[/quote]
,
of positive integers with the following property: all of the terms of the sequence 
have a greatest common divisor 
of natural numbers such that
be an acute scalene triangle. Let 
and 
be two distinct interior points of the segment 
such that 
.
and 
are the feet of the perpendiculars from from 
to the lines 
and 
respectively.
and 
are the feet of the perpendiculars from 
to the lines 
and 
intersect on the line 
be an odd prime and 
be integers so that the integers 
are divisible by 
,
be a point on side 
and 
respectively such that 
is a parallelogram. Points 
and 
are chosen on side 
and 
respectively. Let 
be the circle tangent to segment 
,
past 
past 
decide whether there exist pairwise distinct positive integers 
such that for every 
,
divides 
.
and 
.
such that ![\[f(xy-1)+f(x)f(y)=2xy-1,\]](http://latex.artofproblemsolving.com/8/8/8/888ca39f2b7f8cec6d6426bee28d40eade40a66e.png)
for any real numbers 
and 
,
fractions 
The product of these fractions equals 
,
be a function satisfying
for all integers 
and 
.
and 
for all integers 
Y by Adventure10, Mango247
Let 
be the circumcircle of triangle 
Denote the circle which touches to the sides 
and touches to internally at 
by 
and the circle which touches to the sides 
and touches to internally at 
by 
and the circle which touches to the sides 
and touches to internally at 
by 
Prove that the lines 
are concurrent.
be the circumcircle of triangle 
Denote the circle which touches to the sides 
and touches to internally at 
by 
and the circle which touches to the sides 
and touches to internally at 
by 
and the circle which touches to the sides 
and touches to internally at 
by 
Prove that the lines 
are concurrent.This post has been edited 2 times. Last edited by Kunihiko_Chikaya, Apr 1, 2007, 7:07 PM
is the inversion having 
for pole and 
for constant.
so 
is reflect 
about the bisector of 
is the excircle opposite 
at 
- which is 
.
is reflect 
about the bisector of 
is reflect 
about the bisector of 
and 
is reflect 
about the bisector of 
are same.
is conccurency so use the ceva's theorem ( the trigonometric form ) we have 
is conccurency, too.
is obviously 
Ap,BQ,CR concur in the external centre of similitude of the incentre and circumcentre of ABC.
.
.
be the circumcircle of triangle 
which touches to the sides 
.
and 
of the circles 
(tangent to the lines 
)
(tangent to the lines 
,
)
are concurrently in the isogonal conjugate of the Nagel's point for the triangle 
Let 
be a circle which is tangent internally (externally) to the circle 
define the tangent d to the circle 
.
is a internal (external) bisector of the angle 
.
Let 
be a convex cyclic quadrilateral for which denote 
.
.
The circle ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
,![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
in the points 
(apply the Pascal's theorem to the cyclic hexagon 
,
,
)
,
and 
.
,
are bisectors of the angles 
,
,
and 
.
.
.
obtain 
,
.
,
is the 
,
and 
,
are isogonal conjugate in the angle 
a.s.o. In conclusion, the lines 
,
,
,
,
,
are concurrently in the isogonal conjugate of the Nagel's point for the triangle 
and 
.
and 
,
is the external bisector of the angle 
and the point 
is the middlepoint of the segment ![$[Z_{1}Z]$](http://latex.artofproblemsolving.com/c/7/7/c7726aea9fe1bac96c7dfce15c7393411b261f46.png)
,
is the internal bisector of the angle 
which touches to the sides 
.
and 
of the circles 
(tangent to the lines 
(tangent to the lines 
,
,
are concurrently in the isogonal conjugate of the Gergonne's point for the triangle 
(from your topic about harmonic division, I think). But there's supposed to be another point with the same ratio. How do you know this one is the middlepoint?
is 
is 
