Special Problem Marathon

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SMA_Nahian
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Re: Special Problem Marathon

Unread post by SMA_Nahian » Wed Jun 02, 2021 12:57 pm

FuadAlAlam wrote:
Tue Jun 01, 2021 11:16 pm
Problem 4:

Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that
$$f(x + y) + f(x)^2f(y) = f(y)^3 + f(x + y)f(x)^2$$
for all $x, y \in \mathbb{Z}$.
$\textbf{Solution (P4)}$

Here, the equation of the question is, $$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
Now, for $x=0,y=a$, we get,
$$f(a) + f(0)^2f(a) = f(a)^3+f(a)f(0)^2$$
$$\Rightarrow f(a) = f(a)^3$$
$$\Rightarrow f(a) \in \{-1,0,1\}\ \forall \ a \in \mathbb{Z}$$

Now there is two cases.
$\boxed{\textbf{Case 1:}\ f(a) \in \{-1,1\}\ \forall \ a \in \mathbb{Z}}$
$\underline{\text{Claim :}}$ No matter if $f(x)=1 \text{ or, } f(x) =-1$ this always satisfy the equation given in the question.
$\underline{\text{Proof :}}$ Here, it is trivial that, $f(x)^2= 1$ and $f(x)^3= f(x)$ $ \forall \ x \in \mathbb{Z}$
Therefore,
$$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
$$\Rightarrow f(x+y) + f(y) = f(y)+f(x+y)$$
And, that's always true.
So, for this case, the solution is, for any $x \in \mathbb{Z}$, $f(x)=1$ or $f(x)=-1$
$\boxed{\textbf{Case 2:}\ \text{There is } a\in \mathbb{Z}:f(a) = 0}$
Now there is two subcases.
$\boxed{\textbf{Subcase 1:} \ \text{only } a \text{ is } a=0}$
$\underline{\text{Claim :}}$ For $n\neq 0$,
no matter if $f(n)=1 \text{ or, } f(n) =-1$ this always satisfy the equation given in the question.
$\underline{\text{Proof :}}$ Here, it is trivial that, $f(x)^2= 1$ or $f(x)^2= 0$ and $f(x)^3= f(x)$ $ \forall \ x \in \mathbb{Z}$
For,$f(x)^2 = 1 $ we have already showed that, $$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
$$\Rightarrow f(x+y) + f(y) = f(y)+f(x+y)$$
Which is always true.

And for, $f(x)^2=0$, $$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
$$\Rightarrow f(x+y)=f(y)$$
$$\Rightarrow f(y) = f(y)$$
and this is also true.
So, for this subcasecase, the solution is, for any $x \in \mathbb{Z} \text{ and } x\neq 0$, $f(x)=1$ or $f(x)=-1$ and $f(0)=0$.

$\boxed{\textbf{Subcase 2:} \ \text{There exists } a \text{ such that } a\neq 0 ,f(a)=0}$

For this subcase lets take the integer $a_0:f(a_0) = 0$ and $|a_0|$ is smallest and positive.
Now for, $x=a_0,y= b$, $$f(b+a_0) = f(b)^3$$
$$\Rightarrow f(b+a_0) = f(b)$$
$$\Rightarrow f(b+d\cdot a_0) = f(b) \tag{1}$$
So, it is trivial that for any multiple $d\cdot a_0$ of $a_0$, $f(d\cdot a_0)=0$


$\underline{\text{Claim 1:}}$ There is no other $x:f(x)=0$.
$\underline{\text{Proof :}}$ For the sake of contradiction lets assume that there is a integer $a_1$ such that, $a_1$ is not divisible by $a_0$ and $f(a_1) = 0$
Again, for, $x=a_1,y= b$, $$f(b+d\cdot a_1) = f(b) \tag{2}$$

From equation $(1),(2)$ we get, $$f(x\cdot a_0+y\cdot a_1) =0 \ \forall \ x,y \in \mathbb{Z}$$
From bezout's identity we have, some $x_0,y_0:x_0\cdot a_0+y_0\cdot a_1 = gcd(a_0,a_1)$
So, $f(gcd(a_0,a_1))=0$
As, $|a_0| < |a_1|$ and $a_0$ dosent divide $a_1$, therefore, $$|gcd(a_0,a_1)|<|a_0|$$
That's a contradicts with the minimality of $|a_0|$

So the function is periodic and it repeat at interval of $a_0$

$\underline{\text{Claim 2:}}$ For each period of the function no matter if $f(x)=1 \text{ or, } f(x) =-1$ for $x:a_0\nmid x$ this always satisfy the equation given in the question
$\underline{\text{Proof :}}$ Here $f(x)^2 = 0 \text{ or }1$. For,$f(x)^2 = 1 $ we have already showed that, $$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
$$\Rightarrow f(x+y) + f(y) = f(y)+f(x+y)$$
Which is always true.

And for, $f(x)^2=0$, $$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
$$\Rightarrow f(x+y)=f(y)$$ and this is also true, as, $a_0\mid x$ and the function is periodic and it repeat at interval of $a_0$.

So the solution for this subcase is,
$f(d\cdot a_0) = 0$ for a $a_0\in \mathbb{N}$ and all $f(d\cdot a_0 + c) $ for each $c: a_0\nmid c$ are equal and equal to $+1$ or $-1$ $\blacksquare$
Last edited by SMA_Nahian on Wed Jun 02, 2021 1:54 pm, edited 2 times in total.
"Everything will be okay in the end. If it’s not okay, it's not the end."
——John Lennon

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Anindya Biswas
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Re: Special Problem Marathon

Unread post by Anindya Biswas » Wed Jun 02, 2021 1:02 pm

SMA_Nahian wrote:
Wed Jun 02, 2021 12:57 pm
FuadAlAlam wrote:
Tue Jun 01, 2021 11:16 pm
Problem 4:

Find all functions $f : \mathbb{Z} \rightarrow \mathbb{Z}$ such that
$$f(x + y) + f(x)^2f(y) = f(y)^3 + f(x + y)f(x)^2$$
for all $x, y \in \mathbb{Z}$.
$\textbf{Solution (P4)}$

Here, the equation of the question is, $$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
Now, for $x=0,y=a$, we get,
$$f(a) + f(0)^2f(a) = f(a)^3+f(a)f(0)^2$$
$$\Rightarrow f(a) = f(a)^3$$
$$\Rightarrow f(a) \in \{-1,0,1\}\ \forall \ a \in \mathbb{Z}$$

Now there is two cases.
$\boxed{\textbf{Case 1:}\ f(a) \in \{-1,1\}\ \forall \ a \in \mathbb{Z}}$
$\underline{\text{Claim :}}$ No matter if $f(x)=1 \text{ or, } f(x) =-1$ this always satisfy the equation given in the question.
$\underline{\text{Proof :}}$ Here, it is trivial that, $f(x)^2= 1$ and $f(x)^3= f(x)$ $ \forall \ x \in \mathbb{Z}$
Therefore,
$$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
$$\Rightarrow f(x+y) + f(y) = f(y)+f(x+y)$$
And, that's always true.
So, for this case, the solution is, for any $x \in \mathbb{Z}$, $f(x)=1$ or $f(x)=-1$
$\boxed{\textbf{Case 2:}\ \text{There is } a\in \mathbb{Z}:f(a) = 0}$
For this case lets take the integer $a_0:f(a_0) = 0$ and $|a_0|$ is smallest.
Now for, $x=a_0,y= b$, $$f(b+a_0) = f(b)^3$$
$$\Rightarrow f(b+a_0) = f(b)$$
$$\Rightarrow f(b+d\cdot a_0) = f(b) \tag{1}$$
So, it is trivial that for any multiple $d\cdot a_0$ of $a_0$, $f(d\cdot a_0)=0$
$\underline{\text{Claim 1:}}$ There is no other $x:f(x)=0$.
$\underline{\text{Proof :}}$ For the sake of contradiction lets assume that there is a integer $a_1$ such that, $a_1$ is not divisible by $a_0$ and $f(a_1) = 0$
Again, for, $x=a_1,y= b$, $$f(b+d\cdot a_1) = f(b) \tag{2}$$

From equation $(1),(2)$ we get, $$f(x\cdot a_0+y\cdot a_1) =0 \ \forall \ x,y \in \mathbb{Z}$$
From bezout's identity we have, some $x_0,y_0:x_0\cdot a_0+y_0\cdot a_1 = gcd(a_0,a_1)$
So, $f(gcd(a_0,a_1))=0$
As, $|a_0| < |a_1|$ and $a_0$ dosent divide $a_1$, therefore, $$|gcd(a_0,a_1)|<|a_0|$$
That's a contradicts with the minimality of $|a_0|$

So the function is periodic and it repeat at interval of $a_0$

$\underline{\text{Claim 2:}}$ For each period of the function no matter if $f(x)=1 \text{ or, } f(x) =-1$ for $x:a_0\nmid x$ this always satisfy the equation given in the question
$\underline{\text{Proof :}}$ Here $f(x)^2 = 0 \text{ or }1$. For,$f(x)^2 = 1 $ we have already showed that, $$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
$$\Rightarrow f(x+y) + f(y) = f(y)+f(x+y)$$
Which is always true.

And for, $f(x)^2=0$, $$f(x+y) + f(x)^2f(y) = f(y)^3+f(x+y)f(x)^2$$
$$\Rightarrow f(x+y)=f(y)$$ and this is also true, as, $a_0\mid x$ and the function is periodic and it repeat at interval of $a_0$.

So the solution for this case is,
$f(d\cdot a_0) = 0$ for a $a_0\in \mathbb{N}$ and all $f(d\cdot a_0 + c) $ for each $c: a_0\nmid c$ are equal and equal to $+1$ or $-1$ $\blacksquare$
What if $a_0=0$?
Last edited by tanmoy on Wed Jun 02, 2021 3:46 pm, edited 1 time in total.
Reason: shortening
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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FuadAlAlam
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Re: Special Problem Marathon

Unread post by FuadAlAlam » Wed Jun 02, 2021 1:15 pm

Solution (P5):

Claim-1: $A, B, D, E$ and $A, C, D, F$ are concyclic.
Proof: $\measuredangle BDE = \measuredangle CDE= 90^o - \measuredangle OCD=\measuredangle BAC =\measuredangle BAE$. Thus, $A, B, D, E$ are concyclic. Similarly, $A, C, D, F$ are concyclic.

We use directed length.
It suffices to show that $$BK^2-CK^2=BD^2-CD^2$$
But $BK^2-CK^2=(BK^2-KF^2)-(CK^2-KE^2)=Pow_{(AEF)}B-Pow_{(AEF)}C=BF \cdot BA - CE \cdot CA = BD \cdot BC - CD \cdot CB = BD \cdot BC + CD \cdot BC = BC \cdot (BD+CD)= (BD+DC) \cdot (BD-DC)= BD^2-DC^2=BD^2-CD^2$, as desired.

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Re: Special Problem Marathon

Unread post by FuadAlAlam » Wed Jun 02, 2021 1:23 pm

Anindya Biswas wrote:
Wed Jun 02, 2021 1:02 pm
What if $a_0=0$?
$f(0)=0$ and $f(x) \in \{1, -1\}$ for any $x \in \mathbb{Z} - \{0\}$ should work in this case.

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SMA_Nahian
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Re: Special Problem Marathon

Unread post by SMA_Nahian » Wed Jun 02, 2021 1:48 pm

Anindya Biswas wrote:
Wed Jun 02, 2021 1:02 pm
What if $a_0=0$?
Edited [I hope now it's okay] :oops:
"Everything will be okay in the end. If it’s not okay, it's not the end."
——John Lennon

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SMA_Nahian
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Re: Special Problem Marathon

Unread post by SMA_Nahian » Wed Jun 02, 2021 2:27 pm

$\textbf{Problem 6:}$

Let $ABC$ be a triangle inscribed in circle $\omega$, and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$, and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$. Prove that $O_1$, $O_2$, and the nine-point center of $ABC$ are collinear.
"Everything will be okay in the end. If it’s not okay, it's not the end."
——John Lennon

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Anindya Biswas
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Solution to Problem 06

Unread post by Anindya Biswas » Fri Jun 04, 2021 7:08 pm

SMA_Nahian wrote:
Wed Jun 02, 2021 2:27 pm
Let $ABC$ be a triangle inscribed in circle $\omega$, and let the medians from $B$ and $C$ intersect $\omega$ at $D$ and $E$ respectively. Let $O_1$ be the center of the circle through $D$ tangent to $AC$ at $C$, and let $O_2$ be the center of the circle through $E$ tangent to $AB$ at $B$. Prove that $O_1, O_2$ and the nine-point center of $ABC$ are collinear.
Geo_Nahian.png
Figure
Geo_Nahian.png (125.15KiB)Viewed 5887 times
Definitions and primary observations :
Let $G$ be the centroid of $\triangle ABC$.
Let $h$ be the homothetic transformation centered at $G$ and dialation ratio $-2$.
Let $\Omega=h(\omega), A_1=h(A), B_1=h(B), C_1=h(C), D_1=h(D), E_1=h(E)$.
Let $\omega_9$ be the nine-point circle of $\triangle ABC$ with center $N$. Let $O$ be the center of $\omega$.
$\therefore \omega=h(\omega_9)$
Let $H$ be the orthocenter of $\triangle ABC$. Since $H=h(O)$, we conclude that $H$ is the center of $\Omega$.
$\triangle A_1B_1C_1=h(\triangle ABC)\Rightarrow \triangle ABC$ is the medial triangle of $A_1B_1C_1$.
$\therefore AB\parallel A_1B_1, AC\parallel A_1C_1, BC\parallel B_1C_1$.
Let $P,Q$ be the midpoints of $AC$ and $AB$ respectively.
Let's assume $PB$ intersects $\omega_9$ again at $D_2$, $QC$ intersects $\omega_9$ again at $E_2$.
$\therefore D=h(D_2), E=h(E_2)$.

Claim-1 : $BEC_1D_1, BD_2QE, CDB_1E_1, CE_2PD$ are concyclic.
Proof :
  • $GE\cdot GC_1=-2\cdot GE\cdot GC=-2\cdot GB\cdot GD=GB\cdot GD_1$
    $\therefore BEC_1D_1$ concyclic.
  • $GQ\cdot GE=-\frac12\cdot GC\cdot GE=-\frac12\cdot GD\cdot GB=GD_2\cdot GB$
    $\therefore BD_2QE$ concycic.
Similarly, $CDB_1E_1, CE_2PD$ are concyclic. $\square$

Let's denote the circles $BEC_1D_1, BD_2QE, CDB_1E_1, CE_2PD$ with $\Gamma_2, \gamma_1, \Gamma_1, \gamma_2$ respectively.
Observe that $h(CE_2PD)=C_1EBD_1\Rightarrow h(\gamma_2)=\Gamma_2$. Similarly, $h(\gamma_1)=\Gamma_1$.

Claim-2 : $AC, AB$ are tangent to $\Gamma_1, \Gamma_2$ respectively.
Proof :
$\Omega\cap\Gamma_1=\{B_1, E_1\}$.
$\therefore$ By Reim's theorem on $A_1C$ and $B_1C$, we conclude that $A_1C_1$ and tangent to $\Gamma_1$ through $C$ must be parallel.
Since $AC\parallel A_1C_1, AC$ must be tangent to $\Gamma_1$.
Similarly, $AB$ is tangent to $\Gamma_2$. $\square$
Corollary : $O_1, O_2$ are the centers of $\Gamma_1, \Gamma_2$ respectively.

Claim-3 : $HO_1OO_2$ is a parallelogram.
Proof :
Let $T_1,T_2$ be the centers of $\gamma_1, \gamma_2$ respectively.
Since $\omega, \gamma_2, \Gamma_1$ are coaxial circles, $O,T_2,O_1$ are collinear.
Since $h(OT_2)=HO_2$, we conclude that $OT_2\parallel HO_2\Rightarrow OO_1\parallel HO_2$.
Similarly, $OO_2\parallel HO_1$.
$\therefore HO_1OO_2$ is a parallelogram. $\square$

Since $N$ is the midpoint of $OH$, $N$ must also be the midpoint of $O_1O_2$.
$\therefore O_1,N,O_2$ are collinear. $\blacksquare$
Last edited by Anindya Biswas on Fri Jun 04, 2021 8:15 pm, edited 1 time in total.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Problem 7

Unread post by Anindya Biswas » Fri Jun 04, 2021 7:48 pm

Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that \[\frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\geq\frac32\]
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

rah4927
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Solution to Problem 07

Unread post by rah4927 » Sun Jun 06, 2021 3:52 am


By Cauchy-Schwarz,
$$\sum\limits_{\text{cyc}} \frac{1}{a^3(b + c)} = \sum\limits_{\text{cyc}} \frac{b^2c^2}{a(b + c)} \ge \dfrac{(ab + bc + ca)^2}{2(ab + bc + ca)} \ge \dfrac{3}{2}$$
where the last inequality holds by AM-GM.

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Problem 8

Unread post by rah4927 » Sun Jun 06, 2021 4:13 am

There is a game show with $100$ contestants labelled $1, 2, \ldots, 100$. There are $100$ doors in this game show, each having a distinct hidden number from the set $\{1, \ldots, 100\}$. The contestants are not allowed to communicate with each other during the game. During the game, each contestant's objective is to find the door with their own label behind it. They are allowed to do the following things:

1. Contestant $i$ is allowed to open one door at a time, up to a total of $50$ doors. In particular, this means that they do not have to guess all $50$ doors prior to opening them. If one of the opened doors contains the number $i$ behind it, then contestant $i$ wins.

2. None of the other contestants are privy to what contestant $i$ saw behind the doors. In particular, this means that after the game begins, the contestants cannot communicate with each other.

However, Ipshita, who is in charge of the doors, has bet her friend that all $100$ contestants can win this game. But in order to make this happen, she is allowed to exchange the numbers behind two doors (and she can do this operation exactly once). Devise a strategy for Ipshita and the $100$ contestants so that they can all win the game show.

Note: Ipshita can see the numbers behind the doors, but once she does so she is not allowed to communicate with the contestants.

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