## IMO Marathon

- Phlembac Adib Hasan
**Posts:**1016**Joined:**Tue Nov 22, 2011 7:49 pm**Location:**127.0.0.1-
**Contact:**

### Re: IMO Marathon

Problem 38: In a tennis tournament there are participants from $n$ different countries. Each team consists of a coach and a player whom should settle in a hotel. The rooms considered for the settlement of coaches are different from players' ones. Each player wants to be in a room whose roommates are $\textit{all}$ from countries which have a defensive agreement with the player's country. Conversely, each coach wants to be in a room whose roommates are $\textit{all}$ from countries which don't have a defensive agreement with the coach's country. Find the minimum number of the rooms such that we can $\textit{always}$ grant everyone's desire.

Source: Iran NMO 2014

Source: Iran NMO 2014

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- asif e elahi
**Posts:**183**Joined:**Mon Aug 05, 2013 12:36 pm**Location:**Sylhet,Bangladesh

### Re: IMO Marathon

Is it right?

Country $A$ has a difensive agreement with country $B$ $\rightarrow$ Country $B$ has a difensive agreement with country $A$

Country $A$ has a difensive agreement with country $B$ $\rightarrow$ Country $B$ has a difensive agreement with country $A$

- Phlembac Adib Hasan
**Posts:**1016**Joined:**Tue Nov 22, 2011 7:49 pm**Location:**127.0.0.1-
**Contact:**

- asif e elahi
**Posts:**183**Joined:**Mon Aug 05, 2013 12:36 pm**Location:**Sylhet,Bangladesh

### Re: IMO Marathon

The answer is $n+1$.We proceed by induction.

The base case is trivial.Let this is true for $n=k-1$.Now we prove this for $n=k$.

Tow countries are called friend if one has a difensive agreement with the other.

We choose a country $D$ from the n countries.Let $P$ and $C$ be it's player and coach resp.By the induction hypothesis, the rest players and coaches can stay in $k$ rooms.Let there are $m$ rooms for the players and $k-m$ rooms for coaches.

If there exist a room of players s.t all the countries of the players of that room are friends of $D$,then $P$ can stay in that room.We give another room for $C$.Then all the players and coaches can stay in $m+(k-m)+1=k+1$ rooms and we are done.

Now we consider the other case.So in every room of players,there is at least one player whose country is not a friend of $D$.So there are at least $m$ countries which are not a friend of $D$. This implies,there are at most $(k-1)-m$ countries which are friend of $D$.So by PHP,there is a room of of coaches s.t all the countries of the coaches of that room are not friends of $D$.Then $C$ can stay in that room and another room is given for $P$. So all the players and coaches can stay in $(m+1)+(k-m)=k+1$ rooms and we are done.

To prove $k+1$ is minimal,we consider the case when every $2$ countries are friends.Then all the players can stay in one room and each coach needs a single room.So they need exactly $k+1$ rooms.

So the answer is $n+1$.

The base case is trivial.Let this is true for $n=k-1$.Now we prove this for $n=k$.

Tow countries are called friend if one has a difensive agreement with the other.

We choose a country $D$ from the n countries.Let $P$ and $C$ be it's player and coach resp.By the induction hypothesis, the rest players and coaches can stay in $k$ rooms.Let there are $m$ rooms for the players and $k-m$ rooms for coaches.

If there exist a room of players s.t all the countries of the players of that room are friends of $D$,then $P$ can stay in that room.We give another room for $C$.Then all the players and coaches can stay in $m+(k-m)+1=k+1$ rooms and we are done.

Now we consider the other case.So in every room of players,there is at least one player whose country is not a friend of $D$.So there are at least $m$ countries which are not a friend of $D$. This implies,there are at most $(k-1)-m$ countries which are friend of $D$.So by PHP,there is a room of of coaches s.t all the countries of the coaches of that room are not friends of $D$.Then $C$ can stay in that room and another room is given for $P$. So all the players and coaches can stay in $(m+1)+(k-m)=k+1$ rooms and we are done.

To prove $k+1$ is minimal,we consider the case when every $2$ countries are friends.Then all the players can stay in one room and each coach needs a single room.So they need exactly $k+1$ rooms.

So the answer is $n+1$.

- asif e elahi
**Posts:**183**Joined:**Mon Aug 05, 2013 12:36 pm**Location:**Sylhet,Bangladesh

### Re: IMO Marathon

Problem $39$: Let $ABC$ be a triangle and $\omega $ be it's incircle.$\omega _{A}$ is the circle that passes thgrough $B,C$ and touches $\omega$. Similarly define $\omega _{B},\omega _{C}$. Let $\omega _{B}\cap \omega _{C}=A'$. Similarly define $B',C'$. Prove $AA',BB'$ and $CC'$ concur on $OI$. [$O$ and I is the circumcenter and incenter of $ABC$]

- Fm Jakaria
**Posts:**79**Joined:**Thu Feb 28, 2013 11:49 pm

### Re: IMO Marathon

My solution may contain typing mistakes. Anyway, nice problem.

Solution:

In this solution, I assume some lines are concurrent iff either all are parallel or all have a finite intersection point.

$AA’, BB’, CC’$ are concurrent, say at $X$; as they are respectively radical axises of the pairs $\omega_B, \omega _C ; \omega _A, \omega _C; \omega _A, \omega _B$.

Let us define

$\omega _A \cap \omega = X_A, \omega _B \cap \omega = X_B,

\omega _C \cap \omega = X_C$. Let the tangents $\ell_A, \ell_B, \ell_C$ to $\omega $

at $X_A, X_B, X_C$ form the triangle $Y_AY_BY_C$; with $\ell_b \cap \ell_c = Y_a$; and so on.

We now analyze the radical axises of $\omega _B, \omega _C, \omega$; taken pairwise.

This gives $\ell_B, \ell_C$ and $AA’$ concur; so $A,A',Y_A$ collide. Similar reasoning holds; so $AY_A, BY_B,CY_C$ are respective radical axises of $\omega _B,

\omega _C, \omega _A$ taking pairwise .

Now consider the inversion $\pi$ wrt incenter $I$ and inradius $r$. Let $I_A, I_B,I_C $ be respective excenters beyond $A,B,C$. Suppose the incircle touches $BC,CA,AB$ respectively at $D,E,F$.

Lemma 1:

$X_A, D, I_A$ collide.

Proof: Let the midpoints of $DE,DF$ be respectively $Q,P$. Then $\pi(P) = B$.

As $I_AB \perp IB(=IP)$, we have polar($P$) wrt incircle $\omega $ is $I_AB$.

Similiarly, polar($Q$) wrt $\omega $ is $I_AC$. By La Hire’s theorem, pole($PQ$) wrt $\omega $ is $I_a$.

Denote the circle $DPQ, X_APQ$ respectively by $\Omega _1, \Omega _2$.

Now $\pi$ takes circle $BX_AC$ to $\Omega _2$, $\omega$ to $\omega$ . Since $BX_AC, \omega $ are tangent, so are $\Omega _2$ and $\omega$ .

So $\ell_A$ is the radical axis of $\Omega _2$ and $\omega$. Clearly the homothety centered at D with ratio 1:2 takes $\Omega _1$ to $\omega$; so BC is the radical axis of $\omega _1, \omega$ . Also

$\Omega _1, \Omega _2$ has radical axis $PQ$. So comparing these pairwise radical axises for $\omega , \Omega _1, \Omega _2$; we get that $BC,PQ,\ell_A$ are concurrent. So wrt $\omega$ ; polar($BC$) $= D$, polar($PQ$) = $I_A$, polar($\ell_A$) = $X_A$; are collinear.

We get similar result for triples $X_B,E,I_B ; X_C,F,I_C$.

Lemma 2:

$\triangle$ $I_AI_BI_C$ and $\triangle DEF$ are homothetic with center of homothety on the line $OI$.

.

Proof: Note that $\angle AEF = 90-\angle A/2 = \angle EAI_B$. So

$I_BI_C \parallel EF$. Similar results show that $\triangle I_AI_BI_C, \triangle DEF$ are homothetic ; with center of homothety, say $Z$.

Now $\triangle$ $ABC$ is the orthic triangle of $\triangle$ $I_AI_BI_C$, so

$\triangle I_AI_BI_C$ has orthocenter $I$ and nine-point center $O$. So $OI$ is the Euler line of $\triangle I_AI_BI_C$, so it contains it’s circumcenter, say $R$.

Now $I$ is the circumcenter of $\triangle DEF$, so $Z\in RI = OI$.

Lemma 3: Let $EF \cap X_BX_C = Z_A, FD \cap X_CX_A = Z_B,

DE \cap X_AX_B = Z_C$. Then $Z_A,Z_B,Z_C$ are collinear. Further, $Z_AZ_BZ_C \perp OI$.

Proof: From lemma 1 and lemma 2, it follows that $\triangle DEF, \triangle X_AX_BX_C$ are perspective from the point $Z$. So they are in fact perspective from a line, which must be $Z_AZ_BZ_C$; by Desargues theorem.

Now applying Brocard’s theorem on cyclic quadrilateral $X_BX_CEF$; we get

$\ell$ = polar($Z$) wrt $\omega$ passes through $Z_A$. Similar reasoning yields that

$Z_AZ_BZ_C = \ell$, where by definition $\ell \perp ZI = OI$; using lemma 2.

At this point, we note that wrt $\omega$ ; polar($A$) = $EF$ and

polar($Y_A$) $= X_BX_C$.

By La Hire’s theorem , pole($AY_A$) $= Z_A$. Similiarly; pole($BY_B$) $= Z_B$,

pole($CY_C$)$ = Z_C$.

So we have that pole($Z_AZ_BZ_C$) $= X$. Then $IX\perp Z_AZ_BZ_C$.

From lemma 3; we conclude that $I,O,X$ collide.

Solution:

In this solution, I assume some lines are concurrent iff either all are parallel or all have a finite intersection point.

$AA’, BB’, CC’$ are concurrent, say at $X$; as they are respectively radical axises of the pairs $\omega_B, \omega _C ; \omega _A, \omega _C; \omega _A, \omega _B$.

Let us define

$\omega _A \cap \omega = X_A, \omega _B \cap \omega = X_B,

\omega _C \cap \omega = X_C$. Let the tangents $\ell_A, \ell_B, \ell_C$ to $\omega $

at $X_A, X_B, X_C$ form the triangle $Y_AY_BY_C$; with $\ell_b \cap \ell_c = Y_a$; and so on.

We now analyze the radical axises of $\omega _B, \omega _C, \omega$; taken pairwise.

This gives $\ell_B, \ell_C$ and $AA’$ concur; so $A,A',Y_A$ collide. Similar reasoning holds; so $AY_A, BY_B,CY_C$ are respective radical axises of $\omega _B,

\omega _C, \omega _A$ taking pairwise .

Now consider the inversion $\pi$ wrt incenter $I$ and inradius $r$. Let $I_A, I_B,I_C $ be respective excenters beyond $A,B,C$. Suppose the incircle touches $BC,CA,AB$ respectively at $D,E,F$.

Lemma 1:

$X_A, D, I_A$ collide.

Proof: Let the midpoints of $DE,DF$ be respectively $Q,P$. Then $\pi(P) = B$.

As $I_AB \perp IB(=IP)$, we have polar($P$) wrt incircle $\omega $ is $I_AB$.

Similiarly, polar($Q$) wrt $\omega $ is $I_AC$. By La Hire’s theorem, pole($PQ$) wrt $\omega $ is $I_a$.

Denote the circle $DPQ, X_APQ$ respectively by $\Omega _1, \Omega _2$.

Now $\pi$ takes circle $BX_AC$ to $\Omega _2$, $\omega$ to $\omega$ . Since $BX_AC, \omega $ are tangent, so are $\Omega _2$ and $\omega$ .

So $\ell_A$ is the radical axis of $\Omega _2$ and $\omega$. Clearly the homothety centered at D with ratio 1:2 takes $\Omega _1$ to $\omega$; so BC is the radical axis of $\omega _1, \omega$ . Also

$\Omega _1, \Omega _2$ has radical axis $PQ$. So comparing these pairwise radical axises for $\omega , \Omega _1, \Omega _2$; we get that $BC,PQ,\ell_A$ are concurrent. So wrt $\omega$ ; polar($BC$) $= D$, polar($PQ$) = $I_A$, polar($\ell_A$) = $X_A$; are collinear.

We get similar result for triples $X_B,E,I_B ; X_C,F,I_C$.

Lemma 2:

$\triangle$ $I_AI_BI_C$ and $\triangle DEF$ are homothetic with center of homothety on the line $OI$.

.

Proof: Note that $\angle AEF = 90-\angle A/2 = \angle EAI_B$. So

$I_BI_C \parallel EF$. Similar results show that $\triangle I_AI_BI_C, \triangle DEF$ are homothetic ; with center of homothety, say $Z$.

Now $\triangle$ $ABC$ is the orthic triangle of $\triangle$ $I_AI_BI_C$, so

$\triangle I_AI_BI_C$ has orthocenter $I$ and nine-point center $O$. So $OI$ is the Euler line of $\triangle I_AI_BI_C$, so it contains it’s circumcenter, say $R$.

Now $I$ is the circumcenter of $\triangle DEF$, so $Z\in RI = OI$.

Lemma 3: Let $EF \cap X_BX_C = Z_A, FD \cap X_CX_A = Z_B,

DE \cap X_AX_B = Z_C$. Then $Z_A,Z_B,Z_C$ are collinear. Further, $Z_AZ_BZ_C \perp OI$.

Proof: From lemma 1 and lemma 2, it follows that $\triangle DEF, \triangle X_AX_BX_C$ are perspective from the point $Z$. So they are in fact perspective from a line, which must be $Z_AZ_BZ_C$; by Desargues theorem.

Now applying Brocard’s theorem on cyclic quadrilateral $X_BX_CEF$; we get

$\ell$ = polar($Z$) wrt $\omega$ passes through $Z_A$. Similar reasoning yields that

$Z_AZ_BZ_C = \ell$, where by definition $\ell \perp ZI = OI$; using lemma 2.

At this point, we note that wrt $\omega$ ; polar($A$) = $EF$ and

polar($Y_A$) $= X_BX_C$.

By La Hire’s theorem , pole($AY_A$) $= Z_A$. Similiarly; pole($BY_B$) $= Z_B$,

pole($CY_C$)$ = Z_C$.

So we have that pole($Z_AZ_BZ_C$) $= X$. Then $IX\perp Z_AZ_BZ_C$.

From lemma 3; we conclude that $I,O,X$ collide.

You cannot say if I fail to recite-

the umpteenth digit of PI,

Whether I'll live - or

whether I may, drown in tub and die.

the umpteenth digit of PI,

Whether I'll live - or

whether I may, drown in tub and die.

### Re: IMO Marathon

**Solution $\boxed{39}$**

**Problem**$\boxed{40}$

If $a_1,a_2,a_3,a_4,a_5$ are positive reals bounded below and above by $p$ and $q$ respectively $\left(0<p\le q\right)$, then prove that \[\left(a_1+a_2+a_3+a_4+a_5\right)\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}+\dfrac{1}{a_5}\right)\le 25+6\left(\sqrt{\dfrac{p}{q}}-\sqrt{\dfrac{q}{p}}\right)^2\] and determine the equality cases. Deduce if this bound is strong enough.

**- What is the value of the contour integral around Western Europe?**

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

- asif e elahi
**Posts:**183**Joined:**Mon Aug 05, 2013 12:36 pm**Location:**Sylhet,Bangladesh

### Re: IMO Marathon

Such a lame problem.Nirjhor wrote:

If $a_1,a_2,a_3,a_4,a_5$ are positive reals bounded below and above by $p$ and $q$ respectively $\left(0<p\le q\right)$, then prove that \[\left(a_1+a_2+a_3+a_4+a_5\right)\left(\dfrac{1}{a_1}+\dfrac{1}{a_2}+\dfrac{1}{a_3}+\dfrac{1}{a_4}+\dfrac{1}{a_5}\right)\le 25+6\left(\sqrt{\dfrac{p}{q}}-\sqrt{\dfrac{q}{p}}\right)^2\] and determine the equality cases. Deduce if this bound is strong enough.

WLOG $p\leq a_1\leq a_2\leq a_3\leq a_4\leq a_5\leq q$

Let $f(i,j)=\left(\sqrt{\dfrac{a_i}{a_j}}-\sqrt{\dfrac{a_j}{a_i}}\right)^2$

Easy to prove that $f(i,j)+f(j,k)\leq f(i,k)+2$ for $i\leq j\leq k$

Now

\begin{align}

(\sum a_i)(\sum \frac{1}{a_i}) &=5+\sum_{1\leq i<j\leq 5}{}f(i,j)\\

&= 5+(f(1,2)+f(2,5))+(f(1,3)+f(3,5))+(f(1,4)+f(4,5))+f(1,5)+(f(2,3)+f(3,4))+f(2,4)\\

&\leq 5+20+4f(1,5)+2f(2,4)\\

&\leq 25+6\left(\sqrt{\dfrac{p}{q}}-\sqrt{\dfrac{q}{p}}\right)^2

\end{align}

### Re: IMO Marathon

Post a problem that's not lame.

**- What is the value of the contour integral around Western Europe?**

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

- asif e elahi
**Posts:**183**Joined:**Mon Aug 05, 2013 12:36 pm**Location:**Sylhet,Bangladesh

### Re: IMO Marathon

$\textbf{Problem} \text{ }\boxed{41}$ FInd all primes $p$ for which there exists $n\in \mathbb{N}$ so that

$$

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

$$

[Harder version: Replace $p$ with a general integer $m$]

$$

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

$$

[Harder version: Replace $p$ with a general integer $m$]