[/quote]
,
,
denote the number of its positive divisors (including 1 and itself). Determine all positive integers 
for which there exists a positive integer 
.
let 
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.
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 
be points on the line segments 
and 
respectively such that 
is a parallelogram. Prove that 
.
that satisfy 
for all 
and 
satisfies the following recursion formulas: 
,
.
Y by anantmudgal09, Davi-8191, Jasurbek, donotoven, geometry6, ike.chen, Adventure10, Mango247, Rounak_iitr, and 3 other users
Let 
be a triangle with 
and circumcenter 
. The bisector of 
intersects 
at 
. Let 
be the reflection of with respect to the midpoint of . The lines through and perpendicular to intersect the lines 
and 
at 
and 
respectively. Prove that the quadrilateral 
is cyclic.
be a triangle with 
and circumcenter 
. The bisector of 
intersects 
at 
. Let 
be the reflection of with respect to the midpoint of . The lines through and perpendicular to intersect the lines 
and 
at 
and 
respectively. Prove that the quadrilateral 
is cyclic.This post has been edited 1 time. Last edited by Amir Hossein, Jul 29, 2013, 5:39 PM
Reason: Edited Title and Changed Format of The Problem.
Reason: Edited Title and Changed Format of The Problem.
is midpoint of 
,
be the reflection of 
is isosceles trapezoid. The bisector of 
.
.
is cyclic, so 
- quadrangle 
is cyclic, so points 
lie on the same circle.
,
is isosceles, and so is 
,
;
we infer 
be the intersection of line 
.
is the angle bisector of 
,
are collinear. Then we have 
because all three lines are perpendicular to 
.
,
,
,
.
,
.
.
.
,
.
,
by SAS similarity. Hence, 
.
,![\[\angle{BXC}=\angle{BXD}+\angle{CXD}=\angle{YBE}+\angle{YCE}=180-\angle{BYC}\]](http://latex.artofproblemsolving.com/2/0/8/208ff0a3b85e1dac6b2b3cf408b3c607a6e3b380.png)
,
,
.
,
.
since 
.
.
the midpoint of arc 
and 
.
and 
.
and 
.
are concyclic. and
.
are concyclic.
are concyclic.
and let 
be the midpoint of arc 
the reflection of 
is a isosceles (and hence cyclic) trapezoid. Now 
because they are both isosceles (since 
)
and so 
is cyclic; hence 
so 
is cyclic as well, and we're done.![[asy]/* DRAGON 0.0.9.6
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and 
are similar.
so 
( we have obviously 
)
and 
are similar.Then 
,
so done. 
meets 
and 
at 
and 
intersects circle 
.
.
and ![\[\triangle DXA\sim \triangle GOA\implies \dfrac{DX}{DA}=\dfrac{GO}{GA}\implies DX=\dfrac{DA\cdot GO}{GA}\quad (1)\]](http://latex.artofproblemsolving.com/5/b/6/5b62ca2f094462f65579d40dae9bea9c2dd998a4.png)
![\[Y'E=X'D=\dfrac{BD\cdot DC}{DX}=\dfrac{AD\cdot DG}{DX}=\dfrac{DG\cdot GA}{GO}\]](http://latex.artofproblemsolving.com/7/9/b/79be34058868c6455b6f2c8abec416dc3b96c126.png)
.
and 
.
.
are collinear. Therefore, 
are collinear. So ![[asy]/* File unicodetex not found. */
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Extend 
to meet the circumcircle of 
at 
It is easy to prove that 
bisects 
and as a consequence 
are collinear. Now, note that 
By some easy ratio computations using PoP of 
it follows that 
and similarly 
The rest is simple angle chasing.
is a straight line
as 
.
is a straight line.
is a straight line.
is cyclic
is cyclic.
is cyclic and thus we are done
and combining we have the desired. 
with 
be 
.
is an isosceles trapezoid, meaning it is cyclic.
and 
are isosceles and have the same base angle, so they are similar. Then 
,
is cyclic. Note![\[
XD \cdot TD = AD \cdot ND = BD \cdot CD,
\]](http://latex.artofproblemsolving.com/6/f/7/6f759b2bd11320cdb9a8948f2cef321f39dd648d.png)
implying 
is cyclic, which finishes.
intersect 
again at 
,
.
is cyclic. Let 
,
are collinear, hence 
.
,
,
,
is cyclic. 
we have that 
is a rectangle with 
and 
.
.
.
so we're done.
is a Cyclic Isosceles Trapezoid
is a Cyclic Isosceles Trapezoid
is a Cyclic Rectangle
hence the pairwise radical axes of 
,
and 
don't concur.
cyclic which means 
is cyclic 
be the arc midpoint of minor arc 
Notice that
)
Now, notice that 
via reflection and in turn this means
Thus, we must have that 
.
is parallel to 
So,
But then, this means that 
which in turn implies that we must have 
.
be the intersection of 
.
,
.
and 
which gives us that
and thus 
and thus 
is indeed cyclic.
and let 
is isosceles trapezoid, so it suffices to prove that 
is cyclic. Let 
and let 
be the midpoint of 
.
,
,
is cyclic. Thus it's enough to prove that 
is cyclic. Let 
is isosceles trapezoid, so it suffices to prove that 
is cyclic. Let 
.
,
is the perpendicular bisector of 
.
is tangent of 
,
and 
,
and 
Thus, the desired conclusion is equivalent to 
and 
.
Let 
,
Squaring and using Heron's formula, this is equivalent to 
However, we have 
as desired.
,
.
is cyclic. Indeed, ![\[\measuredangle XAD = \measuredangle OAN = \measuredangle ANO = \measuredangle DYE = \measuredangle DZE.\]](http://latex.artofproblemsolving.com/b/7/a/b7a6bba29cce403e907de6626ad9fd92d6fc18ab.png)
Also similarly observe that 
is trivially cyclic. Hence it now suffices to show 
is cylic, which is trivial as ![\[DX \cdot DZ = DA \cdot DN = DB \cdot DC, \]](http://latex.artofproblemsolving.com/2/5/a/25a71852bced836ac4c14965d565659ab103290a.png)
as desired. 
,
.
is isosceles. Let 
is cyclic.
lies on 
must intersect line 
;
is cyclic.
![[asy]
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be the arc-midpoint of minor-arc 
and since 
,
so it is well-known that 
is the perpendicular bisector of 
Thus, 
is cyclic.
be the midpoint of arc 
.
are concyclic, 
,
.
,
.
be the antipode of 
and 
since 
and 
are both perpendicular to 
,
.
,
concyclic.
.
,
,
.![\[
X=:(p:b^2S_B:c^2S_C)=\left(\frac{p}{p+b^2S_B+c^2S_C},\,\frac{b^2S_B}{p+b^2S_B+c^2S_C},\,\frac{c^2S_C}{p+b^2S_B+c^2S_C}\right)
\]](http://latex.artofproblemsolving.com/8/9/d/89dabcbf64df0c994f70929b0f719c0e573dc950.png)
and let 
.
and 
so![\[
a^2\left(\frac{b^2S_B-c^2S_C}{\sigma}-\frac{b-c}{b+c}\right)+\frac{b^2p}{\sigma}-\frac{c^2p}{\sigma}=0
\]](http://latex.artofproblemsolving.com/1/a/8/1a8487af671cb30302e14afdceebaeb1ed58272f.png)
by the barycentric perpendicular formula. Multiplying by 
gives
Note that![\[
2(b^2cS_B-c^2bS_C)=b^2c(a^2+c^2-b^2)-c^2b(a^2+b^2-c^2)=bc(b-c)(a^2-(b+c)^2)
\]](http://latex.artofproblemsolving.com/3/c/8/3c859cd1a2f56f7734fc2e3cba56a743d9d004ba.png)
so 
and 
.![\[
Y=:(q:b:c)=\left(\frac{q}{q+b+c},\,\frac{b}{q+b+c},\,\frac{c}{q+b+c}\right).
\]](http://latex.artofproblemsolving.com/a/8/1/a810b9210f7fb234df2683602da44a32ffff7352.png)
Then 
so![\[
a^2\left(\frac{b-c}{q+b+c}+\frac{b-c}{b+c}\right)+(b^2-c^2)\frac{q}{q+b+c}=0
\]](http://latex.artofproblemsolving.com/a/9/b/a9be74772222acd7be31da1a5993b02270587378.png)
by the barycentric perpendicular formula. Multiplying by 
gives![\[
a^2(q+2b+2c)+q(b+c)^2=0
\]](http://latex.artofproblemsolving.com/a/2/6/a2667928a034c598e79200223b52e63af76fedf3.png)
so 
and 
.
.
.
,
The value of 
which agrees with our other value of 
