## IMO Marathon

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

Solution $\boxed {19}$:Same as Tahmid.
Solution $\boxed {20}$:Actually $APIE$ is concyclic .Which implies $\angle {A}$=$90$
Problem $19$ seemed harder to me than problem $20$.
Last edited by zadid xcalibured on Tue Jan 22, 2013 8:16 pm, edited 1 time in total.

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### Re: IMO Marathon

$\boxed {20}$

Let perpendicular bisector of $BC$ intersect circle $(ABC)$ at $T$ and $F$ ,
Its wellknown that $A,I,F$ are colinear .
Now $AE||BC$ imply $TF \perp AE$ , Again $IE \perp AE$
So $IE || TF$

Now $\angle PAI=\angle PAF=\angle PTF=\angle PEI$ imply $A,P,I,E$ cyclic .
So $API=90$ .

Note that $\angle IPB+\angle API+\angle C=180$
If $\angle IPB=\angle B$

Then $\angle B+90+\angle C=180$
or $\angle A=90$
$\frac{1}{0}$

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### Re: IMO Marathon

$\boxed {19}$
My Solution . (I did post this because it seemed interesting to me)

Let $l',l_p,l_q$ be the line that is tangent to $\omega$ at $B,P,Q$ respectively.

Now $l' \cap PQ=I , l_p \cap BQ=G, l_q \cap BP=H, l_p \cap l_q=J$ ,
Its well known that $G,H,I$ are colinear.

Let $BJ \cap IG=K,BJ \cap AC=M'$
Note that $G,K,H,I$ is a harmonic division .
And pencil $B(G,K,H,I)$ imply $D,M',C, ∞$ is a harmonic division which imply $M'=M$
$\frac{1}{0}$

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### Re: IMO Marathon

Tahmid Hasan wrote: Problem $20$: In the triangle $ABC$, $\angle B$ is greater than $\angle C$. $T$ is the midpoint of the arc $BAC$ of the circumcircle of $ABC$ and $I$ is the incenter of $ABC$. $E$ is a point such that $\angle AEI=90^\circ$ and $AE\parallel BC$. $TE$ intersects the circumcircle of $ABC$ for the second time in $P$. If $\angle B=\angle IPB$, find the angle $\angle A$.
Source: Iran TST-2008-10.
মোস্ট প্রোবাবলি এটা '১২র মেইন ক্যাম্পের কোন পরীক্ষার চার নাম্বার প্রশ্ন ছিল। ঐ সময় সল্ভ করতে গিয়ে আমি হিসাব ভুল করেছিলাম। $90-\angle A/2-\angle C=\angle B/2+\angle C/2$ লিখে ভুল অ্যাঙ্গেল চেস করেছিলাম তাই মিলছিল না। আর আজকে ফিগার ভুল আঁকলাম। পুরো এক ঘণ্টা নষ্ট।
যাকগে, আমার সমাধান দেই।
Definitions:
$F$, from Nadim vai's proof.
$AF\cap BC=D$.

$\angle EAI=\angle IDB=\angle C+\angle A/2$.
Since $\angle IEA=90$, it follows that $\angle AIE=\angle B/2-\angle C/2$
$\angle TPA=\angle TFA=\angle B/2-\angle C/2$
So $A,P,I,E$ concyclic. The rest is like Nadim vai. So no need to post.
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### Re: IMO Marathon

Problem 21:Let $ABC$ be a triangle with $P$ in its interior(with $BC \neq AC$).The lines $AP,BP,CP$ meet $\odot{ABC}$ again at $K,L,M$.The tangent line at $C$ intersects $AB$ at $S$.Show that from $SC=SP$ it follows that $MK=ML$
Source: IMO 2010-Problem: 4
Last edited by Phlembac Adib Hasan on Wed Jan 23, 2013 8:15 pm, edited 2 times in total.
Reason: to give the problem no. and the source.

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### Re: IMO Marathon

I wish every geometry problems were as easy and as beautiful as this.

Tahmid Hasan
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### Re: IMO Marathon

zadid xcalibured wrote:Let $ABC$ be a triangle with $P$ in its interior(with $BC \neq AC$).The lines $AP,BP,CP$ meet $\odot{ABC}$ again at $K,L,M$.The tangent line at $C$ intersects the $\odot{ABC}$ at $S$.Show that from $SC=SP$ it follows that $MK=ML$
I think think there's a slight mistake with the problem statement- $S$ should be the intersection of $AB$ and the tangent at $C$.
From power of point $SA.SB=SC^2=SP^2 \Rightarrow \angle APS=\angle ABP=\angle AKL \Rightarrow SP \parallel KL$.
$MK=ML \leftrightarrow \angle MLK=\angle MKL$
$\leftrightarrow \angle MLB+\angle BLK=\angle MCL$
$\leftrightarrow \angle MCB+\angle SPL=\angle MCS-\angle LCS$
$\leftrightarrow \angle MCB+\angle SPL=\angle SPC-\angle LBC$
$\leftrightarrow \angle MCB+\angle LBC=\angle SPC-\angle SPL$
$\leftrightarrow \angle LPC=\angle LPC$
which is indeed true.
zadid xcalibured wrote:I wish every geometry problems were as easy and as beautiful as this.
Great problem for learning angle chasing
Phelembac Adib Hasan wrote:আর আজকে ফিগার ভুল আঁকলাম। পুরো এক ঘণ্টা নষ্ট।
আসলে নষ্ট বলা ঠিক হবে না, এমন ভুল আর কখনো হবে না ফলে বড় কোন পরীক্ষায় বেকায়দায় পড়তে হবে না।
Someone else post a new problem.
বড় ভালবাসি তোমায়,মা

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### Re: IMO Marathon

I Love Ratio.
$\frac{SC}{SB}=\frac{SA}{SC}$ $\Longrightarrow \frac{SP}{SB}=\frac{SA}{SP}$
$\triangle{SPB} \sim \triangle{SAP}$ and $\triangle{SCB} \sim \triangle{SAC}$
$\frac{AP}{BP}=\frac{AS}{PS}=\frac{AS}{CS}=\frac{AC}{BC}$
from alternate segment theorem,
$\frac{MK}{MP}=\frac{AC}{AP}$ and $\frac{ML}{MP}=\frac{BC}{BP}$
from this the result follows.

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### Re: IMO Marathon

I propose a new rule that anyone who posts a solution of a problem must certainly post another problem.Or we'll have to wait for someone's grace.
Problem $\boxed{22}$:Let $ABCD$ be a parallelogram.A variable line $l$ passing through the point $A$ intersects the rays $BC$ and $CD$ at points $X$ and $Y$ respectively.Let $K$ and $L$ be the centres of the ex-circles of triangle $\triangle{ABX}$ and $\triangle{ADY}$ touching the sides $BX$ and $DY$ respectively.Prove that $\angle{KCL}$ doesn't depend on the choice of $l$.
Source: IMO Shortlist 2005 G3
Last edited by Phlembac Adib Hasan on Wed Jan 23, 2013 8:44 pm, edited 1 time in total.
Reason: To give the source

Tahmid Hasan
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### Re: IMO Marathon

zadid xcalibured wrote:I propose a new rule that anyone who posts a solution of a problem must certainly post another problem.Or we'll have to wait for someone's grace.
Problem $\boxed{22}$:Let $ABCD$ be a parallelogram.A variable line $l$ passing through the point $A$ intersects the rays $BC$ and $CD$ at points $X$ and $Y$ respectively.Let $K$ and $L$ be the centres of the ex-circles of triangle $\triangle{ABX}$ and $\triangle{ADY}$ touching the sides $BX$ and $DY$ respectively.Prove that $\angle{KCL}$ doesn't depend on the choice of $l$.
Lemma: Let $I_a$ be te $A$-excentre of $\triangle ABC$. Then $\angle AI_aC=\frac 12\angle B,\angle AI_aB=\frac 12\angle C$.
Now $\angle BAK=\frac 12\angle BAX=\frac 12\angle AYD=\angle ALD$. Similarly $\angle AKB=\angle DAL$.
So $\triangle BAK \sim \triangle DLA$[Directly similar]
Hence $\frac{BK}{AD}=\frac{AB}{DL} \rightarrow \frac{BK}{BC}=\frac{CD}{DL} \rightarrow \frac{BK}{CD}=\frac{BC}{DL}$
Since $\angle KBC=\angle CDL(=\frac 12\angle A)$ we conclude $\triangle BKC \sim \triangle DCL$[Directly similar]
So we get $\angle KCB+\angle DCL=180^{\circ}-\frac 12\angle A$ which implies $\angle KCL=180^{\circ}-\frac 12\angle A$ which is indeed independent of $l$.
Sorry, I wasn't home the whole day; so I didn't solve any new problem for which I'm unable to post one.
বড় ভালবাসি তোমায়,মা