Sunaina thinks absurd

  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 the

 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 problem stateme

 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. Note that

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

IMO 2023 P2 Well, IMO 2023 Day 1 problems are out and I thought of trying the geometry problem which was P2.  Problem: Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$. Well, here's my proof, but I would rather call this my rough work tbh. There are comments in the end! Proof Define $A'$ as the antipode of $A$. And redefine $P=A'D\cap (ABC)$. Define $L=SP\cap (PDB)$.  Claim1: $L-B-E$ collinear Proof: Note that $$\angle SCA=\angle SCB-\angle ACB=90-A/2-C.$$ So $$\angle SPA=90-A/2-C\implies \ang

"Let's wait for this exam to get over".. *Proceeds to wait for 2 whole fricking years!  I always wanted to write a book recommendation list, because I have been asked so many times! But then I was always like "Let's wait for this exam to get over" and so on. Why? You see it's pretty embarrassing to write a "How to prepare for RMO/INMO" post and then proceed to "fail" i.e not qualifying.  Okay okay, you might be thinking, "Sunaina you qualified like in 10th grade itself, you will obviously qualify in 11th and 12th grade." No. It's not that easy. Plus you are talking to a very underconfident girl. I have always underestimated myself. And I think that's the worst thing one can do itself. Am I confident about myself now? Definitely not but I am learning not to self-depreciate myself little by little. Okay, I shall write more about it in the next post describing my experience in 3 different camps and 1 program.  So, I got