IMO Marathon

Discussion on International Mathematical Olympiad (IMO)
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zadid xcalibured
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Re: IMO Marathon

Unread post by zadid xcalibured » Tue Jan 22, 2013 8:10 pm

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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Nadim Ul Abrar
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Re: IMO Marathon

Unread post by Nadim Ul Abrar » Tue Jan 22, 2013 8:14 pm

$\boxed {20}$

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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Nadim Ul Abrar
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Re: IMO Marathon

Unread post by Nadim Ul Abrar » Tue Jan 22, 2013 8:50 pm

$\boxed {19}$
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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Phlembac Adib Hasan
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Re: IMO Marathon

Unread post by Phlembac Adib Hasan » Wed Jan 23, 2013 8:58 am

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। :oops:
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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zadid xcalibured
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Re: IMO Marathon

Unread post by zadid xcalibured » Wed Jan 23, 2013 9:56 am

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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zadid xcalibured
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Re: IMO Marathon

Unread post by zadid xcalibured » Wed Jan 23, 2013 10:03 am

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

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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Wed Jan 23, 2013 11:26 am

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. :D
Great problem for learning angle chasing :)
Phelembac Adib Hasan wrote:আর আজকে ফিগার ভুল আঁকলাম। পুরো এক ঘণ্টা নষ্ট। :oops:
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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zadid xcalibured
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Re: IMO Marathon

Unread post by zadid xcalibured » Wed Jan 23, 2013 1:29 pm

I Love Ratio. :mrgreen:
$\frac{SC}{SB}=\frac{SA}{SC}$ $\Longrightarrow \frac{SP}{SB}=\frac{SA}{SP}$
$\triangle{SPB} \sim \triangle{SAP}$ 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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zadid xcalibured
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Re: IMO Marathon

Unread post by zadid xcalibured » Wed Jan 23, 2013 1:59 pm

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

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Tahmid Hasan
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Re: IMO Marathon

Unread post by Tahmid Hasan » Wed Jan 23, 2013 8:43 pm

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. :oops:
বড় ভালবাসি তোমায়,মা

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