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.