[/quote]
,
,
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 
be 
and 
,
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
intersect 
and 
at 
and 
.
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
A scalene triangle 
is inscribed in a circle 
. The bisector of angle 
meets 
at 
. Let 
be the midpoint of the arc 
. The line 
intersects again at 
. Show that the circumcentre of triangle 
coincides with the intersection point of the tangent to at and the line .
is inscribed in a circle 
. The bisector of angle 
meets 
at 
. Let 
be the midpoint of the arc 
. The line 
intersects again at 
. Show that the circumcentre of triangle 
coincides with the intersection point of the tangent to at and the line .
is the foot of the A-external bisector. Thus circle 
with diameter 
is the A-Apollonius circle of 
which is orthogonal to its circumcircle 
Hence, it is enough to show that 
.
.![$\angle DAE= \frac{1}{2} (2 \angle DAE)= \frac{1}{2}( \angle DAE+ \angle NAC + \angle DAE - \angle BAN)= \frac{1}{2}( \angle DAC- \angle BAD)=\frac{1}{2}[\frac{1}{2} \hat CD- \frac{1}{2} \hat BD]= \frac{1}{2} \angle DKE$](http://latex.artofproblemsolving.com/9/1/4/914f2240dd354a1dc2f1be0ba1e23c876cc52404.png)
.
.
meets 
at 
.
and from 
is the A-Apollonius circle of 
.
and 
are perpendicular. We know that circumcenter of 
will be the 
