Simson says... go do geo!
by shiningsunnyday, Dec 10, 2016, 4:02 AM
2014 USA Dec. TST P1 wrote:
Let 
be an acute triangle, and let 
be a variable interior point on the minor arc 
of its circumcircle. Let 
and 
be the feet of the perpendiculars from to lines 
and 
, respectively. Let 
be the intersection of line 
and the perpendicular from 
to 
. Let 
be the line through parallel to 
. Prove that as varies along minor arc , the line always passes through a fixed point.
be an acute triangle, and let 
be a variable interior point on the minor arc 
of its circumcircle. Let 
and 
be the feet of the perpendiculars from to lines 
and 
, respectively. Let 
be the intersection of line 
and the perpendicular from 
to 
. Let 
be the line through parallel to 
. Prove that as varies along minor arc , the line always passes through a fixed point.
be the foot of perpendicular from 
Also let 
intersect the circumcircle again at 
From the diagram, we guess that we have 
a parallelogram. That is, no matter where 
)
passes through 
Since we already have 
it suffices to show 
is a parallelogram.
are both perpendicular to 
Now ![\[\angle{P'BR'} = \overarc{R'P'} = \overarc{BX} = \angle{BAX} \stackrel{\text{YAPX cyclic}}{=} \angle{YPX} \stackrel{PX || BR}{=} \angle{PRB}.\]](http://latex.artofproblemsolving.com/3/8/8/388e2bf46156fdcc0b3f737c36ac4a3f00874772.png)
and in order to show 
we reflect 
across 
It thus suffices to show 
or...
is a parallelogram.
and 
are both isosceles trapezoids, implying 
which combined with 
gives us the desired.
and the conclusion follows.
May 2017
April 2017