## IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: IMO Marathon

FahimFerdous wrote:$Problem 29$:
Fixed points $B$ and $C$ are on a fixed circle $w$ and point $A$ varies on this circle. We call the midpoint of arc $BC$ (not containing $A$) $D$ and the orthocenter of the triangle $ABC$, $H$. Line $DH$ intersects circle $w$ again in $K$. Tangent in $A$ to circumcircle of triangle $ABC$ intersects line $DH$ and circle $w$ again in $L$ and $M$ respectively. Prove that the value of $\frac{AL}{AM}$ is constant.
How can a tangent intersect a circle at two different points? বড় ভালবাসি তোমায়,মা

FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am

### Re: IMO Marathon

Really sorry for the typo. It'd be circumcircle of triangle $ABH$. Forgive me. :-/

Tahmid Hasan
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### Re: IMO Marathon

Tahmid Hasan wrote:
FahimFerdous wrote:Problem $\boxed {28}$:

Trapezoid $ABCD$, with $AB$ parallel to $CD$, is inscribed in circle $w$ and point $G$ lies inside triangle $BCD$. Rays $AG$ and $BG$ meet $w$ again at points $P$ and $Q$, respectively. Let the line through $G$ parallel to $AB$ intersect $BD$ and $BC$ at points $R$ and $S$, respectively. Prove that quadrilateral $PQRS$ is cyclic if and only if $BG$ bisects $\angle CBD$.
My solution: 'If' part: $\angle BQC=\angle BDC=\angle BRG,\angle CBQ=\angle RBG$.
So $\triangle BCQ \sim \triangle BGR$.
Hence $\frac{BC}{BG}=\frac{BQ}{BR} \Rightarrow \frac{BC}{BQ}=\frac{BG}{BR}$.....$(1)$
Since $\angle GBC=\angle QBR$ we conclude from this and $(1),\triangle BCG \sim \triangle BQR \Rightarrow \angle BCG=\angle BQR$.....$(2)$
Simlarly we can prove $\triangle BDQ \sim \triangle BGS \Rightarrow \triangle BDG \sim \triangle BQS \Rightarrow \angle BDG=\angle BQS$.....$(3)$
Now $\angle PCS=\angle PCB=\angle PAB=\angle PGS \Rightarrow PGCS$ is concyclic.
Again $\angle PDR=\angle PDB=\angle PAB=\angle AGR \Rightarrow PGRD$ is concyclic.
Now $\angle SPR=\angle SPG+\angle RPG=\angle SCG+\angle GDR=\angle BCG+\angle BDG=\angle BQR+\angle BQS=\angle SQR$
Hence $PQRS$ is concyclic.
'Only if' part: $CD \parallel RS \Rightarrow \omega,\odot BRS$ are homothetic with centre $B$ which implies they have a common tangent at $B$.
So $PQ,RS,BB$ are tangent at some point $X$ which is the radical centre of $\omega,\odot BRS,\odot PQRS$.
Now $\angle PBX=\angle PQB=\angle PAB=\angle PGX \Rightarrow PGBX$ is concyclic.
So $\angle QPG=\angle XBG \Rightarrow \angle QBA =\angle QAB \Rightarrow QA=QB$
Hence $Q$ lies on the perpendicular bisector of $AB$. Since $AB \parallel CD$, $Q$ lies on the perpendicular bisector of $CD$ too.
So $Q$ is the midpoint of minor arc $CD$ which implies $BG$ bisects $\angle CBD$.
I just figured out the first part of my solution can also be solved the way I solved the second part:
Let $PQ \cap BB=X$. Since $BG$ bisects $\angle CBD$, $Q$ lies on the perpendicular bisector of $AB$.
So $\angle QBA=\angle QAB \Rightarrow \angle QPA=\angle QBX \Rightarrow PGBX$ is cyclic.
Now $\angle PGS=\angle PAB=\angle PBX=\angle PGX \Rightarrow X \in GS$.
$BX$ is tangent to both $\omega,\odot BGS$.
So $XB^2=XS.RS$ and $XB^2=XP.XQ$ hence $XP.XQ=XS.XR$; so $PQRS$ is cyclic.
And @Fahim vai, could you please mention the source? Because I once solved a similar Iran NMO problem which involved the circumcircle of $\triangle AKH$. Can you check? বড় ভালবাসি তোমায়,মা

FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am

### Re: IMO Marathon

Yeah, Tahmid, it's from Iran NMO. Well, as you've already done it, wait to post your solution. If no one posts the solution, then post yours. Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: IMO Marathon

Here's the correct problem statement:
$Problem 29$:
Fixed points $B$ and $C$ are on a fixed circle $w$ and point $A$ varies on this circle. We call the midpoint of arc $BC$ (not containing $A$) $D$ and the orthocenter of the triangle $ABC$, $H$. Line $DH$ intersects circle $w$ again in $K$. Tangent in $A$ to circumcircle of triangle $AKH$ intersects line $DH$ and circle $w$ again in $L$ and $M$ respectively. Prove that the value of $\frac{AL}{AM}$ is constant.
$\angle KAL=\angle KHA, \angle KAL=\angle KDM$.
So $\angle KHA=\angle KDM$.
So $AH \parallel DM$. Since $AH \perp BC$, we get $MD \perp BC$
which implies $M$ is the midpoint of arc $BAC$ and $MD$ is a diameter.
Now $\frac{LM}{AL}=\frac{MD}{AH} \Rightarrow \frac{AM}{AL}=\frac{MD-AH}{AH}$[Dividendo]
so $\frac{AL}{AM}=\frac{AH}{MD-AH}=\frac{2R\cos A}{2R-2R\cos A}=\frac{\cos A}{1-\cos A}$
which is indeed independent of $A$.
Someone else post the next problem.
বড় ভালবাসি তোমায়,মা

FahimFerdous
Posts: 176
Joined: Thu Dec 09, 2010 12:50 am

### Re: IMO Marathon

Problem 30:

In triangle $ABC$, $w$ is its circumcircle and $O$ is the center of this circle. Points $M$ and $N$ lie on sides $AB$ and $AC$ respectively. $w$ and the circumcircle of triangle $AMN$ intersect each other for the second time in $Q$. Let $P$ be the intersection point of $MN$ and $BC$. Prove that $PQ$ is tangent to $w$ iff $OM=ON$.

Tahmid Hasan
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Joined: Thu Dec 09, 2010 5:34 pm

### Re: IMO Marathon

FahimFerdous wrote:Problem 30:

In triangle $ABC$, $w$ is its circumcircle and $O$ is the center of this circle. Points $M$ and $N$ lie on sides $AB$ and $AC$ respectively. $w$ and the circumcircle of triangle $AMN$ intersect each other for the second time in $Q$. Let $P$ be the intersection point of $MN$ and $BC$. Prove that $PQ$ is tangent to $w$ iff $OM=ON$.
Note that $AQ$ is the common chord of $w$ and $\odot AMN$, so $O$ lies on the prendicular bisector of $AQ$.
Let $PQ \cap AC=R$.
Notice that a spiral similarity with centre $Q$ sends $MN$ to $BC$.
So $\triangle QMN \sim \triangle QBC \Rightarrow \angle QNM=\angle QCB \Rightarrow \angle QNP=\angle QCP$
$\Rightarrow QNCP$ is concyclic.
So $\angle QPM=\angle QCN$, but $\angle QCN=\angle QBM$.
Hence $\angle QPM=\angle QBM$
Now $PQ$ is tangent to $w \Leftrightarrow \angle RQA=\angle QBM \Leftrightarrow \angle RQA=\angle QPM \Leftrightarrow AQ \parallel MN$
$\Leftrightarrow QRNM$ is a cyclic trapezoid $\Leftrightarrow O \in$ the perpendicular bisector of $MN$ $\Leftrightarrow OM=ON$. Done!
Someone else post a new problem.
বড় ভালবাসি তোমায়,মা

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### Re: IMO Marathon

Problem 31:A function $f:\mathbb N\to \mathbb N$ satisfies the following conditions:
1. For any two integers $a,b$ with $\gcd(a,b)=1$, $f(ab)=f(a)f(b)$
2. For any two primes $p,q$, $f(p+q)=f(p)+f(q)$
Prove that, $f(3)=3\;$ and $f(1999)=1999.$

Source: France TST 2000

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### Re: IMO Marathon

Oy Adib,are these primes $p$,$q$ distinct???????

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: IMO Marathon

Phlembac Adib Hasan wrote:Problem 31:A function $f:\mathbb N\to \mathbb N$ satisfies the following conditions:
1. For any two integers $a,b$ with $\gcd(a,b)=1$, $f(ab)=f(a)f(b)$
2. For any two primes $p,q$, $f(p+q)=f(p)+f(q)$
Prove that, $f(3)=3\;$ and $f(1999)=1999.$

Source: France TST 2000
Let $P(x,y) \Rightarrow f(xy)=f(x)f(y), Q(x,y) \Rightarrow f(x+y)=f(x)+f(y)$ satisfying the given conditions.
$P(2,3) \Rightarrow f(6)=f(2)f(3), Q(3,3) \Rightarrow f(6)=2f(3)$ which implies $f(2)=2$.
$Q(2,2) \Rightarrow f(4)=4$.
$P(3,4) \Rightarrow f(12)=f(3)f(4)=4f(3)$
$Q(5,7) \Rightarrow f(12)=f(5)+f(7)=f(2)+f(3)+f(2)+f(5)$
$=f(2)+f(3)+f(2)+f(2)+f(3)=6+2f(3)$.
So $4f(3)=6+2f(3) \Rightarrow f(3)=3$.
$Q(2,3) \Rightarrow f(5)=f(2)+f(3)=5$.
$P(2,3) \Rightarrow f(6)=f(2)f(3)=6$
$Q(2,5) \Rightarrow f(7)=f(2)+f(5)=7$
$Q(11,3) \Rightarrow f(14)=f(11)+f(3) \Rightarrow f(2)f(7)=f(11)+f(3) \Rightarrow f(11)=11$.
$Q(17,5) \Rightarrow f(22)=f(17)+f(5) \Rightarrow f(2)f(11)=f(17)+f(5)$.
$Q(59,7) \Rightarrow f(66)=f(59)+f(7) \Rightarrow f(11)f(6)=f(59)+f(7) \Rightarrow f(59)=59$.
Note that $1999$ is a prime, now $Q(1999,7) \Rightarrow f(2006)=f(1999)+f(7)$.
But $f(2006)=f(2)f(1003)=f(2)f(17)f(59)=2.17.59=2006$.
So $f(1999)=1999$.
Someone else post the next problem.
বড় ভালবাসি তোমায়,মা