Since I got tired of sucking at non-geo
by shiningsunnyday, Sep 20, 2016, 7:41 PM
1998 ISL G5 / Lemmas in Oly Geo wrote:
Let
be a triangle with orthocenter
circumcenter
and circumradius
Let
be the reflections of the vertices
across the opposite sides. Prove that they are collinear if and only if 







What on mother earth is this? An if and only if condition involving some obscure
condition and a random collinearity?
Well, no use staring. Let's get our geo kit out and we get, after many failed diagrams, finally a successful one:
![[asy]
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[/asy]](//latex.artofproblemsolving.com/a/a/b/aabdfff2b530baef75775cbcceda32c82d5b1cfa.png)
...where
is defined as the midpoint of 
Hm... this still doesn't give us any lead as to how we should go about dealing with this collinearity condition. Well, what we do know is that we have a lot of reflections and perpendiculars as well as the orthocenter, which make us think of Simson lines, which would allow us to translate this wacky collinearity into a concyclic condition. Since otherwise we'll be sitting around all day, let's extend perpendiculars from
to work backwards to find our triangle whose Simson line is 
![[asy]
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[/asy]](//latex.artofproblemsolving.com/8/a/2/8a21ec805e6916437f8daf58f3fdb08ec66f493a.png)
It appears
is the desired triangle whose Simson line with respect to point
is
Thus the collinearity condition is just another way of saying
is cyclic. But why did I name the three vertices
In fact, it is the case that
is similar to
since
is defined as the line perpendicular to
through
etc.
But is there a specific homothety we can attribute to this transformation? Well... since the homothety's center appears to be the center of
it's not hard to see
the centroid, is the center of the homothety.
![[asy]
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[/asy]](//latex.artofproblemsolving.com/7/7/1/771a1464af96af2a23165611fe8ec4b2c284acd7.png)
Because
bisects
and
bisects
it's easy to see that the B-median of
also passes through the midpoint of
Better, we see that the scale factor of the homothety is
(using the fact the centroid divides a median into
).
Ok. But how do we tie this to the fact
iff
Well we do know that
lies on the line
and so if only we can prove
is in fact homothetic to
using the assumption that
then we we'll be done -
is cyclic iff
is cyclic iff
is collinear.
We have the line
with
as the center of homothety, and so we're really just looking at the Euler line! Moreover, it's well known
Since
we have that
and so
implying
and
are indeed homothetic.
And baby we're home!

Well, no use staring. Let's get our geo kit out and we get, after many failed diagrams, finally a successful one:
![[asy]
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/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/a/a/b/aabdfff2b530baef75775cbcceda32c82d5b1cfa.png)
...where


Hm... this still doesn't give us any lead as to how we should go about dealing with this collinearity condition. Well, what we do know is that we have a lot of reflections and perpendiculars as well as the orthocenter, which make us think of Simson lines, which would allow us to translate this wacky collinearity into a concyclic condition. Since otherwise we'll be sitting around all day, let's extend perpendiculars from


![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
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[/asy]](http://latex.artofproblemsolving.com/8/a/2/8a21ec805e6916437f8daf58f3fdb08ec66f493a.png)
It appears










But is there a specific homothety we can attribute to this transformation? Well... since the homothety's center appears to be the center of


![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(15 cm);
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[/asy]](http://latex.artofproblemsolving.com/7/7/1/771a1464af96af2a23165611fe8ec4b2c284acd7.png)
Because








Ok. But how do we tie this to the fact










We have the line









And baby we're home!
Tidbit
I solved this problem subconsciously at 3 AM as I was dozing off. Now I'm hyped. Time for some NT!
Oh **** English socratic seminar today.
Oh **** English socratic seminar today.
This post has been edited 5 times. Last edited by shiningsunnyday, Oct 6, 2016, 11:41 AM