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.

## Bulgaria 1996

### Bulgaria 1996

"Questions we can't answer are far better than answers we can't question"

- Fm Jakaria
**Posts:**79**Joined:**Thu Feb 28, 2013 11:49 pm

### Re: Bulgaria 1996

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.

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.

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.

the umpteenth digit of PI,

Whether I'll live - or

whether I may, drown in tub and die.

### Re: Bulgaria 1996

WAT?Fm Jakaria wrote: ... We are finished.

**- What is the value of the contour integral around Western Europe?**

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

- Fm Jakaria
**Posts:**79**Joined:**Thu Feb 28, 2013 11:49 pm

### Re: Bulgaria 1996

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.

the umpteenth digit of PI,

Whether I'll live - or

whether I may, drown in tub and die.

### Re: Bulgaria 1996

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

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