Special Problem Marathon

Discussion on Bangladesh National Math Camp
thczarif
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Problem 14

Unread post by thczarif » Sat Jun 19, 2021 12:41 pm

Given a triangle $ABC$ with the circumcircle $\omega$ and incenter $I$. Let the line pass through the point $I$ and the intersection of exterior angle bisector of $A$ and $\omega$ meets the circumcircle of $IBC$ at $T_A$ for the second time. Define $T_B$ and $T_C$ similarly. Prove that the radius of the circumcircle of the triangle $T_AT_BT_C$ is twice the radius of $\omega$.

Md Maruf Hasan
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Re: Problem 14

Unread post by Md Maruf Hasan » Sat Jun 19, 2021 3:05 pm

thczarif wrote:
Sat Jun 19, 2021 12:41 pm
Given a triangle $ABC$ with the circumcircle $\omega$ and incenter $I$. Let the line pass through the point $I$ and the intersection of exterior angle bisector of $A$ and $\omega$ meets the circumcircle of $IBC$ at $T_A$ for the second time. Define $T_B$ and $T_C$ similarly. Prove that the radius of the circumcircle of the triangle $T_AT_BT_C$ is twice the radius of $\omega$.
$\textbf{Solution :}$ Let the external angle bisector of $\angle A$ meet $\omega$ again at $T$ and let $AI \cap \omega=A,M$. Then $T, M$ are the midpoints of arc $BC$ of $\omega$ containing $A$ and not containing $A$ resp.Also $TM$ is a diameter of $\omega$.

Let $TI$ meet $\omega$ again at $T_a$. As $M$ is the center of $\odot(IBC)$ and $MT_a \perp IT_A$, $T_a$ is the midpoint of $IT_A$. Similarly,$T_b, T_c$ are midpoints of $IT_B, IT_C$ resp.So,the homothety centered at $I$ with ratio $2$ takes $\triangle T_aT_bT_c$ to $\triangle T_AT_BT_C$. Hence,the circumradius of $\triangle T_AT_BT_C$ is twice the radius of the circumcircle $\omega$ of $\triangle T_aT_bT_c$.

Md Maruf Hasan
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Re: Special Problem Marathon

Unread post by Md Maruf Hasan » Sat Jun 19, 2021 3:25 pm

$\textbf{Problem 15 :}$ Let $c, d \geqslant 2$ be positive integers.Let $\{a_n\}$ be the sequence which satisfies $a_1=c$, $a_{n+1}=a_n^{d}+c$ for every $n \geqslant 1$. Prove that for any $n \geqslant 2$, there exists a prime number $p$ such that $p \mid a_n$ and $p \nmid a_i$ for $i=1,2,\dots,n-1$.

adnan_me
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Re: Special Problem Marathon

Unread post by adnan_me » Sat Jun 19, 2021 11:25 pm

Md Maruf Hasan wrote:
Sat Jun 19, 2021 3:25 pm
$\textbf{Problem 15 :}$ Let $c, d \geqslant 2$ be positive integers.Let $\{a_n\}$ be the sequence which satisfies $a_1=c$, $a_{n+1}=a_n^{d}+c$ for every $n \geqslant 1$. Prove that for any $n \geqslant 2$, there exists a prime number $p$ such that $p \mid a_n$ and $p \nmid a_i$ for $i=1,2,\dots,n-1$.
For the sake of contradiction assume that $n$ is the smallest integer for which the given condition fails.consider some prime $ p \mid a_n$
let $m$ be the smallest integer such that $p \mid a_m$.
then we have \[a_{m+1}=a_m^d+c=c=a_1 \pmod{p}\] so it follows that \[a_n=a_{n-m} \pmod{p} \Rightarrow p \mid a_{n-m}\]
while $n-im >m$ we can similarly get $ p \mid a_{n-(i+1)m}$.by the minimality of $m$ we have that this process will terminate and we will have $n=mk$ for some $ k \in N$.\\
Redefine $m$ as any divisor of $n$ and redefine $k$ as $k=n/m$
Now assume $p^l || m$ for some prime $p$.we show that $p^l || n$.
we have $a_{m+1}=a_m^d+c \pmod{p^{l+1}}=c=a_1 \pmod{p^{l+1}}$ since $d>1$.so we will get $a_{2m+1}=a_{m+1}=a_1 \pmod{p^{l+1}} $,$a_{3m+1}=a_1 \pmod{p^{l+1}}$ and eventually $a_{(k-1)m+1}=a_1 \pmod{p^{l+1}}$ and $a_n=a_{km}=a_m \pmod{p^{l+1}}$.

So, it follows that $a_n=lcm_{i \mid n, i \not =n}a_i.$
But $a_n=a_{n-1}^d+c \Rightarrow a_n>a_{n-1}^d \Rightarrow a_n>a_i^{d^{n-i}} \Rightarrow a_i<a_n^{\frac{1}{d^{n-i}}}$
\[a_n=lcm_{i \mid n, i \not =n}a_i \le \prod_{i \mid n, i \not =n}a_i\]
\[ <a_n^{\sum_{i \mid n, i \not =n} \frac{1}{d^{n-i}}} \] \[<a_n^{1/2+1/2^2+1/2^3+.....}=a_n\] a contradiction.

adnan_me
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Re: Special Problem Marathon

Unread post by adnan_me » Mon Jun 21, 2021 1:06 pm

$\textbf{Problem 16:}$Find all functions $f : Z \to Z$ with the property that for any surjective function
$g : Z \to Z$, the function $f + g$ is also surjective.

fahim_faiaz_adib
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Re: Special Problem Marathon

Unread post by fahim_faiaz_adib » Mon Jun 21, 2021 4:56 pm

We claim f(a)=s for all a belongs to Z.where s is an integer

Clearly for the function f(a)=s for all $a$ belongs to Z,
f+g function is surjective where g is subjective.

For the sake of contradiction , we consider f(a) $\neq s $ for some a. So eventually we must get some b for which

f(b)-f(b+1) $\neq 0$. let's denote k= f(b)-f(b+1).

Now let define function g(x)=kx.
Clearly g is surjective.

Clearly f(b)+g(b) = f(b+1)+g(b+1).

Hence f+g cannot be surjective.
And we are done.

fahim_faiaz_adib
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Re: Special Problem Marathon

Unread post by fahim_faiaz_adib » Tue Jun 22, 2021 9:52 am

$\textbf{Problem 17:}$
Assume real numbers $a_i,b_i\,(i=0,1,\cdots,2n)$ satisfy the following conditions:
(1) for $i=0,1,\cdots,2n-1$, we have $a_i+a_{i+1}\geq 0$;
(2) for $j=0,1,\cdots,n-1$, we have $a_{2j+1}\leq 0$;
(2) for any integer $p,q$, $0\leq p\leq q\leq n$, we have $\sum_{k=2p}^{2q}b_k>0$.
Prove that $\sum_{i=0}^{2n}(-1)^i a_i b_i\geq 0$, and determine when the equality holds.

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Anindya Biswas
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Re: Special Problem Marathon

Unread post by Anindya Biswas » Tue Jun 22, 2021 10:13 am

fahim_faiaz_adib wrote:
Mon Jun 21, 2021 4:56 pm
Now let define function g(x)=kx.
Clearly g is surjective.
If $k\neq\pm1$ then $g$ is not surjective.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

fahim_faiaz_adib
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Re: Special Problem Marathon

Unread post by fahim_faiaz_adib » Tue Jun 22, 2021 11:33 am

Injective ar surjective gulay felsi

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