Sunaina thinks absurd

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 $k$...

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...

 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...

  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...

  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 ...

 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...

 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...

Yes, I know. This post should have been posted like 2 months ago. Okay okay, sorry. But yeah, I was just waiting for everything to be over and I was lazy. ( sorry ) You know, the transitioning period from high school to college is very weird. I will join CMI( Chennai Mathematical  Institue) for bsc maths and cs degree. And I am very scared. Like very very scared. No, not about making new friends and all. I don't care about that part because I know a decent amount of CMI people already.  What I am scared of is whether I will be able to handle the coursework and get good grades T_T Anyways, here's my EGMO PDC, EGMO, IMOTC and PROMYS experience. Yes, a lot of stuff. My EGMO experience is a lot and I wrote a lot of details, IMOTC and PROMYS is just a few paras. Oh to those, who don't know me or are reading for the first time. I am Sunaina Pati. I was IND2 at EGMO 2023 which was held in Slovenia. I was also invited to the IMOTC or International Mathematical Olympiad Training Cam...