Skip to main content

Summary of 2020

New year post!! Happy New year everyone!!

( Got inspired by SnowPanda's  and anser's blog post).

Okie so first non math post :P. My grammar is really bad.

I feel like 2020 by far has been the most productive year of mine and very unique too ( Just imagine not going to school , Non-dummy school students be like"yeaaahhhhh" ).

The most important thing that happened was my INMO 2020. The most embarrassing yet if this incident wouldn't happen, then I wouldn't realise what hardwork and starving is ( okie sorry too philosophical ).

Anyways here are some of the best experiences:

a. Pr0 people aren't aliens ( exceptions: RG and JIB ): So I joined OTIS this year, and this was such a nice decision. I seriously learnt a lot. 

And before joining OTIS, I thought every person is so pr0 and everybody talks just math and no random thing. And my stereotypical image of Evan was like a strict guy, who only does math and math and math all day and talks only about math   ( I saw nekkoya v2 after I joined OTIS ). but he is really friendly and helpful.

Same goes for SnowPanda and anser. I heard these names before joining OTIS. I was so scared to talk with them at first and now we talk so often. Thanks a lot!! The g-solves we did, I learnt a lot. Seriously. Love u guys.

The spamiferous discord server people!! thanks a lot. 

b. Geometry isn't scary: Yes, it's not!! It's really fun.. trust me. It's fun conjecturing which points are collinear and stuff.. One should try it!! Plus EGMO is OP. It's really nice book!!

c. Hanabi and Anti-chess:  I learnt hanabi because of OTIS. I still remember my first game. Messing up all the finesses :lol:. I bet Evan must have been annoyed. I think I have become a bit better.

And anti-chess!! This is better than chess ( because I mostly loose in chess :P ). Thanks Arpan bhaiya for introducing it to me.

d.MSE: I am using it so frequently and it's a great site! The fact that I love the most is, if I get stuck in some question, I can share my progress there and people post hints to me. Which is way better than getting stuck and directly seeing solutions. I did try to answer few questions too. In all it was great experience!!

e. G-solves FTW: The best thing is I learnt so much, although I couldn't contribute but I learnt a lot .I perhaps g-solved maybe 50 questions . Thanks Anand, ArpanBhaiya, anser ,ATG,Crystal1011,Dapper ,Geometrix, MPF,Mueller.25,NJOY, Psyduck909  , Rohan Bhaiya, Srijon,SnowPanda .

And the Gsolves with Rohan Bhaiya and Aatman dada (MPF) were insane 😂 which only these two did and I was just sitting :P. The amount Rohan bhaiya roast Aatman and vice-versa is insane.

f. Rohan Bhaiya:  I can write a whole big essay about how alien like he is. This guys solve TSTs in minutes. Seriously! But yes he is really helpful and kind . He even accepted my handwritten Psets  !!  

To Rohan Bhaiya: Sorryyyy for disturbing you so much :(

g. Anand: ufff... You ofc needed a special mention and thanks for being such a nice friend. Trust me, if one talks with him without knowing/seeing his resume , one won't even realise that he is talking with a God. He is so down to earth. He cleared so many doubts of mine in AOPS and MSE. One of the main reasons why I got a bit(?) better in geo compared to last year . 

He teaches so nicely!! I learnt Derivatives due to him and Rohan bhaiya. 

In 2020 I studied Olympiad math just for fun and not wrt any competitions. And it turned out to be such a nice thing. Enjoying math is way better than preparing . This year my most memorable moments were G-solves,  hanabi ( I , JIB and MPF) , Integirl competition ( waking up till 12 o'clock just to give the competition and it was so fun ).

Thanks to everyone!! And stay healthy this year.. 

Do write in the comments section about how 2020 was for you (at least write something ! I will be happy to hear your comments ). Follow this blog if you want to see more contest math problems and some non-math posts :P! See you all soon 😊.

Sunaina 💜




Labels:

Comments

  1. Rohan GoyalJanuary 5, 2021 at 6:14 PM

    Lol thanks for mention. Also, please stop apologizing 😂😂. It's not really annoying.

    And the "proness" hype nonsense 😂😂

    Anyway thanks!

    ReplyDelete
  2. AnandJanuary 5, 2021 at 7:26 PM

    OMG! Thanks for the butter 😂

    Well, yeah 2020 was just so fun! I learnt Hanabi this year 😎 (Thanks to Super :P)

    Also, thanks for mentioning me!

    ReplyDelete
    Replies
    1. Sunaina Pati August 22, 2021 at 3:06 PM

      Hehehhehehhehhehehehehhhhehhehhhe You are the best!

      Delete

Post a Comment

Popular posts from this blog

Geometry ( Finally!!!)

 This is just such an unfair blog.  Like if one goes through this blog, one can notice how dominated  Algebra is!! Like 6 out of 9 blog post is Algebra dominated -_- Where as I am not a fan of Algebra, compared to other genres of Olympiad Math(as of now). And this was just injustice for Synthetic Geo. So this time , go geo!!!!!!!!!!!  These problems are randomly from A Beautiful Journey through Olympiad Geometry.  Also perhaps I will post geo after March, because I am studying combi.  Problem:  Let $ABC$ be an acute triangle where $\angle BAC = 60^{\circ}$. Prove that if the Euler’s line of $\triangle ABC$ intersects $AB$ and $AC$ at $D$ and $E$, respectively, then $\triangle ADE$ is equilateral. Solution:  Since $\angle A=60^{\circ}$ , we get $AH=2R\cos A=R=AO$. So $\angle EHA=\angle DOA.$ Also it's well known that $H$ and $O $ isogonal conjugates.$\angle OAD =\angle EAH.$ By $ASA$ congruence, we get $AE=AD.$ Hence $\triangle ADE$ is equilateral....
3 comments
Read more

Problems I did this week [Jan8-Jan14]

Yeyy!! I am being so consistent with my posts~~ Here are a few problems I did the past week and yeah INMO going to happen soon :) All the best to everyone who is writing!  I wont be trying any new problems and will simply revise stuffs :) Some problems here are hard. Try them yourself and yeah~~Solutions (with sources) are given at the end! Problems discussed in the blog post Problem1: Let $ABC$ be a triangle whose incircle $\omega$ touches sides $BC, CA, AB$ at $D,E,F$ respectively. Let $H$ be the orthocenter of $DEF$ and let altitude $DH$ intersect $\omega$ again at $P$ and $EF$ intersect $BC$ at $L$. Let the circumcircle of $BPC$ intersect $\omega$ again at $X$. Prove that points $L,D,H,X$ are concyclic. Problem 2: Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ= 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ ...
Post a Comment
Read more

Just spam combo problems cause why not

This post is mainly for Rohan Bhaiya. He gave me/EGMO contestants a lot and lots of problems. Here are solutions to a very few of them.  To Rohan Bhaiya: I just wrote the sketch/proofs here cause why not :P. I did a few more extra problems so yeah.  I sort of sorted the problems into different sub-areas, but it's just better to try all of them! I did try some more combo problems outside this but I tried them in my tablet and worked there itself. So latexing was tough. Algorithms  "Just find the algorithm" they said and they died.  References:  Algorithms Pset by Abhay Bestrapalli Algorithms by Cody Johnson Problem1: Suppose the positive integer $n$ is odd. First Al writes the numbers $1, 2,\dots, 2n$ on the blackboard. Then he picks any two numbers $a, b$ erases them, and writes, instead, $|a - b|$. Prove that an odd number will remain at the end.  Proof: Well, we go $\mod 2$. Note that $$|a-b|\equiv a+b\mod 2\implies \text{ the final number is }1+2+\dots ...
2 comments
Read more

IMO 2023 P2

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...
1 comment
Read more

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...
Post a Comment
Read more

My experiences at EGMO, IMOTC and PROMYS experience

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...
11 comments
Read more

Challenging myself? [Jan 15-Jan 27]

Ehh INMO was trash. I think I will get 17/0/0/0-1/3-5/10-14, which is def not good enough for qualifying from 12th grade. Well, I really feel sad but let's not talk about it and focus on EGMO rather.  INMO 2023 P1 Let $S$ be a finite set of positive integers. Assume that there are precisely 2023 ordered pairs $(x,y)$ in $S\times S$ so that the product $xy$ is a perfect square. Prove that one can find at least four distinct elements in $S$ so that none of their pairwise products is a perfect square. I will use Atul's sol, cause it's the exact same as mine.  Proof: Consider the graph $G$ induced by the elements of $S$ and edges being if the products are perfect squares. Note that if $xy = a^2$ and $xz = b^2$, then $yz = \left( \frac{ab}{x} \right)^2$, since its an integer and square of a rational number its a perfect square and so $yz$ is an edge too. So the graph is a bunch of disjoint cliques, say with sizes $c_1, c_2, \cdots, c_k$. Then $\sum_{i=1}^k c_i^2 = 2023$, which ...
Post a Comment
Read more

Solving Random ISLs And Sharygin Solutions! And INMO happened!!

Some of the ISLs I did before INMO :P  [2005 G3]:  Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$ Solution: Note that $$\Delta LDK \sim \Delta XBK$$ and $$\Delta ADY\sim \Delta XCY.$$ So we have $$\frac{BK}{DY}=\frac{XK}{LY}$$ and $$\frac{DY}{CY}=\frac{AD}{XC}=\frac{AY}{XY}.$$ Hence $$\frac{BK}{CY}=\frac{AD}{XC}\times \frac{XK}{LY}\implies \frac{BK}{BC}=\frac{CY}{XC}\times \frac{XK}{LY}=\frac{AB}{BC}\times \frac{XK}{LY} $$ $$\frac{AB}{LY}\times \frac{XK}{BK}=\frac{AB}{LY}\times \frac{LY}{DY}=\frac{AB}{DL}$$ $$\implies \Delta CBK\sim \Delta LDK$$ And we are done. We get that $$\angle KCL=360-(\angle ACB+\angle DKC+\angle BCK)=\angle DAB/2 +180-\angle DAB=180-\angle DAB/2$$ Motivation: I took a hint on this. I had other angles but I did...
4 comments
Read more

Introduction

  Hey Everyone!! This is my first Blog post. So let me give a brief introduction about myself. I am Sunaina Pati. I love solving Olympiad math problems,  learning crazy astronomical facts , playing hanabi and anti-chess, listening to Kpop , love making diagrams in Geogebra and  teaching other people maths 😊 . I love geometry , number theory and Combinatorics . I am starting this blog to keep myself a bit motivated in doing studies 😎 . Right now, I am planning to write walkthroughs on some of the best problems I tried over the week which can refer for hints 'cause solutions contain some major spoilers and one learns a lot while solving the problem on his own rather than seeing solutions . Also, there will be some reviews about Kpop songs, study techniques, my day to day lifestyles,exam reviews and ofc some non-sense surprises 😂.  I am planning to  try  posting every week on Sundays or Saturdays ( most probably) ! Though there is no guarantee about when I ...
20 comments
Read more

Some random problems

  I know, I know. Different font indeed. I have deleted a few of my MSE answers. I felt they weren't that good in quality. And a few questions are from my prev aops account which I have deactivated now. I also have posted 10 IOQM types of problems. These can be used while preparing for IOQM. Problem: Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$  are different positive real numbers.  Proof: Note that by AM-GM $$\frac{ab}{c^3}+\frac{bc}{a^3}\ge \frac{2b}{ac}$$ and we also have $$\frac {b}{ac}+\frac{c}{ab}\ge \frac{2}{a}$$. Hence, $$\sum_{cyc}\frac{ab}{c^3}\ge\sum_{cyc}\frac{b}{ac}\ge\sum_{cyc}\frac{1}{a}$$ where everything we got is by applying AM-GM on $2$ terms and then dividing by $2$. USA TST 2007: Triangle $ABC$ which is inscribed in circle $\omega$. The tangent lines to $\omega$ at $B$ and $C$ meet at $T$. Point $S$ lies on ray $BC$ such that $AS$ is perpendicular to $AT$. Points $B_1$ and $C_1...
17 comments
Read more