## IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: IMO Marathon

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 $0$.

Posts: 217
Joined: Thu Oct 27, 2011 11:04 am
Location: mymensingh

### Re: IMO Marathon

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 $a = b$.

Masum
Posts: 592
Joined: Tue Dec 07, 2010 1:12 pm

### Re: IMO Marathon

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
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
Location: Dhaka
Contact:

### Re: IMO Marathon

Sorry for being late to attend the marathon.
For problem 14
Solution:
For problem 13 (Masum Bhai's problem)
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......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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

### Re: IMO Marathon

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.
বড় ভালবাসি তোমায়,মা

sourav das
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
Location: Dhaka
Contact:

### Re: IMO Marathon

Part (a)
Part (b)
Anyone can take my turn
Attachments
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......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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

### Re: IMO Marathon

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.
বড় ভালবাসি তোমায়,মা

SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm

### Re: IMO Marathon

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?
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

sourav das
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
Location: Dhaka
Contact:

### Re: IMO Marathon

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)
Problem (b)
Anyone can take my turn
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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