[/quote]
,
,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
and 
,
,
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.
is an integer, then it is a perfect square.[/quote]
.
is at least rational, so that 
must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs 
for which this happens are 
and 
,
,
The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?
interesting if 
of 
and 
is also a divisor of 
,
and 
.
is prime, 
with 
and 
and circumcenter 
.
and 
be points on the line segments 
respectively such that 
is a parallelogram. Prove that 
.
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
and 
be points on segment 
denote the circle passing through 
denote the circle passing through 
)
.
Y by Davi-8191, centslordm, HWenslawski, son7, ImSh95, Adventure10, Mango247, Rounak_iitr, ItsBesi, and 1 other user
Let 
be a triangle with 
. Let 
and 
be two different points on line 
such that 
and 
is located between and 
. Suppose that there exists an interior point 
of segment 
for which 
. Let the ray 
intersect the circle at 
. Prove that 
.
be a triangle with 
. Let 
and 
be two different points on line 
such that 
and 
is located between and 
. Suppose that there exists an interior point 
of segment 
for which 
. Let the ray 
intersect the circle at 
. Prove that 
.
so by the converse of the Alternate Segment Theorem, 
is tangent to 
.
.
so 
is tangent to 
.
.
is similar to 
so 
so 
is cyclic. Hence 
.
.
.![\[ \angle BQR = \angle DQR=\angle DCR=180-\angle QDR-\angle DCQ=180-(x+2y)-x=180-2x-2y. \]](http://latex.artofproblemsolving.com/5/b/e/5bec4f3d1db057aa0c027a3c12ab3a7fc10264fb.png)
Also, note that 
.
,
,
,
we deduce that 
.
is cyclic, angle chasing yields 
as desired.
.
.
points 
are concyclic.
.
.
.
and 
.
.
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/a/5/0/a50d57e5908c26d1a1247166a8173c13bb2f0968.png)
as 
respectively. Let 
and let 
.
is cyclic. Clearly, 
and 
,
,
.
.
,
is tangent to 
.
.
,
which tells us that 
.
,
be 
let the circle with center 
.
be the circumcircle of 
.
and 
are collinear.
)
and 
as the opposite points of 
followed by a simmetry wrt the bisector 
.
,
and 
be the images of 
and 
are isosceles trapezoids with bases 
and 
,
;
.
.
and 
are now defined as the intersections of one of the diagonals of isosceles trapezoid 
with the axis of the other diagonal; this construsction is obviously simmetrical wrt the axis of 
,
is also an isosceles trapezoid and is therefore cyclic. This concludes the proof.
,
is tangent to 
and 
is cyclic
,
are similar.
implying 
is cyclic quad.
is tangent to the circumcircle so 
is a cyclic quadrilateral so 
.
and 
,
and 
,
.
cut 
at 
,
,
at 
be a point at infinity on line 
,
are concyclic. Therefore 
Also 
Thus, to show that 
,
that satisfy 
,
Also, 
passing through 
Keeping in mind that 
,
is tangent to 
are the midpoints of 
,
is a diameter of 
Therefore 
is harmonic 
is the midpoint of 
,
.
,
,
,
.
which gives 
.
.
is tangent to 
.
is tangent to 
.
is tangent to 
.
is cyclic.
.
and thus 
is tangent to 
and 
.
and 
.
is tangent to 
is tangent to 
with 
it's not too bad.
,
but also 
.
,
is cyclic. Since 
so 
.
,
,
.
.
.
.
since 
.
.
so 
instead of conventionally in terms of 
that ultimately isn't needed if you frame the problem correctly. That being said, the solution is very short when written out.
implies 
.
,
![\[\measuredangle DRC = \measuredangle ARC = \measuredangle ABC = \measuredangle BAC + \measuredangle ACB = \measuredangle BAQ + \measuredangle QBA = \measuredangle BQA = DQA\]](http://latex.artofproblemsolving.com/5/7/0/570f08eca50c8cac9abb20d13d0553d67beb14ce.png)
which implies the required claim.![\[\triangle ABQ \sim \triangle ACB, \triangle PAD \sim \triangle PDC, \triangle AQR \sim \triangle ADC \text{ and } \triangle PAB \sim \triangle PBC\]](http://latex.artofproblemsolving.com/f/6/0/f609887479062282e1b28b582c99f0fc79f228d3.png)
and 
.
.
and 
which implies that 
.
and 
.
.
and 
as well.![\[BQ = \frac{AB\cdot BC}{AC}\]](http://latex.artofproblemsolving.com/7/a/f/7af01e242d889dd7686272345e43f7c374c09b11.png)
Now, we do some length chase.
But then, this means that in fact,![\[BQ = \frac{AB \cdot BC}{AC} = QR\]](http://latex.artofproblemsolving.com/e/e/7/ee76afc26a2be6a7751cb7efc28dd64eec5b4ec6.png)
which was indeed the required result.
,
and now the circumcircle of 
,
is cyclic.
,
.
from 
.
,
.
and 
,
and 
and 
so 
so 
so 
are concyclic. We also have 
so we get 
and we have 
and we have 
and we have 
so it suffices to show that 
(basically 
and now we can use our previous calculations), which is equivalent to 
but we have 
and 
,
,
,
.
,
is cyclic, so 
,
,
are concyclic.
denote directed angles modulo 
.
,
.
.
as desired.
is tangent to 
,
concyclic.
so 
implying 
which gives us that 
giving us that 
and express 
and 
in terms of angles of 
using the cyclicity and tangency fact to get that both angles are equal implying 
(after inscribed angle chase)
QED
are concylic.
and 
.
.
.
and 
implying 
are concylic as desired. 
,
.
and 
by angle sum property, 
.
thus, 
.
as 
and 
.
(since 
and 
.
as desired. 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/e/c/3/ec39a82c37cf691b4d90ff5ef854e4394ef8de72.png)
Firstly we can see that 
Since 
Now the points 
are concyclic points too. By Angle Chasing we get, 
Also, 
and 
are concyclic points.
is an isosceles triangle or, 
We get some angle condition that 
And 
Therefore, 
