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

Problems with meeting people!

Yeah, I did some problems and here are a few of them! I hope you guys try them! Putnam, 2018 B3 Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n - 1$, and $n-2$ divides $2^n - 2$. Proof We have $$n|2^n\implies n=2^a\implies 2^a-1|2^n-1\implies a|n\implies a=2^b$$ $$\implies 2^{2^b}-2|2^{2^a}-2\implies 2^b-1|2^a-1\implies b|a\implies b=2^c.$$ Then simply bounding. USAMO 1987 Determine all solutions in non-zero integers $a$ and $b$ of the equation $$(a^2+b)(a+b^2) = (a-b)^3.$$ Proof We get $$ 2b^2+(a^2-3a)b+(a+3a^2)=0\implies b = \frac{3a-a^2\pm\sqrt{a^4-6a^3-15a^2-8a}}{4}$$ $$\implies a^4-6a^3-15a^2-8a=a(a-8)(a+1)^2\text{ a perfect square}$$ $$\implies a(a-8)=k^2\implies a^2-8a-k^2=0\implies \implies a=\frac{8\pm\sqrt{64+4k^2}}{2}=4\pm\sqrt{16+k^2}. $$ $$ 16+k^2=m^2\implies (m-k)(m+k)=16.$$ Now just bash. USAMO 1988 Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1...

How to prepare for RMO?

"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...

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...

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...

Symmetric Polynomials #week 6

Well... I haven't seen much symmetric polynomials in Olympiads, but still I am learning, because I found them cute. And I am basically using this blog as my notes :P What are symmetric polynomials? One can understand this with examples. If we are considering over 3 variables, $x_1,x_2,x_3$ then $$\sum_{sym}x_1^2\cdot x_2^3\cdot x_3=x_1^2\cdot x_2^3\cdot x_3+x_1^2\cdot x_3^3\cdot x_2+x_2^2\cdot x_1^3\cdot x_3+x_2^2\cdot x_3^3\cdot x_1+x_3^2\cdot x_1^3\cdot x_2.$$ See? $3!$ terms! Let's take one more example with again over 3 variables, $x_1,x_2,x_3$ then $$\sum_{sym}x_1^2\cdot x_2^2= x_1^2\cdot x_2^2+x_1^2\cdot x_3^2+x_2^2\cdot x_1^2+x_2^2\cdot x_3^2+x_3^2\cdot x_1^2+x_3^2\cdot x_2^2$$ Wait.. why 2 times ? So basically what happens in symmetrictric sums, is we go through all $n!$ possible permutations. So, here we have $a^2\cdot b^2\cdot c^0$ as like the "general" form type, right? Now, list down all the $3!=6$ permutations of $x_1,x_2,x_3$, and put them in the gene...

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...

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...

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...

Sunaina thinks absurd

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