Let $ABC$ be a triangle with $90^\circ \ne \angle A \ne 135^\circ$. Let $D$ and $E$ be external points to the triangle $ABC$ such that $DAB$ and $EAC$ are isoscele triangles with right angles at $D$ and $E$. Let $F = BE \cap CD$, and let $M$ and $N$ be the midpoints of $BC$ and $DE$, respectively. Prove that, if three of the points $A$, $F$, $M$, $N$ are collinear, then all four are collinear.
Problem
Source: 2009 MMO Problem #2
Tags: geometry, circumcircle, geometric transformation, rotation, geometry unsolved
Draftsman
13.09.2011 18:57
I don't think is true. Perhaps there's a wrong info. in the problem, can you check it?
anonymouslonely
14.09.2011 19:56
I obtained that it's true. if $ M,N,F $ are collinear then $ BC||DE $ and that means $ AB=AC $. if $ A,M,F $ are collinear then using areas we'll obtain $ N $ is on this line. the other two cases are similar to this one.
Atlas_ha78
17.12.2013 17:28
Can you post your solution? Thanks
MMEEvN
18.12.2013 06:48
The first two cases where $M,N,F$ being collinear and $A,M ,F$ being collinear are easy to prove .Hence I will do $A,N,M$ being collinear. First I will prove that $MD = ME$ .this can be proved very easily by cos rule to $\Delta MBD$ and $\Delta MEC$ .But I will use a different method. Let the foot of perpendicular from $A$ on $BC$ be $H$.Easy to observe that $AHBD$ and $AHCE$ are cyclic..Let the circumcircle of $HDE$ intersect $BC$ again at $R$.Now we can easily see that $RD=RE$ ..Consider the composition of rotation centered at $D$ by $90$ and centered at $E$ by $90$.This takes $B$ to $C$ .Since the sum of rotation is $180$ and since it takes $B$ to $C$ it is a rotation by $180$ around the midpoint of $BC$ that is $M$ .Let us rotate $R$ by $90$ around $D$ and let the image be $R '$ .it is easy to observe that $RDR'E$ is a square .Now rotate $R'$ around $E$ .Which will take $R'$ back again to $R$ .Hence under the rotation $R \to R' .\Rightarrow R$ should coincide with $M$. Hence $ME=MD$ Now if $A,N, M$ are collinear $MN \perp DE \Rightarrow AD =AE \Rightarrow AB=AC$ .The rest is trivial.
lemon94
19.07.2019 17:34
Is proving AF||MN suffice to prove the proposition?