## IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm

### Re: IMO Marathon

asif e elahi wrote:$\textbf{Problem} \text{ }\boxed{41}$ FInd all primes $p$ for which there exists $n\in \mathbb{N}$ so that
$$p|n^{n+1}-(n+1)^n$$
This is true for all odd prime $p$.Take $n=(p-1)^{2}$
"Questions we can't answer are far better than answers we can't question"

rah4927
Posts: 108
Joined: Sat Feb 07, 2015 9:47 pm

### 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

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm

### Re: IMO Marathon

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

rah4927
Posts: 108
Joined: Sat Feb 07, 2015 9:47 pm

### Re: IMO Marathon

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

$\textbf{Problem}$ $\boxed{43}$ 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

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm

### 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"

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm

### 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 $2014$
"Questions we can't answer are far better than answers we can't question"

rah4927
Posts: 108
Joined: Sat Feb 07, 2015 9:47 pm

### Re: IMO Marathon

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 $2014$
My solution uses a useful idea that is handy in increasing the number of prime factors of this type of expression.

$\textbf{Lemma:}$ $a^{p^k}+b^{p^k}$, 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

rah4927
Posts: 108
Joined: Sat Feb 07, 2015 9:47 pm

### 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

rah4927
Posts: 108
Joined: Sat Feb 07, 2015 9:47 pm

### Re: IMO Marathon

$\boxed{\text{Problem 47}}$ In a $100\times 100$ board, you put right angled isosceles triangles in some unit squares, such that every side (including the ones on the boundary) of each unit square is a leg of some isosceles triangle. What is the maximum number of unit squares with no triangles ?

Source : ARO (not sure)
Last edited by rah4927 on Sun Aug 28, 2016 8:52 pm, edited 2 times in total.