AoPS Academy
, let 
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer such that 
.
let 
be the foot of the altitude from 
to the side 
and 
, 
, 
be the incenter, -excenter, and 
-excenter, respectively. Denote by 
and 
the other intersection points of the circle 
with the lines 
and 
, respectively. Prove that 
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
. If this is still too hard assume in 
that 
is a prime.
is divided into rectangles with two cells. Two rectangles are called adjacent if they share a segment of length 1 or 2. In each rectangle the amount of adjacent with it rectangles is written.
be the orthocenter of acute triangle 
, let 
be the foot of the altitude from to 
, and let 
be the reflection of across . Suppose that the circumcircle of triangle 
intersects line at two distinct points 
and 
. Prove that is the midpoint of 
.
can be placed tangent to side 
of 
such that the first circle is tangent to 
, subsequent circles are externally tangent to each other, and the last is tangent to 
. Similarly, 
circles of radius 
can also be placed along in this manner. The inradius of is 
, where and are relatively prime positive integers. Find 
.
, where 
. Each of the 
subsets of 
is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set 
, we then write 
for the number of subsets of 
that are blue.
and 
of , ![\[ f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2). \]](http://latex.artofproblemsolving.com/1/8/1/18184e9196e6c73601653d85f1cfd6d9798a8e43.png)
. Let its incircle be tangent to at , while the 
mixtilinear incircle be tangent to 
at 
. Let the tangent to at intersect at . Prove that 
is a cyclic quadrilateral.
be the touch point of mixtlinear-excircle.Let 
be the foot of the A-Nagell cevian,and let 
be the reflection of over the perpendicular bisector of ..Than 
are collinear. with power 
and compose with the reflection over -angle bisector.Notice that 
and 
and 
and so we're done.
. Let 
. Let be the reflection of in the midpoint of . Let the isogonal ray to 
intersect at 
. Let be the reflection of in the perpendicular bisector of . Then 
are collinear. as the foot of 
proves your claim. Also note that setting as the foot of -altitude immediately solves 
.
so pretty nice.
be the point where the -excircle of 
touch 
.
problem 
, we find that 
and 
are isogonals w.r.t. to 
. So, 
is the center of the spiral similarity that takes 
to 
. Also, from a well known lemma, we have 
.
. But, 
. So, 
is the center of the spiral similarity that takes 
to 
. Thus, 
, and we are done!
