Circles $C_{1}$ and $C_{2}$ are tangent to each other at $K$ and are tangent to circle $C$ at $M$ and $N$. External tangent of $C_{1}$ and $C_{2}$ intersect $C$ at $A$ and $B$. $AK$ and $BK$ intersect with circle $C$ at $E$ and $F$ respectively. If AB is diameter of $C$, prove that $EF$ and $MN$ and $OK$ are concurrent. ($O$ is center of circle $C$.)
Problem
Source: Iran 2002
Tags: geometry, circumcircle
grobber
10.04.2004 07:43
First of all, we have to remark that the circumcircle of KEF has the same radius as C. This can be observed with the aid of an inversion of pole K and power -KA*KF. This inversion turns the line AB into the circle KEF and it leaves circle C alone (it's invariant). It turns circles C1 and C2 into 2 parallel lines which are both tangent to the images of C and AB, which are C and the circle KEF respectively. This means that 2 common tangents of C and KEF are parallel, so their radii are equal. That was just an observation; no let's get to the problem: We use the same inversion further on. We have to show that the line KO and the circles KAB and KE'F' are concurrent, where X' is the image of X by the inversion. E' and F' are the points of tangency between C and 2 parallel lines, so they're the extremities of a diameter of C, and the conclusion follows easily: Let R be the radius of C. Then the powers of O with respect to the circles KAB and KE'F' are equal (both equal to R<sup>2</sup>), so O is on the radical axis of the 2 circles, so it's on KT where T is the other point where the 2 circles intersect (the first such point being K). I hope there's nothing wrong with it... Note that the condition revealed in the first paragraph, that the radii of KEF and C are equal, means that the angle <AKB is equal to 135 deg, or, in other words, OE is perpendicular to OF (this is another one of those useless observations ).
livetolovemath030894
05.01.2010 20:15
Can you post Problem's figure???????
pisgood
06.04.2017 06:18
Hate to bring this up again after 13 years, but I can't figure out what this problem's saying. How can $AB$ be the diameter of $C$ if $AB$, $C$, $C_1$, and $C_2$ are all tangent to each other?
vsathiam
11.10.2017 06:15
Tiny diagram is attached- should give gist of problem's diagram. (E, F are on circle)
Attachments:
PARISsaintGERMAIN
11.10.2017 21:49
Let $N'=KN\cap C, M'=MK\cap C, X=EF\cap MN, Y=MB\cap EN'$ Important lemma: $N'M'$ is diameter of circle $C$ Pascal's theorem on $M'N'EABM,$ $M'N'\cap AB=O$ $N'E\cap BM=Y$ $EA\cap MM'=K$ is collinear. Pascal's theorem on $MNN'EFB,$ $MN\cap EF=X$ $M'N'\cap FB=K$ $N'E\cap BM=Y$ is collinear. Thus, $(K,O,X,Y)$ is collinear. Also, using pascal on other hexagon, $Z=FM'\cap NA$ is also in that line.