INMO Scores and Results

Heya! INMO Results are out! Well, I am now a 3 times IMOTCer :D. Very excited to meet every one of you! My INMO score was exactly 26 with a distribution of 17|0|0|0|0|9, which was a fair grading cause after problem 1, I tried problem 6 next. I was hoping for some partials in problem 4 but didn't get any.

I am so so so excited to meet everyone! Can't believe my olympiad journey is going to end soon..

I thought to continue the improvement table I made last year! ( I would still have to add my EGMO performance and also IMO TST performance too)

2018-2019[ grade 8]: Cleared PRMO, Cleared RMO[ State rank 4], Wrote INMO

2019-2020[ grade 9]: Cleared PRMO, Cleared RMO[ State topper], Wrote INMO ( but flopped it)

2020-2021[grade 10]: Cleared IOQM, Cleared INMO [ Through Girl's Quota]

2021-2022[grade 11]: Wrote EGMO 2022 TST[ Rank 8], Qualified for IOQM part B directly, Cleared IOQM-B ( i.e INMO) [Through general quota],

2022-2023 [grade 12]: Wrote EGMO 2023 TST [ Rank 2], Made EGMO Team and is IND2, Qualified for INMO directly, Cleared INMO [ 12th grade category i.e Top 12 kids in 12th grade], EGMO Silver, Wrote IMOTSTS

Comments

Amit March 7, 2023 at 1:45 AM
I saw the results. There were two from Assam. Was the other Anand?
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Shreeansh HotaMarch 17, 2023 at 8:58 PM
I also got 26 in INMO in class 12. We barely managed to qualify. :P
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TestJanuary 17, 2024 at 4:01 PM
How to get selected in oromys please give some good suggestions!!!
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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

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 will post , so if you are

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"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

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. Problem: A convex quadrilateral $

Reflecting on past

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Bio is Love..

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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 didn't r

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 2n\equiv n(2n+1

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

Sunaina thinks absurd

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