[/quote]
,
be an equilateral triangle. Let 
be a point inside of it such that the square root of the distance of 
such that if any 
elements are removed from the set 
,
be a triangle. On side 
take points 
and 
such that 
take points 
and 
and on side 
take points 
such that 
Let 
and 
are concurrent.
lie the following points: 
and 
on 
,
on 
,
on 
.![\[
P = AM\cap BN,\quad
R = KM\cap LN,\quad
S = KN\cap LM,
\]](http://latex.artofproblemsolving.com/7/e/a/7ea90a0f1261ed7445bbac58fb0cdd76810259f3.png)
and let the line 
meet 
.
,
are collinear.
.
intersects 
and 
and 
,
of triangle 
lies on the circle 
and 
coincide.
be a triangle and 
and 
be two points on 
,
such that 
and let 
.
to the interior angle bisector of 
intersect 
in 
.
.
,
be its incenter. Incircle touches sides 
at 
,
to 
, respectively. Rays 
intersect 
at 
,
touches incircle at 
.
.
.
and 
are equal.
and circumcircle 
.
be the midpoints of sides 
,
,
and let 
be the intersections of line 
with line 
and line 
,
be the midpoint of arc 
of 
.
bisects 
.
let 
and 
,
and 
.
and 
,
.
.
be the projection of 
.
intersects the circumcircle of 
.
and 
intersect 
and 
respectively. Prove that the lines 
,
,
are concurrent.
Y by Danielzh, BS2012, EpicBird08, bachkieu
I got an email about new FedEx shipment earlier today. I never ordered anything and was pretty confused but it caught my interest because it shipped out of Elgin, IL, which is only ~15 miles from the place where MOP is taking place and was shipped directly to my name and the email I signed up to AMCs with (which I don't use for much other things).
This is a very stupid question and it might be a coincidence but did any other AoPSers waiting on MOP email receive this ;-;
This is a very stupid question and it might be a coincidence but did any other AoPSers waiting on MOP email receive this ;-;
This post has been edited 1 time. Last edited by Mathandski, Apr 18, 2025, 9:00 PM
