IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
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Masum
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Re: IMO Marathon

Unread post by Masum » Fri Dec 14, 2012 5:34 pm

আমি এই দোষে অভিযুক্ত হইতে চাই না। তাই ম্যারাথনের নিয়ম এই যে, এক সমস্যা দুই দিন যাবত সমাধান না হইলে উহা বাদ দিয়ে পরবর্তী সমস্যাতে চলিয়া যাইতে হইবে।
বি।দ্র। আমি গুরুচণ্ডালী দোষে দূষিত হইয়া গিয়াছি।
One one thing is neutral in the universe, that is $0$.

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zadid xcalibured
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Re: IMO Marathon

Unread post by zadid xcalibured » Wed Dec 19, 2012 3:40 pm

To revive the marathon here goes a problem.
Problem $14$:Let $a$ and $b$ be positive integers such that $a^n +n$ divides $b^n +n$ for
every positive integer $n$. Show that $a = b$.

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Masum
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Re: IMO Marathon

Unread post by Masum » Wed Dec 19, 2012 4:27 pm

Nice problem. BTW, some more hints for the previous problem. Note that if we set $y=0,f(x^2)=0$. So, target $f(x^2-y^2)$ so that $x^2-y^2=z^2$ and one of $f(x)f(y)$ must be $0$. For example, choose $x=a+1,y=2a$. Write the equation as \[f(ab)=f(\dfrac{a+b}2)f(\dfrac{a-b}2)\]
One one thing is neutral in the universe, that is $0$.

sourav das
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Re: IMO Marathon

Unread post by sourav das » Tue Jan 15, 2013 7:21 pm

Sorry for being late to attend the marathon.
For problem 14
Solution:
Take prime $p$ such that $p$ is co-prime to both $a,b$. Then take $n=p-1$ and use Fermat's little theorem to prove $p|a-b$. But there are infinite $p$ primes co-prime to $a-b$. implies $a=b$.

P.S.:: My solution is identical to official solution. I couldn't find any different solution. You are welcome to give another one.
For problem 13 (Masum Bhai's problem)
Solution:
Let $P(x,y)$ be the statement. $P(1,0)$ implies $f(1)=f(1)f(0)$..........(K)
Consider cases:
i)$f(0)=0$
Then, $f(1)=0$ . Now, $P(x,x-1)$ implies $f(2x-1)=f(x-1)f(x)................(*)$
Simple induction to prove $f(2n-1)=0$ for all $n\in \mathbb N$ as one of $x,x-1$ is odd.
Also $f(4n)=f(n-1)f(n+1)$................(**) .So, $f(4)=0$. We will use induction to prove $f(4n)=0$. Now note that either $n-1,n+1$ both odd or one of them is divisible by $4$. It implies $f(4n)$ must be 0..........(a).
Now, if $f(4n-2)$ is $0$ for all $n\in \mathbb N$ then $f(n)=0$ for all $n$..........(Solution-1)

Otherwise, let $f(4k-2)$ is non-zero for some $k \in \mathbb N$. Then, $P(4k-2,4k'-2)$ implies $f(4k'-2)=0$ for $k'<k$ (using (a) because here $x^2-y^2$ is divisible by 4) and $P(4k''-2,4k-2)$ implies $f(4k''-2)=0$ for $k''>k$ using (a). It implies $f(4k-2)$ is non-zero for some $k\in \mathbb N$ and $f(n)=0$ for all other $n$...........(Solution-2)

ii)$f(0)$ is non-zero but $f(1)=0$
Same way in case(i), prove that $f(2n-1)=0$, for all $n\in \mathbb N$
Now since $(2n+1)^2-(2n-1)^2=(n+2)^2-(n-2)^2$; Using $P$ we will get $f(2n+1)f(2n-1)=f(n+2)f(n-2)$........(***)
So, $f(4)=0$ using (***) and same way in case (i), $f(4n)=0$ for all $n\in \mathbb N$.
Now, $P(4n-2,0)$ implies $f((4n-2)^2)=f(4n-2)f(0)$. But it implies $f(4n-2)=0$ from previous conditions.
So $f(0)$ is non zero and $f(n)=0$ for all $n\in \mathbb N$..............(Solution-3)

iii)Both $f(0),f(1)$ are non-zero. But $f(2)=0$
Then $f(0)=1$ using $K$. Use (*) to get $f(3)=0$ and $f(5)=0$. Again use induction to get $f(2n+1)=0$ for all $n\in \mathbb N$. Use (***) setting $n=2$ to get $f(4)=0$. Again use induction to get $f(4n)=0$ for all $n\in \mathbb N$. Now $P(4n-2,0)$ implies $f((4n-2)^2)=f(4n-2)=0$ as $4|(4n-2)^2$.
So $f(0)=1$, $f(1)$ non-zero and $f(n)=0$ for all other $n\in \mathbb N$............(Solution-4)

iv)$f(0),f(1),f(2)$ are non-zero. Then $f(0)=1$.Let $f(1)=a,f(2)=b$.
Use (*) to get $f(3)=ab,f(5)=ab^2$ and use (**) to get $f(4)=a$ But using (***)
$f(4)=f(3)f(5)$ implies $a=a^2b^3$ and so $a=b=1$
Now use (*) in (***) to get $f(n+2)f(n-2)=f(n-1)f(n)^2f(n+1)$ and use induction to get $f(n)=1$ for all $n \in \mathbb N_0$..................(Solution-5)

(Done)

Check: $x^2-y^2$ is divisible by $4$ or it is an odd number (Then one of $x,y$ is odd). Now it's easy to check every solution is a perfect match.

P.S.: I'm tired with long case studies but I always try to give full solution. Many people try to skip many parts and fall into traps. I suggest everyone to give complete solution. If you find any bug , inform me. I hope for any other smart solution.
I request everyone to post another problem and be regular again.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Tue Jan 15, 2013 8:11 pm

Problem $15$: Let $ABC$ be a non-isosceles triangle. $(I)$, the incircle of $\triangle ABC$ touches $BC,CA,AB$ at $D,E,F$ respectively.The line through $E$ perpendicular to $BI$ cuts $(I)$ again at $K$.The line through $F$ perpendicular to $CI$ cuts $(I)$ again at $L$. $J$ is midpoint of $KL$.
(a).Prove that $D,I,J$ are collinear.
(b). $B,C$ are fixed points, $A$ moves such that $\frac{AB}{AC}=k$ with $k$ constant. $IE,IF$ cut $(I)$ again at $M,N$ respectively. $MN$ cuts $IB,IC$ at $P,Q$ respectively.Prove that the perpendicular bisector of $PQ$ goes through a fixed point.
Source: Vietnam 2013-3.
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sourav das
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Re: IMO Marathon

Unread post by sourav das » Wed Jan 16, 2013 5:13 pm

Part (a)
$K,L$ are the reflection of ($E$ on $BI$), ($F$ on $CI$) respectively. But $D$ is the reflection of ($E$ on $CI$) and ($F$ on $BI$). So, $\angle EDF = \angle KFD= \angle KLD$ and also $\angle FDE= \angle LED= \angle LKD$ which implies $DL=DK$ and so $DI$ is the perpendicular bisector of $KL$. So, $D,I,J$ collinear.
Part (b)
Let $AI$ meets $BC$ at $L$. Then $L$ is fixed. Let $W$ is the midpoint of $PQ$ and perpendicular bisector intersect $BC$ at $J$. Define $CI\cap EF =O, BI\cap EF =H$ . Angle chase will imply $\angle FOI= \frac {1}{2}\angle B$ . So, $IFOB$ cyclic. So, $BO \perp CO$. Similarly we can prove that $CH \perp BH$ . So $BC$ is the diameter of circle $BCHO$. If $Z,R$ are midpoints of $BC,HO$ respectively, then $ZR$ is perpendicular to $HO$ . But note that $(Q,N,M,W,P)$ are reflection of $(O,F,E,R,H)$ with respect to $I$. But as $ZR||IL||WZ$ and $IW=IR$, so $L$ is the midpoint of $JZ$ . As $L,Z$ are fixed, so $J$ the fixed point.
Anyone can take my turn
Attachments
Capture.PNG
Part(b)
Capture.PNG (57.32 KiB) Viewed 2853 times
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Wed Jan 16, 2013 9:55 pm

Problem $16$: Let $ABC$ be an acute triangle. $(O)$ is circumcircle of $\triangle ABC$. $D$ is on arc $BC$ not containing $A$. Line $\ell$ moves through $H$ ($H$ is the orthocenter of $\triangle ABC$) cuts $\odot ABH,\odot ACH$ again at $M,N$ respectively.
(a). Find $\ell$ such that the area of $\triangle AMN$ is maximum.
(b). $d_{1},d_{2}$ are the lines through $M$ perpendicular to $DB$, the line through $N$ perpendicular to $DC$ respectively. $d_{1}$ cuts $d_{2}$ at $P$. Prove that $P$ moves on a fixed circle as $\ell$ varies.
Source: Vietnam 2013-6.
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SANZEED
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Re: IMO Marathon

Unread post by SANZEED » Thu Jan 17, 2013 1:06 am

Problem:$16a$
The area of $\triangle AMN$ will be maximum if $\ell$ is parallel to $BC$.
Proof:Let $\ell'$ be another line passing through $H$ and intersecting the circles $ABH,ACH$ at $M',N'$ respectively,while we are assuming that $\ell \parallel BC$. Now $\angle AMN=\angle AMH=\angle AM'H=\angle AM'N'$ and $\angle ANM=\angle ANH=\angle AN'H=\angle AN'M'$. So $\triangle AMN\sim \triangle AM'N'$.
Now $\frac{\triangle AMN}{\triangle AM'N'}=\frac{AM^{2}}{AM'^{2}}$ so the area of $\triangle AMN$ will be maximum if $AM$ is maximum. And $AM$ is maximum if it is a diameter of circle $ABH$. If so, then $\angle AHB=90^{\circ}$ which means that $\ell \parallel BC$.

I am still in problem with part b. Is the question written here correct? :oops: :oops:
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

sourav das
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Re: IMO Marathon

Unread post by sourav das » Thu Jan 17, 2013 1:24 pm

Tahmid Hasan wrote:Problem $16$: Let $ABC$ be an acute triangle. $(O)$ is circumcircle of $\triangle ABC$. $D$ is on arc $BC$ not containing $A$. Line $\ell$ moves through $H$ ($H$ is the orthocenter of $\triangle ABC$) cuts $\odot ABH,\odot ACH$ again at $M,N$ respectively.
(a). Find $\ell$ such that the area of $\triangle AMN$ is maximum.
(b). $d_{1},d_{2}$ are the lines through $M$ perpendicular to $DB$, the line through $N$ perpendicular to $DC$ respectively. $d_{1}$ cuts $d_{2}$ at $P$. Prove that $P$ moves on a fixed circle as $\ell$ varies.
Source: Vietnam 2013-6.
Solution:
Problem (a)
Same as SANZEED
Problem (b)
Let $l'_1\perp DB, l'_2\perp DC$ through $H$ intersecting circle $ACH,ABH$ at $F,E$ respectively.
Let $m_1\cap m_2=A'$ where $m_1,m_2$ are perpendicular to $DB,DC$ from $C,B$ respectively.So $D$ is the orthocenter of $A'BC$. Now $A'D||AH,A'B||EH,A'C||FH$ . Also note that $\angle A'=180-\angle D= \angle A$ . Now using cyclic property, $90-\angle A = \angle AEH= \angle AFH$ . But $\angle A'BD= \angle A'CD = 90 -\angle A$
. That's why $\triangle A'CB$ and $\triangle HEF$ are homothetic.

Note that, radius of circle $AHB,AHC$ are equal (Use sine law to prove they are equal to radius of circle $ABC$)
Now, let $ME\cap NF = P'$. Then angle chase will imply $\angle MP'N= \angle A$. Again as $MP\perp EA$ and $NP \perp FA$ so, $\angle MPN= \angle A$. It implies that, $\angle EMP=\angle FNP$...(i). Also $\frac{MP}{NP}=\frac{sin\angle MNP}{sin \angle NMP}= \frac{sin \angle MHE}{ sin \angle NHF}$ (As,$HE||A'B||PN$ and $HF||A'C||PM$)
But $\frac{sin \angle MHE}{ sin \angle NHF}=\frac{ME}{NF}$ because of two equal radius.

So,$\frac{MP}{NP}=\frac{ME}{NF}$....(ii)
Using (i),(ii); a spiral similarity with center $P$ and angle $\angle A$ sends ($M,E$) to ($N,F$)
That means $\angle EPF = \angle A$. So, locus of $P$ is circle $EAF$
(Proved)
Anyone can take my turn
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Thu Jan 17, 2013 5:06 pm

Problem $17$: Let $ABCD$ be a trapezoid with parallel sides $AB>CD$. Points $K$ and $L$ lie on the line segments $AB$ and $CD$, respectively, so that $\frac {AK}{KB}=\frac {DL}{LC}$. Suppose that there are points $P$ and $Q$ on the line segment $KL$ satisfying $\angle APB=\angle BCD$ and $\angle CQD=\angle ABC$. Prove that the points $P,Q,B$ and $C$ are concyclic.
Source: IMO 2006-G2.
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