Let $k$ and $k'$ be concentric circles with center $O$ and radius $R$ and $R'$ where $R<R'$ holds. A line passing through $O$ intersects $k$ at $A$ and $k'$ at $B$ where $O$ is between $A$ and $B$. Another line passing through $O$ and distict from $AB$ intersects $k$ at $E$ and $k'$ at $F$ where $E$ is between $O$ and $F$. Prove that the circumcircles of the triangles $OAE$ and $OBF$, the circle with diameter $EF$ and the circle with diameter $AB$ are concurrent.
Problem
Source: Mediterranean Mathematics Competition 2005, Problem 2
Tags: geometry, circumcircle, geometry proposed
Ashegh
07.01.2006 11:15
i think inversion will help... i will think about it and will post the solution... but im not sure with it by inversion...
Ashegh
07.01.2006 20:24
yes yes my guess was true.... but u know,we wont prove the problem with inversion.we just use it to find the trusts in the problem. $lemma$:( very nice and usefull) let $(C1)$and $(C2)$ be two circles which intersects each other in points $A$ and $B$. suppose a variable line goes through $B$ and intersects $(C1)$ and $(C2)$ in points$C$ and $D$,respectivly. then: the triangles $AO1O2$ and $ADC$ are simmilar to each other. its clear ,but we prove it very simply to do our best. $\angle AO1O2=\angle ACD$ and $\angle AO2O1=\angle ADC$. the lemma will also be true ,if $B$ is betwin$C$ and $D$.or if $D$ is betwin $B$ and $C$. then: lets prove our problem by the lemma. name the intersection of the circumcirclesof hthe triangles $OAE$ and $OFB$,$M$. before that we use inversion. an inversion with the center $O$ and the radius $DE.DF$ ,shows us that the angle betwin the circumcircles of the triangles $OAE$ and $OFB$,is $90$. because:the lines $A'E'$ and $B'F'$ which are the inversive shapes of the circles $OAE$ and $OFB$, are prependicular to each other.and it shows us that the angle betwin the circumcircles of the triangles $OAE$ and $OFB$,is $90$. then acording to the lemma,$AB$ goes through one of the intersections of the two circles.(i mean $O$). and we have:the triangles $MAB$ and $MO1O2$ are simmilar.then it shows that $\angle AMB=90$ by a simmilar way we can claime that $\angle EMF=90$. then it shows that:a circle with the diameter $FE$ goes through $M$. and a circle with the diameter $BA$ goes through $M$. and the problem is proved...
Yimin Ge
07.01.2006 20:26
ashegh wrote: i think inversion will help... i will think about it and will post the solution... but im not sure with it by inversion... Well, I have no idea how inversion (on a circle - as you probably meant) works (yes I know, I'm an uneducated geometrician), however I solved it by a mere angle-chasing (unspectacular but easy and stupit )
Let $T$ be the point of intersection of the circumcircles of $OAE$ and $OBF$. Let $\angle AOE = \alpha$ and $\angle BOF = \beta$. We have $\alpha+\beta = 180$. $\triangle OAE$ and $\triangle OBF$ are isosceles, thus $\angle OEA = \angle OAE = 90-\frac{\alpha}{2}$ and $\angle OFB = \angle OBF = 90-\frac{\beta}{2}$.
$BFTO$ is cyclic, hence $\angle OTF = 180 - \angle OBF = 90+\frac{\beta}{2}$.
Also, $OATE$ is cyclic, hence $\angle OTE = \angle OAE = 90-\frac{\alpha}{2}$.
Thus, $\angle FTE = \angle OTF - \angle OTE = 90+\frac{\beta}{2} - 90 + \frac{\alpha}{2} = 90$, hence T lies on the circle with diameter $EF$.
Similary, we'll show that $\angle ATB = 90$:
$BOTF$ is cyclic, hence $\angle OTB = \angle OFB = 90-\frac{\beta}{2}$.
$AOET$ is cyclic, hence $\angle OTA = \angle OEA = 90-\frac{\alpha}{2}$.
Thus $\angle ATB = \angle OTA + \angle OTB = 90-\frac{\alpha}{2} + 90 - \frac{\beta}{2} = 90$, hence T lies on the circle with diameter $AB$.
$\Longrightarrow q.e.d.$
Ashegh
07.01.2006 20:31
the best thing is to solve easy problems easy... the thing that u did...