Shortlist 2022 C1

  Today we shall try IMO Shortlist 2022 C1. A \pm 1-sequence is a sequence of 2022 numbers a_1, \ldots, a_{2022}, each equal to either +1 or -1. Determine the largest C so that, for any \pm 1-sequence, there exists an integer k and indices 1 \le t_1 < \ldots < t_k \le 2022 so that t_{i+1} - t_i \le 2 for all i, and\left| \sum_{i = 1}^{k} a_{t_i} \right| \ge C.
We claim that the answer is \boxed{506}. 506 is the upper bound. Just consider the sequence +1,-1,-1,+1,+1,-1,-1,+1\dots,-1,-1,+1,+1,-1.
Here 1, -1, -1, 1 is repeated 505 times and 1,-1 is concatted to it. Now,our sequence would be a_1,a_3,a_4,a_5,a_7,\dots which on summing would give 506. And clearly, this would give the upper bound. Now, we show that 506 is attainable by every sequence. WLOG there are at least 1011 positive numbers in the sequence. Then we choose +1 whenever we can. Let the sequence be c_1,b_1,\dots, c_n,b_n where c_i are ...

IMO Shortlist 2021 C1

 I am planning to do at least one ISL every day so that I do not lose my Olympiad touch (and also they are fun to think about!). Today, I tried the 2021 IMO shortlist C1.  (2021 ISL C1) Let S be an infinite set of positive integers, such that there exist four pairwise distinct a,b,c,d \in S with \gcd(a,b) \neq \gcd(c,d). Prove that there exist three pairwise distinct x,y,z \in S such that \gcd(x,y)=\gcd(y,z) \neq \gcd(z,x). Suppose not. Then any 3 elements x,y,z\in S will be (x,y)=(y,z)=(x,z) or (x,y)\ne (y,z)\ne (x,z). There exists an infinite set T such that \forall x,y\in T,(x,y)=d, where d is constant. Fix a random element a. Note that (x,a)|a. So (x,a)\le a.Since there are infinite elements and finite many possibilities for the gcd (atmost a). So \exists set T which is infinite such that \forall b_1,b_2\in T (a,b_1)=(a,b_2)=d.
Note that if (b_1,b_2)\ne d then we get a contradiction as we get a set satisfying the proble...

Some Geometry Problems for everyone to try!

 These problems are INMO~ish level. So trying this would be a good practice for INMO!  Let ABCD be a quadrilateral. Let M,N,P,Q be the midpoints of sides AB,BC,CD,DA. Prove that MNPQ is a parallelogram. Consider \Delta ABD and \Delta BDC .Note that NP||BD||MQ. Similarly, NM||AC||PQ. Hence the parallelogram. In \Delta ABC, \angle A be right. Let D be the foot of the altitude from A onto BC. Prove that AD^2=BD\cdot CD. Note that \Delta ADB\sim \Delta CDA. So by similarity, we have \frac{AD}{BD}=\frac{CD}{AD}.
In \Delta ABC, \angle A be right. Let D be the foot of the altitude from A onto BC. Prove that AD^2=BD\cdot CD. Let D\in CA, such that AD = AB.Note that BD||AS. So by the Thales’ Proportionality Theorem, we are done! Given \Delta ABC, construct equilateral triangles \Delta BCD,\Delta CAE,\Delta ABF outside of \Delta ABC. Prove that AD=BE=CF. This is just congruence. N...