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be the set of all 
-tuples 
such that each 
is a subset of 
. Let 
denote the number of elements of the set 
. Find![\[ \sum_{(A_1, \ldots, A_n)\in F} |A_1\cup A_2\cup \cdots \cup A_n| \]](http://latex.artofproblemsolving.com/a/0/8/a0802b66ee9ec20913bbbf2c4a2ba546ff0e0dd3.png)
for which 
is a perfect square.
be a point in the plane of 
, and 
a line passing through . Let 
be the points where the reflections of lines 
with respect to intersect lines 
respectively. Prove that are collinear.
and 
be positive integers. Prove that there exists a positive integer 
such that for every odd integer 
, the digits in the base-
representation of 
are all greater than .
, the sides have integers lengths and 
. Circle 
has its center at the incenter of . An excircle of is a circle in the exterior of that is tangent to one side of the triangle and tangent to the extensions of the other two sides. Suppose that the excircle tangent to 
is internally tangent to , and the other two excircles are both externally tangent to . Find the minimum possible value of the perimeter of .
Where is Fermat point, 
is its isogonal conjugate and 
and 
are circumcenter and orthocenter of 
the second fermat point? Because if it is, then it's not true. As far as I know, If 
are the first and second Fermat points, and 
are the first and second isodynamic points (just think that 
is the isogonal conjugate of 
), then 
.
s.t. 
is a circle having the segment bounded by the foot of the 
-bisector and the foot of the exterior -bisector. This circle is called the -Apollonius circle. Similarly we define the 
and 
-Apollonius circles. It can be shown that these three circles are coaxal, and the two points through all three of them pass are called the isodynamic points of the triangle 
.
that is fermat point and and are citcumcenter and orthocenter.
, 
, 
are the reflections of a point P in the sides BC, CA, AB of a triangle ABC, then the circumcenter of the triangle 
is the isogonal conjugate Q of the point P with respect to the triangle ABC, and we have 
, 
and 
.
, 
, 
.
, 
and 
. Together with , , (from Lemma 3), this yields YZ || EL, ZX || LD and XY || DE. Hence, the triangles XYZ and DEL are homothetic. Since the triangle DEL is equilateral (Lemma 2), it follows that the triangle XYZ is also equilateral. Hence, instead of saying that the circumcenter of the triangle XYZ is the point F, we can simply claim that the center of the triangle XYZ is the point F.
be the image of the point J in the homothety h. Then, since a homothety maps lines to parallel lines, we have 
. But 
(since the point X is the reflection of the point J in the line BC). Thus, 
. But the point D is the center of the equilateral triangle constructed outwardly on the side BC of triangle ABC, and hence lies on the perpendicular bisector of this side BC. Thus, the line 
, passing through D and being perpendicular to BC, must be the perpendicular bisector of this side BC. In other words, the point lies on the perpendicular bisector of the side BC. Similarly, the same point lies on the perpendicular bisectors of the other two sides of triangle ABC. And this shows that our point coincides with the circumcenter O of triangle ABC. Hence, the image of the point J in the homothety h is the point O.
also is a point in triangle that:![\[ <F'BC=FBA , <F'CB=FCA\]](http://latex.artofproblemsolving.com/5/8/9/589cf07ac938e8cf8b1222f7776ecc7dccd69a41.png)
but I remember there are some much harder problems about the Fermat and isodynamic points.
interesting collinearities, the so-called FNI collinearities. For a list of these collinearities, you can look at my Hyacinthos message #6602, or you can also consult four Forum Geometricorum papers by the late Lawrence S. Evans (paper 1, paper 2, paper 3, paper 4). also is a point in triangle that:
