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]

Yeyy!! I am being so consistent with my posts~~ Here are a few problems I did the past week and yeah INMO going to happen soon :) All the best to everyone who is writing! I wont be trying any new problems and will simply revise stuffs :) Some problems here are hard. Try them yourself and yeah~~Solutions (with sources) are given at the end! Problems discussed in the blog post Problem1: Let $ABC$ be a triangle whose incircle $\omega$ touches sides $BC, CA, AB$ at $D,E,F$ respectively. Let $H$ be the orthocenter of $DEF$ and let altitude $DH$ intersect $\omega$ again at $P$ and $EF$ intersect $BC$ at $L$. Let the circumcircle of $BPC$ intersect $\omega$ again at $X$. Prove that points $L,D,H,X$ are concyclic. Problem 2: Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ= 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ ...

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 ...

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 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...

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...

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 ...

Some random problems

I know, I know. Different font indeed. I have deleted a few of my MSE answers. I felt they weren't that good in quality. And a few questions are from my prev aops account which I have deactivated now. I also have posted 10 IOQM types of problems. These can be used while preparing for IOQM. Problem: Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$ are different positive real numbers. Proof: Note that by AM-GM $$\frac{ab}{c^3}+\frac{bc}{a^3}\ge \frac{2b}{ac}$$ and we also have $$\frac {b}{ac}+\frac{c}{ab}\ge \frac{2}{a}$$. Hence, $$\sum_{cyc}\frac{ab}{c^3}\ge\sum_{cyc}\frac{b}{ac}\ge\sum_{cyc}\frac{1}{a}$$ where everything we got is by applying AM-GM on $2$ terms and then dividing by $2$. USA TST 2007: Triangle $ABC$ which is inscribed in circle $\omega$. The tangent lines to $\omega$ at $B$ and $C$ meet at $T$. Point $S$ lies on ray $BC$ such that $AS$ is perpendicular to $AT$. Points $B_1$ and $C_1...

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...

Sunaina thinks absurd

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