Bulgaria 1996

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tanmoy
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Bulgaria 1996

Unread post by tanmoy » Thu Feb 26, 2015 8:52 pm

The circles $k_{1}$ and $k_{2}$ with respective centers $O_{1}$ and $O_{2}$ are externally
tangent at the point $C,$ while the circle $k$ with center $O$ is externally tangent to $k_{1}$
and $k_{2}$. Let $l$ be the common tangent of $k_{1}$ and $k_{2}$ at the point $C$ and let $AB$ be the
diameter of $k$ perpendicular to $l$. Assume that $O_{2}$ and $A$ lie on the same side of $l$.
Show that the lines $AO_{1},BO_{2},l$ have a common point.
"Questions we can't answer are far better than answers we can't question"

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Fm Jakaria
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Re: Bulgaria 1996

Unread post by Fm Jakaria » Fri Feb 27, 2015 9:13 am

Let $C,Y,Z$ be pairwise touchpoints of $k_1,k_2,k$. Note that the circle $CYZ$ with center $I$(say) is the incircle of triangle $O_1O_2O$.Here $C,Y,Z$ are the points where incircle touches the sides. So $CI$ is the common tangent of $k_1,k_2$; so is both perpendicular to $O_1O_2$ and $AB$. Then $O_1O_2$ and $AB$ are parallel. Let $M$ be the midpoint of $O_1O_2$; and $CY \cap AB = E$, $CZ \cap AB = D$.
Note that triangles $CYO_2, EYO$ are similiar; so because of $O_2YC$ is an isoceles triangle with $O_2Y = O_2C$, so is $OYE$ with $OY = OE$. So in fact $E \equiv A$. Similiarly $D \equiv B$. Let $O_1A \cap O_2B = J$.
Because $O_1O_2AB$ is a trapezium; $MO$, the midline between parallel sides of it, contains $J$.
Apply Pappu's theorem to the ordered collinear triples: $O_1,C,O_2$; $A,O,B$ . We get that $J \in YZ$.
So we get that $J = OM \cap YZ$. Now our last step follows from applying Zhao's Lemma on triangle $OO_1O_2$: saying that $CI, YZ$ and the median $OM$ are concurrent. This gives that $J \in CI$. We are finished.
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the umpteenth digit of PI,
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Nirjhor
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Re: Bulgaria 1996

Unread post by Nirjhor » Fri Feb 27, 2015 12:27 pm

Fm Jakaria wrote: ... We are finished.
WAT? :shock:
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.


Revive the IMO marathon.

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Fm Jakaria
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Re: Bulgaria 1996

Unread post by Fm Jakaria » Fri Feb 27, 2015 8:14 pm

As mentioned in the solution, $IC$ is the line $l$.
You cannot say if I fail to recite-
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.

Tahmid
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Re: Bulgaria 1996

Unread post by Tahmid » Sat Feb 28, 2015 2:39 am

let , $k_{1},k_{2}$ touches $k$ at points $X,Y$ respectively. and $l\cap AB=Z$
so, $AX,BY,CZ$ are concurrent at the orthocentre of triangle $ABC$
by ceva's theorem we have $\frac{AZ}{ZB}=\frac{r_{1}}{r_{2}}$

now let, $AO_{1}\cap BC=R$ ; $BO_{2}\cap AC=S$
then , $\frac{CS}{SA}\cdot \frac{AZ}{ZB}\cdot \frac{BR}{RC}=\frac{r_{2}}{AB}\cdot \frac{r_{1}}{r_{2}}\cdot \frac{AB}{r_{1}}=1$

$\therefore l,AO_{1},BO_{2}$ are concurrent

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