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### Re: IMO Marathon

Posted: Fri Nov 16, 2012 7:30 am
Tahmid Hasan wrote:Problem $8$:Let $ABC$ be a triangle such that $AB \neq AC$. The
internal bisector lines of the angles $ABC$ and $ACB$ meet the opposite
sides of the triangle at points $B_0$ and $C_0$, respectively, and the
circumcircle $ABC$ at points $B_1$ and $C_1$, respectively. Further, let
$I$ be the incentre of the triangle $ABC$. Prove that the lines
$B_0C_0$ and $B_1C_1$ meet at some point lying on the parallel through
$I$ to the line $BC$.
Source:Romania TST-2010-Exam-3-Problem-2.
Problem 8.PNG (46.91 KiB) Viewed 3289 times
Let $AI \cap \circ ABC =H, C_1H \cap AB=F, B_1H \cap AC=G$

Now $F,G,I$ are colliner using pascal's theorem .

$AIGB_1$ is cyclic.

So $\angle B_1IG=\angle CAB_1=\angle B_1BC$.
that imply $FG || BC$ .

As $A.I,H$ are colliner ,we can say $B_1C_1,B_0C_0,FG$ concur using
Desargues .

(wow !! i used Desargues for the first time .. yey)

### Re: IMO Marathon

Posted: Fri Nov 16, 2012 8:06 am
Problem $9$
A circle intersects sides $BC,CA,AB$ of $\triangle ABC$ at
two points for each side in the following order: $(D_1 , D_2 ), (E_1 , E_2 )$ and $(F_1 ,F_2)$.
Line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, $E_1F_1$ and $E_2D_2$ intersectat point $M$,
$F_1D_1$ and $F_2E_2$ intersect at point $N$.
Prove that $AL,BM$ and $CN$ are concurrent.

source: China Mathematical Olympiad'05, Day-1,Problem 2 .

### Re: IMO Marathon

Posted: Fri Nov 16, 2012 12:26 pm
Nadim Ul Abrar wrote:$AIGH$ is cyclic.
এইটা কেমনে সম্ভব?
I think you meant to say $AIGB_1$

### Re: IMO Marathon

Posted: Fri Nov 16, 2012 1:10 pm
yap .. thanks .

### Re: IMO Marathon

Posted: Fri Nov 16, 2012 3:18 pm
Nadim Ul Abrar wrote:Problem $9$
A circle intersects sides $BC,CA,AB$ of $\triangle ABC$ at
two points for each side in the following order: $(D_1 , D_2 ), (E_1 , E_2 )$ and $(F_1 ,F_2)$.
Line segments $D_1E_1$ and $D_2F_2$ intersect at point $L$, $E_1F_1$ and $E_2D_2$ intersectat point $M$,
$F_1D_1$ and $F_2E_2$ intersect at point $N$.
Prove that $AL,BM$ and $CN$ are concurrent.

source: China Mathematical Olympiad'05, Day-1,Problem 2 .
Let $BC \cap MN=P,CA \cap NL=Q,AB \cap LM=R$.
Our requirement is equivalent to proving $\triangle ABC,\triangle LMN$ are point perspective.
By Desargues' theorem our requirement shifts to proving they are line perspective ie. $P,Q,R$ are collinear.
Applying Pascal's theorem on cyclic hexagon $D_1D_2E_2F_2E_1F_1$ we get $P,M,N$ are collinear.
So $P= D_1D_2 \cap E_1F_2$. Similarly we get $Q=E_1E_2 \cap D_2F_1,R=F_1F_2 \cap D_1E_2$.
Apllying Pasacal's theorem on cyclic hexagon $D_1E_2E_1F_2F_1D_2$ we get $P,Q,R$ are collinear.

### Re: IMO Marathon

Posted: Fri Nov 16, 2012 3:23 pm
Problem $10$:Let $ABC$ be a triangle with $AB = AC$ . The angle bisectors of $\angle CAB$ and $\angle ABC$ meet the sides $BC$ and $CA$ at $D$ and $E$ , respectively. Let $K$ be the incentre of triangle $ADC$. Suppose that $\angle BEK = 45^{\circ}$. Find all possible values of $\angle CAB$.
Source:IMO-2009-4

### Re: IMO Marathon

Posted: Fri Nov 16, 2012 7:23 pm
Solution 10 :
Let $AD \cap BE=F$

Using sine rule on $\triangle KFE ,\triangle KEA ,\triangle KFA$
We can write this equation .

$\displaystyle \frac {sin{(45+\frac{A}{4})}}{sin {\frac{3A}{4}}}=\frac {cos{\frac{A}{2}}}{sin {45}}$ .
$\displaystyle \rightarrow \frac{sin {\frac{A}{4}}+cos {\frac{A}{4}}}{sin{\frac{A}{4}}cos{\frac{A}{2}}+cos{\frac{A}{4}}sin{\frac{A}{2}}}=2cos{\frac {A}{2}}$
$\displaystyle \rightarrow sin {\frac{A}{4}}+cos {\frac{A}{4}}=sin{\frac{A}{4}}[2cos^2{\frac{A}{2}}]+cos{\frac{A}{4}}[2sin{\frac{A}{2}}cos{\frac {A}{2}}]$
$\displaystyle \rightarrow sin {\frac{A}{4}}+cos {\frac{A}{4}}=sin{A+\frac{A}{4}}+sin{\frac{A}{4}}$
$\displaystyle \rightarrow cos {\frac{A}{4}}=cos{90-(A+\frac{A}{4})}$

So $\frac{A}{4}=90-(A+\frac{A}{4}) \rightarrow A=60$

$\frac{A}{4}=-(90-(A+\frac{A}{4})) \rightarrow A=90$

(Edited)

### Re: IMO Marathon

Posted: Fri Nov 16, 2012 8:06 pm
Nadim Ul Abrar wrote:Solution 10 :

Using sine rule on $\triangle KFE ,\triangle KEA ,\triangle KFA$
We can write this equation .

$\displaystyle \frac {sin{(45+\frac{A}{4})}}{sin {\frac{3A}{4}}}=\frac {cos{\frac{A}{2}}}{sin {45}}$ .
$\displaystyle \rightarrow \frac{sin {\frac{A}{4}}+cos {\frac{A}{4}}}{sin{\frac{A}{4}}cos{\frac{A}{2}}+cos{\frac{A}{4}}sin{\frac{A}{2}}}=2cos{\frac {A}{2}}$
$\displaystyle \rightarrow sin {\frac{A}{4}}+cos {\frac{A}{4}}=sin{\frac{A}{4}}[2cos^2{\frac{A}{2}}]+cos{\frac{A}{4}}[2sin{\frac{A}{2}}cos{\frac {A}{2}}]$
$\displaystyle \rightarrow sin {\frac{A}{4}}+cos {\frac{A}{4}}=sin{A+\frac{A}{4}}+sin{\frac{A}{4}}$
$\displaystyle \rightarrow cos {\frac{A}{4}}=cos{90-(A+\frac{A}{4})}$

So $\frac{A}{4}=90-(A+\frac{A}{4}) \rightarrow A=60$

$\frac{A}{4}=-(90-(A+\frac{A}{4})) \rightarrow A=90$
Please denote $F$

### Re: IMO Marathon

Posted: Sun Nov 18, 2012 3:55 pm
Problem 11 .
Let quadrilateral $ABCD$ be inscribed in a circle. Suppose lines $AB$ and $DC$
intersect at $P$ and lines $AD$ and $BC$ intersect at $Q$. From $Q$, construct the two tangents $QE$
and $QF$ to the circle where $E$ and $F$ are the points of tangency. Prove that the three points
$P, E, F$ are collinear.

Source : China MO 1997.

### Re: IMO Marathon

Posted: Sun Nov 18, 2012 9:27 pm
Solution $\boxed {11}$
It's well known that if $AC \cap BD = K$, then $\triangle KPQ$ is a self polar triangle, which implies pole of $Q$ (which is $EF$) goes through $P$. So, $P, E, F$ are collinear.