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be an acute scalene triangle with circumcenter 
, and let 
be on line 
such that 
. The circle with diameter 
intersects the circumcircle of 
at two points 
and 
, where 
. Points 
, 
, 
, 
are defined analogously.
, 
, 
are concurrent.
, 
, 
are concurrent on the Euler line of triangle . [/list]Evan Chen
such that for any real numbers 
and 
,
Proposed by Dorlir Ahmeti, Kosovo
with centroid , such that 
. Let 
and 
be the second points of intersection of lines 
and 
with the circumcircle of triangle , respectively. Prove that the line 
is tangent to the Euler circle of triangle .
such that 
, for all 
.
is a set of positive integers.)
and 
be inter section of 
with 
and 
. Then 
and 
hence 
So we are done!
be the second-intersections of 
with respectively.
and so 
.
. This indeed,gives 
.
be the midpoint of the side 
.
and 
be such that 
.
intersect with 
and 
intersect with 
at 
.
is cyclic.
are intersection point of 
.![\[ AL.AL_1=AY.AB=AP.AC , \]](http://latex.artofproblemsolving.com/4/5/9/45959120e78003f1f1c87f42598ee7244fed9065.png)
![\[ CL.CL_1=CX.CB=CQ.CA \]](http://latex.artofproblemsolving.com/b/5/6/b5612bb9876e1286c6deac60fb331cc54d7c75a2.png)
and ![\[ \vec{ML}.\vec{ML_1}=0 .\]](http://latex.artofproblemsolving.com/6/5/4/654ea60bd2b474681c060d080d561fe8a1783120.png)
is cycl
is cycle.
and 
.
concycli![$] = \angle CQX [\because Q, X, B, A$](http://latex.artofproblemsolving.com/d/7/3/d7313d176f76804494c027b92125c8e7dfba6eeb.png)
![$]$](http://latex.artofproblemsolving.com/0/7/1/07174a92e49be0bb0db1beef2bb6c245a8b212f9.png)
concycli
and we've 
concyclic.![$[QED]$](http://latex.artofproblemsolving.com/7/9/f/79f768c32001fa5cb49e13629878a11b3c62b159.png)
intersects 
,
if 
or 
or 
or 
.
or 
or 
.
.
and 
where 
.
and 
.
.
.
satisfies the circle equation and it does. We are done.
and 
.
and 
.
.
and 
.
intersect 
and 
.
and 
.
and 
,
.
and 
.
with roots 
and 
.
and 
.
satisfies these same conditions, so this is 
is a root of 
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and 
are similar.
cyclic and 
cyclic, so 
and 
,
.
,
,
denote the circles 
and 
respectively. Define a function 
as: ![\[f(P) = \text{Pow}(P, \Omega)-\text{Pow}(P, \omega)\]](http://latex.artofproblemsolving.com/4/b/a/4ba9d088885570c040918a0692a676b0d179271f.png)
Since 
lies on the radical axis of 
we have 
.
since 
.
is the average 
and 
due to the aforementioned tricx, we obtain ![\[f(M) = AP \cdot (AC/2) = MQ \cdot MA\]](http://latex.artofproblemsolving.com/e/5/1/e5172b900b62fe94626ba1b5c6e7b5b0517237e4.png)
since 
(from 
)
.![\[f(M) = \text{Pow}(M, \Omega) - \text{Pow}(M, \omega) = \text{Pow}(M, \Omega) - (-MQ \cdot MA)\]](http://latex.artofproblemsolving.com/e/3/1/e31426ccac526b8c500b35be615ef19cb3ef61a9.png)
implying 
,
.