IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
Tahmid Hasan
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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?
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FahimFerdous
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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?
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FahimFerdous
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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
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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.
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FahimFerdous
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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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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.
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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
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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.
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