,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
be the foot of the altitude from 
to the side 
,
,
be the incenter, 
-
and 
the other intersection points of the circle 
with the lines 
and 
,
.
such that![\[(a^2-b^2)(b^2-c^2) = abc.\]](http://latex.artofproblemsolving.com/7/1/8/7188261c40a45f5e15533100432d36abbf515efd.png)
.
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 
be the foot of the altitude from 
,
intersects line 
and 
.
.
can be placed tangent to side 
of 
such that the first circle is tangent to 
,
.
circles of radius 
can also be placed along 
,
.
,
.
subsets of 
is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set 
,
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)
Y by codyj, Davi-8191, Abdollahpour, itslumi, jhu08, mathematicsy, megarnie, Lamboreghini, Adventure10, Mango247, Rounak_iitr, Shadow6885, NicoN9, and 7 other users
Let 
and 
be on segment 
of an acute triangle 
such that 
and 
. Let 
and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .
Proposed by Giorgi Arabidze, Georgia.
and 
be on segment 
of an acute triangle 
such that 
and 
. Let 
and 
be the points on 
and 
, respectively, such that is the midpoint of 
and is the midpoint of 
. Prove that the intersection of 
and 
is on the circumference of triangle .Proposed by Giorgi Arabidze, Georgia.
and 
which are symmetric to 
and 
,
and 
,
.
and 
are parallel.
be the intersection of 
be points on 
such that 
and 
and 
are similar, then 
and 
are similar. Then 
and 
are similar, then 
and 
are similar. Then 
is a parallelogram. Take 
is parallel to 
and 
.
and 
,
so 
and 
and 
so in the triangle 
we have ![\[\frac{\sin(x-A)}{\sin(x)}=\frac{AC}{AB}\]](http://latex.artofproblemsolving.com/e/0/a/e0a847df323278506b00883a5d252f2c3cc34890.png)
and by the same way ![\[\frac{\sin(y-A)}{\sin(y)}=\frac{AB}{AC}\]](http://latex.artofproblemsolving.com/5/5/3/5538c403fa6f3417655b9b5b8b06d74da013aada.png)
![\[\sin(x-A)\sin(y-A)=\sin(x)\sin(y)\]](http://latex.artofproblemsolving.com/e/8/1/e81c6ab921fae057c0cefcd3f0bd8f39ca0c120c.png)
so 
and therefore ![\[\cos(x+y-2A)=\cos(x+y)\]](http://latex.artofproblemsolving.com/b/7/2/b72b67a7257ec66db6ec4dc99d04a8725c93b053.png)
which gives ![\[x+y=A\]](http://latex.artofproblemsolving.com/3/9/0/3909c3b8e840d629a69eb7da870cb712bb5a74ed.png)
and if ![\[<BRC+<BCR=A\]](http://latex.artofproblemsolving.com/9/9/4/99463ee90671b4a0056d7745b25ba43294524b8b.png)
is cycle
,
,
respectively.
is the double of 
,
is the double of 
.
has the same area as 
and 
respectively.
,
.
be the circumcenter of 
.
,
.
,
.
be the midpoints of 
.
.
lies on circle 
,
with diameter 
.
lie on 
,
and 
are collinear, i.e 
is on 
.
,
and 
are collinear, so by converse of Pascal's Theorem, 
which is precisely circle 
,
be the intersection between 
and 
.
and 
.
,
and 
are similar, where we get 
(because both 
and finally:
.
is cyclic.
is cyclic as well. This ends the proof.
,
,
.
, 
and 
so the triangles 
and 
are similar hamotetic (with ratio 2) .On the other 
so we have 
lies on the circumcircle of the triangle 
.
be similar to 
.
is similar to 
and we are done!
.
.
.
.
so there exists a homothety mapping 
to 
,
we have 
,
.
.
is 
which clearly lies on the circumcircle.
From similar triangles 
and 
we have 
Combined with the fact that 
we deduce that 
implying 
Let 
We have that 
We then have 
Similarly, 
From sine rule in 
and from sine rule in 
Combining them, we see that![\[\sin^2(\theta - \angle BAC) = \sin^2 \theta \implies \cos (2(\theta - \angle BAC)) + 1 = \cos^2 \theta \\ \implies 2 \theta = \angle BAC \implies \theta = \dfrac{\angle BAC}{2}.\]](http://latex.artofproblemsolving.com/4/6/c/46c111dede65fac5a72358f01f29f3bce84b0360.png)
we have that 
so 
is cyclic, as desired. 
cyclic, so it suffices to prove 
.
so it suffices to prove 
.
is similar to 
similarity. Clearly angles 
.
,
,
as desired.
such as 
is 
be tangent at 
we have 
tangents 
.
or 
.
or 
are collinear. Similarly, we have 
are collinear
with 
.![\[(AM; PP_{\infty}) \overset{B}{=} (A, BM \cap (ABC); C, B) \implies K \in BM.\]](http://latex.artofproblemsolving.com/7/e/a/7eac08e0b95fc6bcf9ea8de8d3c654fffc280d21.png)
,
.
.
.
.![\[\angle AQC=180-\angle CAQ-\angle C=180-\angle B-\angle C=\angle A=\angle BCT,\]](http://latex.artofproblemsolving.com/4/b/e/4beb5e5549419656924300328427b2dfaa8cb097.png)
so 
,![\[-1=(AX;BC)\overset{B}{=}(AX;TY)\overset{C}{=}(AN';P_{\infty, CT}Q),\]](http://latex.artofproblemsolving.com/7/6/b/76b998ba4911ceec7d11a3c6e7807600d9f9c393.png)
so 
,
,
.
and 
.
and 
,
Therefore, 
be the reflection of 
is a parallelogram. By the definitions of 
.![\[PABM \sim ACBK, \qquad QACN \sim ABCJ.\]](http://latex.artofproblemsolving.com/7/5/1/7516788eef95600fd87f2981c8bc3d5a2a2f2945.png)
Now, letting 
as desired.
,
and thus 
.
.
,
Consider the tangent at 
Hence, 
which finishes.
Observe that from our conditions 
But 
so by SAS 
so 
is cyclic. Hence, 
so 
the reflection of 
and 
are similar, so triangles 
and 
are similar, implying that 
.
is a parallelogram, we should have:![\[ \angle ABT + \angle ACT = \angle ABM + \angle ACN = \angle CAC’ + 180^{\circ} - \angle CAC’ = 180^{\circ} \]](http://latex.artofproblemsolving.com/7/f/5/7f55d990bd0c68210e762ac3473749724d50ab08.png)
Thus providing that 
so then its immediate that 
and similarly 
.
and 
.
by parameterizing 
whicih gives 
.
be 
.
,
,
.
and 
.
and 
.
and 
.
,
.
and 
.
.
which clearly is true so 
.
with the segment 
is the intersection of 
,
.
Therefore 
and we are done.
