[/quote]
,
,
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 
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 
and 
is a parallelogram. Prove that 
.
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
Y by Adventure10, Mango247, and 1 other user
Let 
be a triangle with circumcenter 
and incenter 
. The points 
and 
on the sides 
and 
respectively are such that 
and 
. The circumcircles of the triangles 
and 
intersect at 
. Prove that 
.
be a triangle with circumcenter 
and incenter 
. The points 
and 
on the sides 
and 
respectively are such that 
and 
. The circumcircles of the triangles 
and 
intersect at 
. Prove that 
.This post has been edited 1 time. Last edited by Amir Hossein, Jul 29, 2013, 5:40 PM
Reason: Edited Title and Changed Format of The Problem.
Reason: Edited Title and Changed Format of The Problem.
.
intersect 
and 
respectively. Let circumcircle of triangle 
intersect line 
at 
.
is point on 
.
is cyclic, and 
,
and 
are congruent, so 
triangles 
and 
are congruent 
.
,
.
lies on circumcircle (call it 
)
,
.
lies beetwen 
,
as ![\[\angle SIL=\angle PDB=\angle PEC=\angle ILP,\]](http://latex.artofproblemsolving.com/1/9/d/19dfc957f7c641e8aff2573803a2410038793044.png)
so quadrangle 
is cyclic, but points 
lie on the circle with center 
be arbitrary points on the sides of 
and 
the reflections of 
of 
fixed points on lines 
(
)
is cyclic(circle 
.
meets 
(
for previously described 
where 
.
be the power of 
with respect to 
.
)
,
,
,
is linear in 
and cyclic. But then![\[ -(P,\omega)\sum{a} = \sum yca(BF) + zba(CE) = \sum xbc(AE+AF) = abc\sum x = abc \]](http://latex.artofproblemsolving.com/b/6/7/b67e8fabffead03bb6bd163103d6bbea96e2eea1.png)
gives 
,
is the Miquel point of 
,![\[ - (P',\omega) = y'(AF\cdot AB) + z'(AE\cdot AC) \]](http://latex.artofproblemsolving.com/f/9/a/f9a397c519edef402763a89b66e44cbfe2e3ea4a.png)
yields![\[ -(P',\omega)\sum{x} = \sum x'(yBF\cdot BA+zCE\cdot CA) = -(P,\omega)\sum{x'}, \]](http://latex.artofproblemsolving.com/9/d/2/9d2e0da9a4b0444cae90fcb67a85e37862508606.png)
as desired. (In the original problem, we have 
,
.
be the feet from 
.
with Miquel point 
.
and 
as linear functions in 
(or 
;
is the (directed) angle of rotation 
.
.![\[ \overline{m} = \frac{ \sum_{\text{cyc}} d(ab - ac)}{\sum_{\text{cyc}} d (b-c)},\]](http://latex.artofproblemsolving.com/6/c/7/6c70ed23fe75ca9fe96126884a3d991ffc38bc82.png)
![\[{|m| = \frac{ \sum_{\text{cyc}} d(ab - ac)}{\sum_{\text{cyc}} d (b-c)} \cdot \frac{\sum_{\text{cyc}} \overline{d} (b-c)}{ \sum_{\text{cyc}} \overline{d}(ab - ac)}}.\]](http://latex.artofproblemsolving.com/c/f/9/cf90516196c33fdc41c12f754c86410d0c86cb61.png)
,
and 
cancel out; it then remains to check that the denominators are equal, which is not hard using 
be any points on 
and P an arbitrary point on the same plane.
meets the circle 
at 
and 
are defined similarly. If 
,
are concyclic.
be the intersection of 
-
be the reflection of 
,
are the points that are translated constant length from 
denote the lines 
,
,
,
.
.
clockwise by 
and get 
.
,
,
passes through 
meet at some point 
.
,
or quadrilateral 
is cyclic. Similarly, 
is cyclic, implying 
.
,
and 
analogously), say it intersect the angle bisector of A at X, and the circle of center O, radius OI (circle 
)
.
,
and 
analogously. Now, we have that BFYD is cyclic. But <BYP = < IYP = 180 - <IXP = <AXP = <AFP = 180-<AFP and therefore BFPYD is cyclic pentagon, so 
.
and so 
.
so OQ=OI, which is what we wanted. 
and suppose that 
points 
s.t. 
But by the lemma above 
So 
are concyclic. Denote this circle as 
to be point on 
is the reflection of line 
in line 
It can be seen that 
So that 
since 
are concyclic. So 
are reflection of lines 
in lines 
respectively. Let 
since 
and 
bisects 
we know that 
and so 
So 
are both reflection of 
in 
hence 
Similarly 
and 
.
it implies that 
so 
are cyclic. Hence 
and by the definition above 
and 
are limit cases.
.
is intersection point of perpendiculars to 
are collinear and 
.
be circumcircles of the triangles 
,
,
and 
be diameters of the 
,
are similar and analoguosly ... 
the triangles 
are similar.
with coeffitsiyent 
,
and 
.
)
and 
are the perpendicular bisectors of 
,
,
.
:
.
,
,
,
.
,
.
,
.
and 
.
lie on 
,
thus 
.
.
hence 
Throughout the solution 
denotes reflection over arbitrary line 
Denote by 
by 
intersection of 
with perpendicular to 
Define 
analogously.
concur at 
is cyclic. By length condition 
moreover 
so 
and all similar statements we deduce that 
are concyclic, and clearly 
concur at 
But 
and cyclically, so 
and all analogous equalities hold. Thus 
and we are done.
.
,
be the reflection of 
,
,
.
.
,
.
,
meet 
again at 
.
.
we have 
.
,
(this is the key step of 2006 G9).
.
Thus, 
,
intersects 
.
and 
are congruent, done!
at 
respectively.
at 
It's known that if any circle passing through 
intersects 
at 
respectively then 
But 
so in fact 
cyclic, similarly 
cyclic.
so 
centered at 
done. 
and others follow analogously. Let 
It follows 
with oriented lengths (i.e. exactly one of 
lie on segments 
,
This completes the proof. 
are concurrent on 
.
and 
.
are same, we need to show 
are concurrent on 
meet 
.
have that option so we may prove the following claim.
is cyclic.
so 
.
,
then 
so 
.
is cyclic so 
.
.
as the reflections of 
.
in 
in 
the 2 points must coincide. Therefore, 
which means 
is concyclic and the rest can be seen easily.
be the reflections of 
lies on a circle with radius of 
be the midpoints of 
.
so 
and the intersection of 
and 
all lie on one circle. 
so the intersection of 
(intersection with 
all intersect on the smaller circle
.
be the projections of 
,
and 
.
be the antipodes of 
on their respective circles, then it follows by taking a coordinate system along the sides of the triangle that the antipodes 
of 
are the reflections of 
of 
onto 
,
,
.
be the reflection of 
.
.
is the desired intersection point. Notice that the final reflection is a reflection over the vector 
.
.
,
.
.
as desired.
,
be the feet of the altitude from 
,
,
,
be the reflection of ![\[AF'+AE'=AB+AC-BF'-CE'=AB+AC-AF-AE=AB+AC-BC=AB'+AC'\]](http://latex.artofproblemsolving.com/c/1/7/c1799bc6194e3bec36700dffabd5b488896e5051.png)
so 
.
so 
,
is cyclic. Note that 
since ![\[\angle ID''D=\angle ID'B=\angle IE'C=\angle IE''E\]](http://latex.artofproblemsolving.com/a/8/9/a89f3e81075f365e90423dbc36faaa0761612421.png)
so 
and 
intersect on the circle 
.
also concurs with those. Clearly, 
and 
denote the power of point 
that 
,
.
(using the midpoint 
of arc 
and see it equals 
)
(analogously).
is cyclic we invert wrt 
,
are collinear using the facts that 
.
be the circle with center 
,
and 
,![\[AS=\frac{b+c}{2\cdot \cos\frac{\alpha}{2}}\]](http://latex.artofproblemsolving.com/c/e/4/ce457d7acad77d5b13d37663cbbfac638f957428.png)
Proof : Let 
,![\[AM=AN=\frac{b+c-a}{2}\]](http://latex.artofproblemsolving.com/0/1/2/012b9ca197a037ed6d0da46e6415c329d7d71d19.png)
It follows that 
.
,![\[R=SI=\frac{a}{2\cdot \sin\widehat{BIC}}=\frac{a}{2\cdot \sin\left(90+\frac{\alpha}{2}\right)}=\frac{a}{2\cdot \cos\frac{\alpha}{2}}\]](http://latex.artofproblemsolving.com/3/9/4/39416cc1623262427d5b2aa31dc972008f64d083.png)
The conclusion follows by summing the expressions for 
,
are concurrent at a point 
,
are concurrent at a point 
be the point where line ![\[BX=\frac{BD+BF}{2\cdot \cos \frac{\beta}{2}}=\frac{CA}{2\cdot \cos\frac{\beta}{2}}\]](http://latex.artofproblemsolving.com/e/8/c/e8cd0ca25668e3e1e7a5544ab6141da3de482bff.png)
However, note that if we let ![\[R_{(AIC)}=\frac{CA}{2\cdot \sin\widehat{AIC}}=\frac{CA}{2\sin\left(90+\frac{\beta}{2}\right)}=\frac{CA}{2\cdot \cos\frac{\beta}{2}}\]](http://latex.artofproblemsolving.com/c/4/c/c4ce4cec36691ccc96d7f40aed4f1d75e843ad48.png)
It follows that 
,
.![\[\angle YPX=\angle DPX-\angle DPY=180-\frac{\angle B}{2}-\frac{\angle C}{2}=90+\frac{A}{2}=180-\angle YIX,\]](http://latex.artofproblemsolving.com/9/6/3/9636bb79b286a784ee5401e63f80c574e3c2bd0d.png)
hence 
,
cyclic, where 
b
![\[Pow(A,\Omega)=AI\cdot AX=2R\cdot \sin(\alpha/2)\cdot 4R \cdot \sin (\beta/2) \sin (\gamma/2)\]](http://latex.artofproblemsolving.com/2/5/4/25491f9891b34c98268707ecb8ad72c9dc4d3e96.png)
, which is symmetric in 
on 
and 
.
is constant. Note that ![\[ L = \frac{c}{a+c} \cdot A + \frac{a}{a+c} \cdot C, \]](http://latex.artofproblemsolving.com/7/9/9/799b90c65122efe96eee5958d17b309321ff52d7.png)
so we have by linearity on the function 
that 
which is indeed constant.
on 
so 
is cyclic. It remains to show that 
.
)
.![\[ \text{Pow}(B, (MIN)) = BM \cdot BI = \frac{b/2}{\cos(\angle B/2)} \cdot \frac{r}{\sin(\angle B/2)} = r \cdot \frac{b}{\sin(\angle B)} = 2Rr = \text{Pow}(B, \omega), \]](http://latex.artofproblemsolving.com/e/d/f/edf3d22fdafc32c302f74c64814818413b10f566.png)
so 
all lie on a circle centered at 
as well). Let 
.
As 
,
which is clear as 
is similar to 
since they are both isosceles triangles with vertex angle 
.
is cyclic. Invert around 
on 
and similarly for the others. We have 
Since 
is just 
,
is clearly also 
is the circle centered at 
