![\[ x + y^2 = p^m \]](http://latex.artofproblemsolving.com/e/c/f/ecf553036f22327dd457dc06611b6bc1b0ef072c.png)
![\[ x^2 + y = p^n \]](http://latex.artofproblemsolving.com/5/b/d/5bd672934a395aabd6865ca4b50d8da75c0fe07e.png)
For 
positive integers and 
being a prime.
and 
.
Equality holds when 
or 
,
,
and 
lie on a circle 
such that 
.
at 
and 
.
intersect again at 
.
Equality holds when 
Y by nguyendangkhoa17112003, Adventure10, and 3 other users
Let 
be an acute triangle. The points 
and 
are midpoints of 
and 
respectively, and 
is an altitude of . The circumcircles of 
and 
meet in 
where 
. Prove that 
passes through the midpoint of 
.
V. Filimonov
be an acute triangle. The points 
and 
are midpoints of 
and 
respectively, and 
is an altitude of . The circumcircles of 
and 
meet in 
where 
. Prove that 
passes through the midpoint of 
.V. Filimonov
and 
.![$[MN]$](http://latex.artofproblemsolving.com/1/6/2/162e03b9cf481dcfb3d5bdf078be84feab5d2f6e.png)
lies on the radical axis (which happens to be 
at 
then 
,
Also, it is interesting to notice that if the circumcircles of 
meet each other (for the second time) at 
is a symmedian.
Very nice solution solution .Thank you for this and the explanation.
is cyclic.
is the midpoint of ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
,
and 
are symmetric w.r.t to the perpendicular bisector of 
and 
be the circumcircles of triangle 
and 
so 
.
.
.
meets 
cuts 
and 
respective at 
and 
.
and 
we get: 
is in the axis radical of 
pass through 
is the circumcircle of 
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be the midpoint of 
and 
Note that 
(where we have used the facts that 
and 
)
Similarly, 
the power of 
are equal, so we're done. 
and 
denote the circumcircles of 
lies on 
.
.
and 
.
,
,
,
.
