rah4927 wrote: ↑Sun Jun 06, 2021 4:13 amThere 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.
Special Problem Marathon
- Anindya Biswas
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"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- Anindya Biswas
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Problem 09
Let $n\geq3$ be a fixed positive integer. A function $f:\mathbb{R}^2\to\mathbb{R}$ has the property that for any points $P_1, P_2,\cdots,P_n$ which are vertices of a regular $n$-gon, we have \[\sum_{i=1}^{n}f(P_i)=0\]
Prove that $f(P) = 0$ for all points $P\in\mathbb{R}^2$.
Prove that $f(P) = 0$ for all points $P\in\mathbb{R}^2$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
Re: Problem 09
Let the vertices of a regular $n$ gon be $\{A_{1,k}\}_{k=1}^{n}$ in clockwise order.Anindya Biswas wrote: ↑Sun Jun 06, 2021 3:59 pmLet $n\geq3$ be a fixed positive integer. A function $f:\mathbb{R}^2\to\mathbb{R}$ has the property that for any points $P_1, P_2,\cdots,P_n$ which are vertices of a regular $n$-gon, we have \[\sum_{i=1}^{n}f(P_i)=0\]
Prove that $f(P) = 0$ for all points $P\in\mathbb{R}^2$.
Now fix the point $A_{1,1}$ in its place and rotate the polygon by $\theta= \left(\frac{360}{n} \right)^\circ$ and let the new points be defined as $\{A_{2,k}\}_{k=1}^{n}$ such that $A_{1,i} \to A_{2,i}$. Now rotate the polygon total $n-1$ times by angle $\theta$ and define those polygons as $\{A_{i,j}\}_{j=1}^{n}$ where it is the $i$ polygon. Now we can easily check that,
\[\sum_{j=1}^{n}f(A_{i,j})=0 \quad \forall i\]
And the polygons $\{A_{i,j}\}_{i=1}^{n}$ where $2 \leq j \leq n$ are also regular with center $A_{1,1}$. So,
\[\sum_{i=1}^{n}f(A_{i,j})=0 \quad \forall j \ s.t. 2 \leq j \leq n\]
Now, we have that,
\begin{align*}
\sum_{i=1}^{n}\sum_{j=1}^{n}f(A_{i,j}) &= nf(A_{1,1})+\sum_{j=2}^{n}\sum_{i=1}^{n}A_{i,j}\\
\Rightarrow nf(A_{1,1})&=0\\
\Rightarrow f(A_{1,1})&=0
\end{align*}
So, $f(P)=0$ for all $P \in \mathbb{R}^2$
Problem 10
Define a function $f: \mathbb N \to \mathbb N$ by $f(1) = 1$, $f(n+1) = f(n) + 2^{f(n)}$ for every positive integer $n$. Prove that $f(1), f(2), \cdots, f(3^{2013})$ leave distinct remainders when divided by $3^{2013}$.
- FuadAlAlam
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Problem 11
In a $\triangle ABC$, $D$ and $E$ are respectively on $AB$ and $AC$ such that $DE\parallel BC$. $P$ is the intersection of $BE$ and $CD$. $M$ is the second intersection of $(APD)$ and $(BCD)$, $N$ is the second intersection of $(APE)$ and $(BCE)$. $w$ is the circle passing through $M$ and $N$ and tangent to $BC$. Prove that the lines tangent to $w$ at $M$ and $N$ intersect on $AP$.
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Solution of Problem 11
Let $AP \cap BC=L$
$LM\cap AC=F$
$LM \cap DE=K$
$LN\cap AB=Q$
$LN\cap DE=J$
Let $\measuredangle$ denote directed angle.
$\underline{Claim:}$ Circle $MNL$ is tangent to $BC$ at $L$ that is $w$ is tangent to $BC$ at $L$.
$\underline{Proof:}$ Clearly $M,N$ are the centers of spiral similarity sending $AP$ to $BC$ and $CB$ respectively.
$\measuredangle BPN = \measuredangle BLN = \measuredangle EJN \Rightarrow AJNPE$ is cyclic.
Similarly $ADPKM$ is cyclic.
$\therefore LN.LJ=LP.LA=LK.LM \Rightarrow JNKM$ is cyclic.
$\measuredangle LMN=\measuredangle KJL=\measuredangle BLN$
$\therefore$Circle $ MNL$ is tangent to $BC$ at $L$ $\blacksquare$
$\underline{Claim:}$ $AL$ bisects $JK$.
$\underline{Proof:}$ We can easily show that $L$ is the midpoint if $BC$.
$\measuredangle FAN= \measuredangle BLN = \measuredangle LMN \Rightarrow AMNF$ is cyclic
Similarly $AMNQ$ is cyclic $\Rightarrow AMNFQ$ is cyclic.
$\measuredangle FQN=\measuredangle FAN= \measuredangle EJN \Rightarrow QF\parallel DE \parallel BC$
Since $BL=LC$, by lenght chasing we can show $JD=EK$
But $AL$ bisects $DE \Rightarrow AL$ bisects $JK$ $\blacksquare$
Now $\triangle LJK \sim \triangle LMN$
$LA$ is the $L$-median of $\triangle LJK$
$\therefore LA$ is the $L$-symmedian of$\triangle LMN$
$\therefore$ Tangents to $w$ at $M,N$ intersect on $AP$.
$LM\cap AC=F$
$LM \cap DE=K$
$LN\cap AB=Q$
$LN\cap DE=J$
Let $\measuredangle$ denote directed angle.
$\underline{Claim:}$ Circle $MNL$ is tangent to $BC$ at $L$ that is $w$ is tangent to $BC$ at $L$.
$\underline{Proof:}$ Clearly $M,N$ are the centers of spiral similarity sending $AP$ to $BC$ and $CB$ respectively.
$\measuredangle BPN = \measuredangle BLN = \measuredangle EJN \Rightarrow AJNPE$ is cyclic.
Similarly $ADPKM$ is cyclic.
$\therefore LN.LJ=LP.LA=LK.LM \Rightarrow JNKM$ is cyclic.
$\measuredangle LMN=\measuredangle KJL=\measuredangle BLN$
$\therefore$Circle $ MNL$ is tangent to $BC$ at $L$ $\blacksquare$
$\underline{Claim:}$ $AL$ bisects $JK$.
$\underline{Proof:}$ We can easily show that $L$ is the midpoint if $BC$.
$\measuredangle FAN= \measuredangle BLN = \measuredangle LMN \Rightarrow AMNF$ is cyclic
Similarly $AMNQ$ is cyclic $\Rightarrow AMNFQ$ is cyclic.
$\measuredangle FQN=\measuredangle FAN= \measuredangle EJN \Rightarrow QF\parallel DE \parallel BC$
Since $BL=LC$, by lenght chasing we can show $JD=EK$
But $AL$ bisects $DE \Rightarrow AL$ bisects $JK$ $\blacksquare$
Now $\triangle LJK \sim \triangle LMN$
$LA$ is the $L$-median of $\triangle LJK$
$\therefore LA$ is the $L$-symmedian of$\triangle LMN$
$\therefore$ Tangents to $w$ at $M,N$ intersect on $AP$.
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Problem 12
Given an acute triangle $ABC$, point $D$ is chosen on the side $AB$ and a point $E$ is chosen on the extension of $BC$ beyond $C$. It became known that the line through $E$ parallel to $AB$ is tangent to the circumcircle of $\triangle ADC$. Prove that one of the tangents from $E$ to the circumcircle of $\triangle BCD$ cuts the angle $\angle ABE$ in such a way that a triangle similar to $\triangle ABC$ is formed.
- Anindya Biswas
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Re: Problem 12
NujhatDisha wrote: ↑Wed Jun 16, 2021 4:31 pmGiven an acute triangle $ABC$, point $D$ is chosen on the side $AB$ and a point $E$ is chosen on the extension of $BC$ beyond $C$. It became known that the line through $E$ parallel to $AB$ is tangent to the circumcircle of $\triangle ADC$. Prove that one of the tangents from $E$ to the circumcircle of $\triangle BCD$ cuts the angle $\angle ABE$ in such a way that a triangle similar to $\triangle ABC$ is formed.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- Anindya Biswas
- Posts:264
- Joined:Fri Oct 02, 2020 8:51 pm
- Location:Magura, Bangladesh
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Problem 13
Let $k$ be a fixed positive integer. Let $\mathbb{N}$ be the set of all positive integers. A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the condition \[\sum_{i=1}^{k}f^i(n)=kn\] For all $n\in\mathbb{N}$. Here $f^i(n)$ is the $i$-th iteration of $f$. Prove that $\forall n\in\mathbb{N}, f(n)=n$.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
Re: Problem 13
We will prove this by induction.Anindya Biswas wrote: ↑Thu Jun 17, 2021 11:42 pmLet $k$ be a fixed positive integer. Let $\mathbb{N}$ be the set of all positive integers. A function $f:\mathbb{N}\to\mathbb{N}$ satisfies the condition \[\sum_{i=1}^{k}f^i(n)=kn\] For all $n\in\mathbb{N}$. Here $f^i(n)$ is the $i$-th iteration of $f$. Prove that $\forall n\in\mathbb{N}, f(n)=n$.
Claim 1: $f$ is injective.
If $f(a)=f(b)$ for some $a,b \in \mathbb{N}$ then $f^i(a)=f^i(b)$ for all $i\in \mathbb{N}$. So,
\[ka=\sum_{i=1}^kf^i(a)=\sum_{i=1}^kf^i(b)=kb \Rightarrow a=b\]
Claim 2: $f(1)=1$
If for some $i$ $f^i(1)>1$ then,
\[\sum_{i=1}^kf^i(1) >k\]
So, $f^i(1)=1$ for all $i$ so, $f(1)=1$.
Claim 3: $f(i)=i$ for all $i\in \mathbb{N}$
Lets assume that $f(i)=i$ for all $1 \leq i \leq m$. Now, $f(m+1) \nleq m$ since $f$ is injective. If $f(m+1) \neq m+1$ then $f(m+1) >m+1$. And so, $f^i(m+1) \nleq m$ for any $i$ because of injectivity. Now,
\[\sum_{i=1}^kf^i(m+1) > k(m+1)\]
which is a contradiction. So, $f(m+1)=m+1$. $ \blacksquare$