Definitely no geos [Feb 1-Feb 7]

Okay I did way more problems.. it's just I am too lazy to type up. And Ig no more posts before EGMO cause I am too much busy for boards :( bai people

Problem 1.6: Prove that the number of answers for $|a_1|+|a_2|+...+|a_k|≤n$ is equal to the number of answers for $|b_1|+|b_2|+...+|b_n|≤k$, where $a_i,b_i$ are integers.

Proof: Let $f(n,k)$ be the number of solutions to $|a_1|+\dots+|a_n|\le k$.

Note that $$f(n,k) = f(n-1,k)+f(n-1,k-1)+[f(n-1,k-1)+2f(n-1,k-2)+\dots+2f(n-1,0)]$$ $$=f(n-1,k)+f(n-1,k-1)+f(n,k-1).$$

Similarly, we get $$f(k,n)=f(k-1,n)+f(k-1,n-1)+f(k,n-1).$$

Now, we can simply induct on $a+b$ and show $f(a,b)=f(b,a)$.

Done!

EGMO 2012 P2

Let $n$ be a positive integer. Find the greatest possible integer $m$, in terms of $n$, with the following property: a table with $m$ rows and $n$ columns can be filled with real numbers in such a manner that for any two different rows $\left[ {{a_1},{a_2},\ldots,{a_n}}\right]$ and $\left[ {{b_1},{b_2},\ldots,{b_n}} \right]$ the following holds:$$\max\left( {\left| {{a_1} - {b_1}} \right|,\left| {{a_2} - {b_2}} \right|,...,\left| {{a_n} - {b_n}} \right|} \right) = 1$$

Proof: We have the answer as $2^n$. We go with induction. For $n=1$ it is true. Say it is true for $n=k$. We will show that it is true for $n=k+1$, the max possible value of $m$ is $2^{k+1}$. Suppose not then $m>2^{k+1}$.

So there exists $3$ rows in the $m\times k+1$ table such that the first $k$ elements are same (by PHP). So we consider the $k+1$ th column of the three rows ( $\{a_i\}, \{b_i\}, \{b_i\}$). We have $a_i=b_i=c_i$ for $i\in[k]$. Since $$\max(|a_1-b_1|,\dots,|a_{k+1}-b_{k+1}|)=1.$$

We get $$|a_{k+1}-b_{k+1}|=1$$ $$|b_{k+1}-c_{k+1}|=1$$ $$|a_{k+1}-c_{k+1}|=1$$ which is impossible to happen as $a_{k+1},b_{k+1},c_{k+1}$ are pairwise distinct.

Equality case construction: write $0,1,\ldots 2^n-1$ in binary form and with each digit in one cell in each row.

AIME 2019 P6

In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$, for some fixed $b \geq 2$. A Martian student writes down

$$3 \log(\sqrt{x}\log x) = 56$$ $$\log_{\log (x)}(x) = 54$$

and finds that this system of equations has a single real number solution $x > 1$. Find $b$

Proof: Let $\log_bx=c$.

So $\log_cx=54\implies x=c^{54}$. Also $$3\log(\sqrt{x}\log x) =56\implies 3\log(\sqrt{x})+3\log(\log x) =56\implies 3c/2+3\log_bc=56.$$

But $$3\frac{log_xc}{\log_xb}=3\times \frac{1}{54}\times c=\frac{c}{18}\implies 3c/2+c/18=56$$ $$\implies c=36\implies x=36^{54}\implies b=216.$$

EGMO 2019 P1

Find all triples $(a, b, c)$ of real numbers such that $ab + bc + ca = 1$ and

$$a^2b + c = b^2c + a = c^2a + b.$$

Proof: We begin with homogenization. We get $$a^2b+c[ab+bc+ca]=b^2c+a[ab+bc+ca]\implies c^2(b+a)=c(b^2+c^2)$$

So $c=0$ or $a^2+b^2=c(a+b)$.

If $c=0$ then $a^2+b=a=b\implies a=b=\pm 1$. Note that $a,b\ne 0$ as $ab=1$.

If $c\ne 0$ ( So assume all are non zero). Then $$c(a+b)=a^2+b^2$$ $$a(c+b)=c^2+b^2$$ $$b(a+c)=a^2+c^2$$ $$\implies 2ab+2bc+2ca=2a^2+2b^2+2c^2$$ which by am-gm $\implies a=b=c$ but $ab+bc+ca=1\implies a=b=c=\pm \frac{1}{\sqrt{3}}$.

USAMO 2006 P4

Find all positive integers $n$ such that there are $k \geq 2$ positive rational numbers $a_1, a_2, \ldots, a_k$ satisfying $a_1 + a_2 + \ldots + a_k = a_1 \cdot a_2 \cdots a_k = n.$

Proof: Answer is all positive integers except $1,2,3,5$.

For $n=1,2,3$, we use AM-GM, $$\frac{a_1+\dots+a_k}{k}\ge \sqrt[k]{a_1\dots a_k}\implies n^{k}\ge k^k\cdot n\implies n^{k-1}\ge k^k\implies n\le 3.$$

For $n$, $5^{k-1}\ge k^k$. So $k\le 2$. Let $a_1+a_2=a_1a_2=5$, then solving we get:

$$a_1,a_2=\frac{5\pm\sqrt5}2$$which are both irrational.

For $n$=prime $>7$, we have $\frac{n}{2}, \frac{1}{2}, 1, \dots$.

For $n=$ composite, say $n=a\cdot b$. So consider $n=a\cdot b\cdot 1 \cdot 1\dots$.

Serbia 2014

Let $ABCD$ be a quadrilateral such that $\angle BCA + \angle CAD = 180^{\circ}$ and $\overline{AB} = \overline{AD} + \overline{BC}$. Prove that$$\angle BAC + \angle ACD = \angle CDA$$

Proof: Define $X=AD\cap CB$. Since $$\angle BCA+\angle CAD=180\implies \angle BCA=180-\angle CAD=\angle XAC\implies XA=XC.$$

Now, consider point $B'$ on $AX$ such that $AB'=BC$. Note that $AB'BC$ is an isosceles triangle. Also, we have $B'A=BC\implies B'D=AB$. So $\angle BAC=\angle B'CA$. So enough to show $B'D=B'C$. But $B'C=AB=B'D$. Done!

Bulgaria EGMO TST 2021 Problem 1

On the side $AB$ of a triangle $ABC$ is chosen a point $P$. Let $Q$ be the midpoint of $BC$ and let $CP$ and $AQ$ intersect at $R$. If $AB + AP = CP$, prove that $CR = AB$.

Proof: We redefine the points. Define $A'$ as the reflection of $A$ over $Q$ and $B'$ be the reflection of $B$ over $A$. Define $R$ such that $CA'=CR$ and $R\in AQ$. Define $CR\cap AB=P$. Note that $$\angle PRA=\angle CRA'=\angle AA'C=\angle A'AP\implies AP=PR.$$

But $$AB'=AB=A'C=RC\implies PB'=PC\implies PA+AB=CP.$$

And we are done.

Comments

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....

Problems I did this week [Jan8-Jan14]

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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 ...

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...

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...

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...

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...

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 ...

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 ...

Let's complex bash Part 1

I have to learn complex bash. And almost everyone knows that I am notes taking girl so thought why not make a post on complex bash ( so that I don't get emotionally demotivated lol).😇 There wasn't any need for learning complex bash, but it was in my dream checklist i.e " To learn a bash." And since I am not loaded with exams, I think it's high time to learn Bash and new topics. Also if anyone from the "anti-bash" community is reading, sorry in advance and R.I.P. Notes:- 1. Complex numbers are of the form $z=a+ib,$ where $a$ and $b$ are real numbers and $i^2=-1.$ 2. In polar form, $z=r(\cos \theta+~~i\sin\theta)=~~re^{i\theta},$ where $r=~~|z|=~~\sqrt{a^2+b^2},$ which is called the magnitude. 3. Here we used euler's formula i.e $\cos \theta+~~i\sin\theta=~~e^{i\theta}.$ 4. The $\theta $ is called the argument of $z,$ denored $\arg z.$ ( $\theta$ can be considered in $\mod 360$ and it is measured anti-clockwise). 5. The complex conjugate of $z$ is ...

Sunaina thinks absurd

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