This is true for all odd prime $p$.Take $n=(p-1)^{2}$asif e elahi wrote:$\textbf{Problem} \text{ }\boxed{41}$ FInd all primes $p$ for which there exists $n\in \mathbb{N}$ so that

$$

p|n^{n+1}-(n+1)^n

$$

## IMO Marathon

### Re: IMO Marathon

"Questions we can't answer are far better than answers we can't question"

### Re: IMO Marathon

**Problem ${42}$**

The number $N$ is the product of $k$ different primes ($k \ge 3$). A and B take turns writing composite divisors of $N$ on a board, according to the following rules. One may not write $N$. Also, there may never appear two coprime numbers or two numbers, one of which divides the other. The first player unable to move loses. If $A$ starts, who has the winning strategy?

Source: St.Petersburg 1997

### Re: IMO Marathon

Too easy.

"Questions we can't answer are far better than answers we can't question"

### Re: IMO Marathon

You are also supposed to provide the next problem.tanmoy wrote:Too easy.

$\textbf{Problem}$ $\boxed{43}$ There are $n$ points in the plane such that no three of them are

collinear. Prove that the number of triangles, whose vertices are chosen from these n points and

whose area is $1$, is not greater than $\dfrac23(n^2-n)$ .

Source: Iran

### Re: IMO Marathon

rah4927 wrote:$\textbf{Problem}$ $\boxed{43}$ There are $n$ points in the plane such that no three of them are

collinear. Prove that the number of triangles, whose vertices are chosen from these n points and

whose area is $1$, is not greater than $\dfrac23(n^2-n)$ .

Source: Iran

Last edited by tanmoy on Fri Aug 19, 2016 8:35 pm, edited 2 times in total.

"Questions we can't answer are far better than answers we can't question"

### Re: IMO Marathon

$\boxed{\textbf{Problem 44}}:$Let $k$ be a nonzero natural number and $m$ an odd natural number.Prove that there exists a natural number $n$ such that the number $m^{n}+n^{m}$ has at least $k$ distinct prime factors.

$\textbf{Source}:$ Romania TST $2014$

$\textbf{Source}:$ Romania TST $2014$

"Questions we can't answer are far better than answers we can't question"

### Re: IMO Marathon

My solution uses a useful idea that is handy in increasing the number of prime factors of this type of expression.tanmoy wrote:$\boxed{\textbf{Problem 44}}:$Let $k$ be a nonzero natural number and $m$ an odd natural number.Prove that there exists a natural number $n$ such that the number $m^{n}+n^{m}$ has at least $k$ distinct prime factors.

$\textbf{Source}:$ Romania TST $2014$

$\textbf{Lemma:}$ $a^{p^k}+b^{p^k}$, where $p$ is an odd prime , has at least $k$ distinct prime factors.

$\textbf{Proof:}$ Take out all the common divisors of $a$ and $b$ and assume that $(a,b)=1$. Inducting on $k$, we get that $a^{p^k}+b^{p^k}$ has at least $k-1$ distinct prime factors since $a^{p^{k-1}}+b^{p^{k-1}}|a^{p^k}+b^{p^k}$ . So it suffices to prove that $G=\left(\dfrac{a^{p^k}+b^{p^k}}{a^{p^{k-1}}+b^{p^{k-1}}},a^{p^{k-1}}+b^{p^{k-1}}\right)=1$. This is true since $\left(\dfrac{a^n+b^n}{a+b},a+b\right)|n$ , where $n$ is odd and $(a,b)=1$. Since $G|p$, but $\dfrac{a^{p^k}+b^{p^k}}{a^{p^{k-1}}+b^{p^{k-1}}}$ is greater than $p$, so it must have another prime factor different from $p$ and we are done.

Now, let $n=p^{p^k}$ for some prime not dividing $m$ and we are done.

$\boxed{\textbf{Problem 45}}$ The isosceles triangle $\triangle ABC$, with $AB=AC$, is inscribed in the circle $\omega$. Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$, and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$, respectively.

Prove that as $P$ varies, the circumcircle of triangle $\triangle PI_BI_C$ passes through a fixed point.

Source: USAJMO

- nahin munkar
**Posts:**81**Joined:**Mon Aug 17, 2015 6:51 pm**Location:**banasree,dhaka

### Re: IMO Marathon

rah4927 wrote: $\boxed{\textbf{Problem 45}}$ The isosceles triangle $\triangle ABC$, with $AB=AC$, is inscribed in the circle $\omega$. Let $P$ be a variable point on the arc $\stackrel{\frown}{BC}$ that does not contain $A$, and let $I_B$ and $I_C$ denote the incenters of triangles $\triangle ABP$ and $\triangle ACP$, respectively.

Prove that as $P$ varies, the circumcircle of triangle $\triangle PI_BI_C$ passes through a fixed point.

Source: USAJMO

**OK.Here is my solution:**

By scaled diagram, we let $ M_A $ is the fixed point where $ M_A $ is the midpoint of arc $ \stackrel{\frown}{BC} $. We extend $ M_AP $ to $ L $ & $ I_CP $ to $ K $. As, $ PA $ bisects $ \angle I_BPI_C $ & $ \triangle APM_A $ is right angled (get from ABC is isosceles & $ M_A $ is midpoint of arc $ BC $),we get $ \angle I_BPL= \angle KPL $. So,$ M_AP $ bisects $\angle I_BPI_C $ externally...

**(1).**

Now,we define two points $ M_B $ , $ M_C $ as the midpoint of arc $ AC,AB $ resp. Here,$ P,I_C,M_B $ & $ P,I_B,M_C $ both collinear from well-known incenter lemma. Now we get again using it, $ M_BI_C=M_BA=M_CA=M_CI_B $...@. & It's obvious that, $ M_AM_B=M_AM_C $...(b). & it can be seen easily, $ \angle M_AM_CI_B = \angle M_AM_BI_C $....(c). SO,from @,(b),(c), $ \triangle M_AM_CI_B \cong \triangle M_AM_BI_C $ . So, $ M_AI_B=M_AI_C. $ From (1), we get, $ M_A $ is the midpoint of arc $ I_BI_C $ of $ \bigodot PI_BI_C $ (here just the angle is externally bisected of a well-known lemma).Now, we can conclude as $ M_A $ lies on the circle.

**SO,**$ \bigodot PI_BI_C $

**passes through a fixed point**.....

**(proved)**

**# Mathematicians stand on each other's shoulders**.

*~ Carl Friedrich Gauss*

### Re: IMO Marathon

$\boxed{\textbf{Problem 46}}$ Let $\leftarrow$ denote the left arrow key on a standard keyboard. If one opens a text editor and types the keys "ab$\leftarrow$ cd $\leftarrow \leftarrow$ e $\leftarrow \leftarrow$ f", the result is "faecdb". We say that a string $B$ is reachable from a string $A$ if it is possible to insert some amount of $\leftarrow$'s in $A$, such that typing the resulting characters produces $B$. So, our example shows that "faecdb" is reachable from "abcdef".

Prove that for any two strings $A$ and $B$, $A$ is reachable from $B$ if and only if $B$ is reachable from $A$.

Source : USA TSTST

Prove that for any two strings $A$ and $B$, $A$ is reachable from $B$ if and only if $B$ is reachable from $A$.

Source : USA TSTST

### Re: IMO Marathon

Source : ARO (not sure)

Last edited by rah4927 on Sun Aug 28, 2016 8:52 pm, edited 2 times in total.