Sunaina thinks absurd

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...

INMO Scores are out!! I am now a two times INMO awardee :) I got 16|0|1, so 17 in total! Yes, 16 in P1 T_T. I was thinking I would lose marks because of the way I wrote.  Lemme tell ya'll what happened that day but first I should share a few thoughts I had before the exam. My thoughts Honestly, my preparation for INMO was bad. In fact, I should say I didn't work hard at all. As I have said earlier, I had lost all my hopes for INMO and Olympiads as a whole after EGMO TSTs happened.  Art by Jelena Janic EGMO TSTs i.e European Girl's Mathematical Olympiad Team selection Tests 2022.  Literally my thoughts after EGMO TSTs I feel very ashamed to share but I got 1 mark in my EGMO TSTs. Tests in which I literally gave my whole life. I did so many ISLs ( like SO MANY), I mocked EGMO 2021 TST where my score was 28/42 and I perfected Day 2. 1 mark in the TST just showed my true potential. There are way better people than me in olys. A friend even said to me, "If I wouldn't...

Hey everyone! This is probably not the best post I have written but I thought something is better than nothing :P. So everyone who is in the Olympiad community have seen blogposts of pro-people describing how they qualified some fancy Olympiads.  But very few people talk about sad stuff. And yes INMO is now officially 2 weeks away.. Let me talk about a few things though: EGMO TSTs: I feel very ashamed to share but I got 1 mark in my EGMO TSTs. Tests in which I literally gave my whole life. I did so many ISLs ( like SO MANY), I mocked EGMO 2020 TST where my score was 28/42 and I perfected Day 2. 1 mark in the TST just showed my true potential. There are way better people than me in olys. A friend even said to me, "If I wouldn't have got into the team, I would have left math". I felt my whole hard work wasted. I have realised that hard work to some extent helps you in Olys but in the end, you really have to be smart( Not dumb like me).  My mental state after EGMO TSTs was ...

Continuing from  here . Book Referred: David Burton, Elementary Number Theory Also, thanks to Pranjal Bhaiya and Ritam bhaiya for taking lectures in NT in EGMO Camp and Sophie! We state and prove LTE first. $$v_p(x^n-y^n)=v_p(x-y)+v_p(n), n|x-y, n\not\mid x.$$ Walkthough: We use induction on $v_p(n).$ [ We show for $v_p(n)=0, v_p(n)=1$ and then use induction. 1. We show it for $v_p(n)=0.$ That is show $v_p(x^n-y^n)=v_p(x-y)$ To show this is true, $$v_p(\frac{x^n-y^n}{x-y})=v_p(x^{n-1}+yx^{n-2}+y^2x^{n-3}+\dots+y^{n-1})=0.$$ As $x\equiv y\pmod p.$ So,  $$x^{n-1}+yx^{n-2}+y^2x^{n-3}+\dots+y^{n-1}\equiv nx^{n-1}\pmod p. $$ And $p\not\mid nx^{n-1}$ 2. We show it for $v_p(n)=1.$ That is show $v_p(x^n-y^n)=v_p(x-y)+1$ To show this is true, $$v_p(\frac{x^n-y^n}{x-y})=v_p(x^{n-1}+yx^{n-2}+y^2x^{n-3}+\dots+y^{n-1})=1.$$ As $x\equiv y\pmod p\implies x=y+pk$ So,  $$x^{n-1}+yx^{n-2}+y^2x^{n-3}+\dots+y^{n-1}\pmod {p^2} $$ $$\equiv (pk+y)^{n-1}+(pk+y)^{n-2}y+(pk+y)^{n-3}y^2+\dots+y^{...

Continuing from the previous post from  here . Another memory lane of theorem I want to prove/try! Book Reffered: David Burton, Elementary Number Theory Problem 1: $$p_n<2^{2^{n-1}}$$ Proof: Note that $$p_{n+1}\le p_1p_2\dots p_n+1\le 2\cdot \dots 2^{2^{n-1}}=2^{2^{n-1}}$$ Problem 2: If the $n>2$ terms of the arithmetic progression $$p,p+d, p+2d,\dots,p+(n-1)d$$ are all primes then the common difference $d$ is divisible by every prime $q<n.$ Proof: If not then there exists a $q$ such that $(d,q)=1, q<n.$ But then we can get a $r$ such that $p\equiv -rd\mod q.$ Problem 3: Let $p_n$ denote the $n$ th prime. For $n>3$ show that $$p_n<p_1+p_2+\dots p_{n-1}.$$ Proof: Use induction and Bertrand's postulate. We get that $$p_{n+1}<2p_n<p_1+p_2+\dots p_{n-1}+p_n.$$ I should try to prove FLT and Wilson on my own too! But lemme state them in problems format. Anyways.. Problem4: If $n=a^2+b^2=c^2+d^2$ then $$n=\frac{(ac+bd)(ac-bd)}{(a+d)(a-d)}$$ Proof:  Note ...

I thought to revise David Burton and try out some problems which I didn't do. It's been almost 2 years since I touched that book so let's see! Also, this set of problems/notes is quite weird since it's actually a memory lane. You will get to know on your own! I started with proving a problem, remembered another problem and then another and so on! It was quite fun cause all these questions were the ones I really wanted to solve! And this is part1 or else the post would be too long. Problem1: Prove that for $n\ge 1$  $$\binom{n}{r}<\binom{n}{r+1}$$ iff $0\le r\le \frac{n-1}{2}$ Proof: We show that $\binom{n}{r}<\binom{n}{r+1}$ for $0\le r\le \frac{n-1}{2}$ and use the fact that $$\binom{n}{n-r}=\binom{n}{r}$$ Note that $\binom{n}{r}= \frac{n!}{r!(n-r)!}, \binom{n}{r+1}=\frac{n!}{(r+1)!(n-r-1)!}$  Comparing, it's enough to show that $$\frac{1}{n-r}<\frac{1}{r+1}\text{ or show } n-r>r+1$$ which is true as $0\le r\le \frac{n-1}{2}$ Problem2: Show that the exp...

Heyoo!!! Birthday FEs!!!!!! $11$ FEs!! Also I would be posting solutions to RG's FE handout, I am done with 10 prs :P!! Problem: Find all functions $f :\Bbb R \rightarrow \Bbb R$ such that $$2f (x) - 5f (y) = 8, \forall x, y \in \Bbb R$$ Solution: $$2f(x)-5f(y)=8$$ $$\implies 2f(x)-5f(x)=8$$ $$\implies f(x)=\frac{-8}{3}, \text{ a constant function }$$ We did this in Rohan Bhaiya's FE class..Oh btw the EGMO camp is sooo niceee! I am loving it!! It's such a big deal to be able to train and attend the camp with EGMO team members! Problem: Find all functions $f :\Bbb R \rightarrow \Bbb R$ such that $$f (x) + xf (1 -x) = x, \forall x\in \Bbb R.$$ Solution: $$f(x)+xf(1-x)=x$$ $$f(1-x)+(1-x)f(x)=1-x$$ This is actually in the linear equations in two variable form! $$x+ay=a$$ $$y+bx=b$$ Anyways,  $$f(x)+xf(1-x)=x$$ $$xf(1-x)+f(x)(1-x)x=(1-x)x$$ $$ \implies f(x)(x-x^2)-f(x)=-x^2\implies f(x)=\frac{-x^2}{x-x^2-1}=\frac{x^2}{x^2-x+1}$$ But verifying, this doesn't work. Problem: ...

Continuing from the previous post ( which is  here  ). This is a bit short post. I think this short post is not at all helpful in olys but then let's do it cause I find them interesting :P. Here are my notes.. Book referred: Daniel A Marcus Graph theory Chapter I: Independence Sets: Independent vertex set:  An independent vertex set in a graph is a set of vertices that does not include any adjacent vertices. Independent edge set:  An independent edge set in a graph is a set of edges that does not include any edges that share an endpoint. Maximal Independent sets: Largest set possible. $\alpha =$ the number of vertices in a maximal independent set $\alpha ' =$ the number of edges in a maximal independent edge set Theorem1: In a graph with $n$ vertices and $e$ edges, a) $\alpha'$ is less than $n/2$ b) if $d_1,d_2,\dots $ be the degress in a maximal independent vertex set, then the sum $d_1+d_2+\dots $ is less than $e.$ Proof: Note that every edge will have distinct v...