Sunaina thinks absurd

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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

EGMO 2024 P2

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 incircle and $B

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 the

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 problem stateme

Some Geometry Problems for everyone to try!

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

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