[/quote]
,
,
be the projection from 
onto 
be a point on the extension of 
past 
.
be on the perpendicular bisector of 
such that 
are on the same side of ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
and 
,
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
intersect 
and 
at 
intersect 
.
.
and 
.
.
be the point on 
.
by 
,
for every positive integer 
.
leave distinct remainders when divided by 
.
of distinct positive integers such that there exists a positive integer 
for which the number of divisors of 
and of 
are equal.
such that ![\[f(2x+y) + f(x+f(2y)) = f(x)f(y) - xy\]](http://latex.artofproblemsolving.com/0/6/2/0623b7c328af6d088cfc5dd288afe3e17a94ff29.png)
for all reals 
and 
.
and 
be two sequences of integers such that 
and 
for all 
.
divides 
for an infinity of integers 
such that 
for all 
are coprime. Consider the sequence 
given by 
.
such that 
is even and 
is an integer. Prove that the numerator in its simplest form of 
is an odd integer greater than 
.
be an integer and let 
be 
.
such that the n subsets 
are also disjoint.
s
satisfies 
.
,
.
such that for all irrational numbers 
and 
,
where 
Y by Adventure10, Mango247
In an acute scalene triangle 
the bisector of the acute angle between the altitudes 
and 
meets the sides 
and 
at 
and 
respectively. The bisector of the angle 
intersects the segment joining the orthocenter of and the midpoint of 
at point 
. Prove that , , , lie on a circle.
the bisector of the acute angle between the altitudes 
and 
meets the sides 
and 
at 
and 
respectively. The bisector of the angle 
intersects the segment joining the orthocenter of and the midpoint of 
at point 
. Prove that , , , lie on a circle.
- diametrically opposite to 
in the circle 
,
- its orthocenter. Obviously, 
and 
are symmetrical about the angle bisector 
,
( 
is a parallelogram ).
,
( 1 ), so 
( 2 ). But 
,
is angle bisector, hence 
,
and 
,
is cyclic.
,
; 
; 
,
,
,
,so 
is concylic
and 
.
be the reflection of 
across point 
and 
,
is a parallelogram. Furthermore, 
,
is cyclic. But 
,
is a diameter of 
.
is isogonal to 
and 
.
gives us 
by AA similarity. Therefore, 
.
bisects 
,
,
.
bisects 
,
.
,
.
,
is cyclic, as desired. //
and 
are coaxial. Let 
be the midpoints of the arcs 
and 
.
we conclude 
so 
From angle bisector theorem, we have 
we conclude that 
as desired.
denote the intersection of circle with diameter 
.
.
the antipode of 
bisects the segment 
.
as well. Now, observe that 
.![\[C_1P/PA = HC_1/HA = HA_1/HC = A_1Q/QC\]](http://latex.artofproblemsolving.com/2/3/f/23f80303ef884e03871a49246c3288b0ef827d3b.png)
which implies that 
.
are all concylic. First, observe that from the spiral sim, 
is cyclic. Next, remark that 
due to 
.![\[\angle PXH = \angle PXC_1+\angle C_1XH = \angle W'XB+\angle C_1XH = \angle HXW+\angle C_1XH = \angle C_1XW = \angle C_1BW = \angle PBR\]](http://latex.artofproblemsolving.com/0/b/2/0b23b3f178024e0e4521fe8dc72edd6438f48a5c.png)
where we used 
due to the pairs 
being antipodes in 
,
and thus in particular points 
are concylic, as required.
