[/quote]
,
be real numbers satisfying
be a fixed square and 
a variable point on segment 
.
is constructed such that 
is on segment 
and 
.
be the intersection point of lines 
and 
.
be a positive integer. In an 
board, two opposite sides have been joined, forming a cylinder. Determine whether it is possible to place 
.
.
denote the number of positive integers less than 
for which the equation 
has at least 
solutions in 
that can be written in the form
where 
are pairwise coprime positive integers such that 
,
,
.
,
,
be the midpoints of diagonals 
and 
,
and that 
,
.
.
or 
tile.
be an integer. Dominoes are placed on an 
such that each row and each column has a value of 
.
be a monic polynomial with integer coefficient. And suppose there exist 4 distinct integer 
such that 
.
be the circumcircle of triangle 
,
inmixtilinear circle is tangent to 
at 
respectively. 
and 
and 
is the intersection of 
and 
and 
and 
.
Prove 
Find maximum 
are roots of 
,
by 
grid of squares coloured in the same way as a standard checkerboard. Find the total number of ways to place 
counters on white squares so that each square contains at most one counter and no two counters are in diagonally adjacent white squares.
such that 
and :
Y by Amir Hossein, Davi-8191, a28546, nguyendangkhoa17112003, AlastorMoody, DashTheSup, pog, superagh, centslordm, mathematicsy, HWenslawski, Adventure10, Mango247, MS_asdfgzxcvb
In an acute triangle 
the points 
and 
are the feet of the altitudes through 
and 
respectively. The incenters of the triangles 
and 
are 
and 
respectively; the circumcenters of the triangles 
and 
are 
and 
respectively. Prove that 
and 
are parallel.
the points 
and 
are the feet of the altitudes through 
and 
respectively. The incenters of the triangles 
and 
are 
and 
respectively; the circumcenters of the triangles 
and 
are 
and 
respectively. Prove that 
and 
are parallel.This post has been edited 1 time. Last edited by Amir Hossein, Jul 29, 2013, 5:38 PM
Reason: Edited Title and Changed Format of The Problem.
Reason: Edited Title and Changed Format of The Problem.
,
and 
respectively. Triangles 
then ![\[\frac{QD}{BQ}=\frac{AP}{PE}=\frac{AM}{MB},\]](http://latex.artofproblemsolving.com/b/e/4/be4720ae74d4f19dacb7fdb9bc67b788741d126b.png)
thus points 
is cyclic. So, ![\[\angle AI_{1}I_{2}+\angle ABI_{2}=\angle AI_{1}F+\angle FI_{1}I_{2}+\frac{1}{2}\angle CBA=\frac{\pi}{2}+\frac{1}{2}\angle CBA+\angle BMI_{2}+\frac{1}{2}\angle CBA=\angle MI_{2}B+\angle BMI_{2}+\frac{1}{2}\angle CBA=\pi.\]](http://latex.artofproblemsolving.com/a/9/0/a900116336902f2352ed83244ccb38a82ea21b5f.png)
is cyclic, so circumcircles of triangles 
and 
intersect at incenter 
of triangle 
.
is incenter of triangle 
,
and 
is orthocenter of triangle 
and 
then we need to prove that 
now let the incircle of 
at 
respectively then we will prove that
is perpendicular bisector of 
,if we prove this then 
so 
is similar t
be a point on 
such that 
is perpendicular bisector of 
then we have
so 
then we are done
and the circumcircle of 
.
.
.
.
.
.
by the double angle formulas, since 
and the analogous for 
,
is cyclic and 
.
and 
.
.
.
and since 
we see that 
,
.
,
.
,
and therefore 
is isosceles, so that 
.
is 
,
.
and we're done.
.
,
is homothetic to 
.
,
and 
,
.
,
,
are concyclic.
meet at 
,
.
bisects 
meet at incenter 
are collinear which is immediate from the radical axis theorem regarding 
hence we are done. Q.E.D
.
H
.
.
and 
are equal.Hence 
.
.
.
.
and the result follows....
Note that 
and 
intersect at 
By some easy trig bash, we obtain![\[AI_1 \times II_2= BI_1 \times IB_2= 2 \times \left( 2R \sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2} \right) ^2,\]](http://latex.artofproblemsolving.com/b/1/c/b1cd870a75b73c8b5696459161dd75a3ba0bd51f.png)
is the circumradius of 
so by PoP it follows that 
and 
It follows that 
so it suffices to show that 
which is simple angle chasing.
is perpendicular to 
and 
and 
where 
the bisector of 
are pairwise similar.
thus 
is cyclic.
.
.
and 
thus 
.
q.e.d.
be the incenter of 
.
.
is cyclic.
is the second intersection point of the circles 
are the common chords of the circles 
.
are bisectors they intersect at 
passes from 
.
(obvious) and we conclude that 
be the incenters of 
,
on 
are inscribed in the circle of diameter 
,
Therefore, 
It follows that ![\[IA \cdot II_1 = IA(IA - I_1A) = IA^2(1 - \cos A) = 2 \cdot IA^2 \cdot \sin^2 \frac{A}{2}.\]](http://latex.artofproblemsolving.com/7/9/c/79cc9aca64af33899d7138e12788af8881a6bd00.png)
Then if 
is the projection of 
,
Similarly, 
,
are concyclic.
are concyclic and 
are concyclic. Then since 
is the common chord of 
,
In particular, it is sufficient to show that 
Then since 
,
,
and the radius of the incircle of triangle 
.![\[ r_a (b \cos A + c \cos A + 2R \cos A \sin A) = \sin A \cdot \cos^2 A \cdot bc \]](http://latex.artofproblemsolving.com/0/0/7/007f9ce857a46f3f4b310ec4a7a1d176da9995d9.png)
and obviously ![\[ r(a+b+c) = \sin A \cdot bc \implies \frac{AI_1}{AI}=\frac{r_a}{r} = \cos A \]](http://latex.artofproblemsolving.com/c/6/e/c6e0be070f27eac14d717c9be6f136c0349e00ee.png)
Evidently 
,![\[ AI (AI - AI_1) = BI (BI - BI_2) \]](http://latex.artofproblemsolving.com/e/c/f/ecf5709902cea6ab714a6ec5a2d18f8117cc5f61.png)
With our work above this rearranges to ![\[ AI^2 (1-\cos A) = BI^2 (1-\cos B) \]](http://latex.artofproblemsolving.com/d/d/1/dd1c7a7ee2bbfebe178c95e038da731db01d7495.png)
true as 
.
,
.![\[ \angle AIH + \angle I_2I_1I = 90 \]](http://latex.artofproblemsolving.com/0/a/2/0a24ab392390ebe1242d98b49b0e18ff35c57c63.png)
Using cyclic 
and standard angle formulas ![\[ 180-\angle AIC + \angle IBA = 90 \]](http://latex.artofproblemsolving.com/f/d/3/fd3790f1dfea18b00bbeee916df0188638fca171.png)
![\[ 180-(90+\frac{B}{2})+\frac{B}{2} = 90 \]](http://latex.artofproblemsolving.com/6/f/9/6f9f738f759abc34f8ca04e8689acaf99b088dd7.png)
We are done.
![[asy]
unitsize(3cm);
pair A, B, C, D, E, F, I, I1, I2, I3, O1, O2, K, P;
A = dir(70);
B = dir(200);
C = dir(-30);
D = foot(A, B, C);
E = foot(B, A, C);
F = foot(C, A, B);
I = incenter(A, B, C);
I1 = incenter(A, E, F);
I2 = incenter(B, D, F);
I3 = incenter(C, D, E);
O1 = circumcenter(A, C, I1);
O2 = circumcenter(B, C, I2);
P = IP(Line(C, I, 10), Line(A, B, 10));
K = IP(Line(C, I, 10), Line(I1, I2, 10));
draw(A--B--C--A);
draw(D--E--F--D);
draw(A--I, blue);
draw(B--I, blue);
draw(C--P, green);
draw(I1--I2--I3--I1, orange + linewidth(1));
draw(circumcircle(A, B, I1), red + linewidth(0.8));
draw(circumcircle(B, C, I2), red + linewidth(0.8));
draw(circumcircle(C, A, I3), red + linewidth(0.8));
draw(O1--O2);
clip(CR((0,0),1.75));
dot("$A$", A, W);
dot("$B$", B, S);
dot("$C$", C, S);
dot("$D$", D, S);
dot("$E$", E, E);
dot("$F$", F, W);
dot("$I$", I, E);
dot("$I_1$", I1, E);
dot("$I_2$", I2, S);
dot("$I_3$", I3, S);
dot("$O_1$", O1, E);
dot("$O_2$", O2, S);
dot("$P$", P, W);
dot("$K$", K, W);
[/asy]](http://latex.artofproblemsolving.com/6/d/5/6d540bd1945e9e581c1896d765037997d72adb62.png)
be the incenters of triangles 
and let 
.
and 
are the bisectors of 
and 
Due to properties of the orthic triangle,
Hence,
We have,
By the Law of Sines, we find that
with the last equality following from the half-angle formula. Again, by the Law of Sines,
Therefore,
Hence, 
and 
are also cyclic.
and 
is 
.
.
.
Also,
Hence,
Thus, 
.
.![\[\left(\frac{IA}{IB}\right)^2 = \frac{II_2/IB}{II_1/IA} \iff \frac{\sin^2 \frac B2}{\sin^2 \frac C2} = \frac{1-\cos B}{1-\cos C}\]](http://latex.artofproblemsolving.com/e/5/9/e599770d388253ae65193ebc6363f1bd2b98097a.png)
as the incenter of 
,
,
and 
)
and 
,
Similarly for 
is independent of vertex. Now note that
Thus 
,
,
.
as well, proving the result.
.
,
are cyclic quadrilaterals.
.
,
.
,
and now 
and 
.
.
.
.
.
and 
,
,
.
.
.
which is what we wanted to prove. We are ready.
index.
are cyclic.
and 
are cyclic. So, 
and, similarly, 
.
.
and 
.
.
which means 
.
which means 
but 
.
.
which means 
which means 
.
at 
and 
,
and 
.
,
and 
are both cyclic quadrilaterals
and 
,
,
,
are similar, so 
is perpendicular to the bisector of 
.
and 
to meet 
and 
and 
,
,
.
,
is cyclic. Furthermore, since 
,
.
is perpendicular to the bisector of 
is cyclic.
be the line through 
is perpendicular to the bisector of 
,
and 
,
to 
(This is a well-known property of the orthic triangle.) This spiral similarity sends 
so 
This implies that 
Hence 
We will repeatedly use this fact, which we call (*) for convenience.
be the incenter of the triangle.
are concyclic.
since 
is cyclic. Also, 
so 
Therefore, 
However, 
so 
concluding our proof of the claim.
and the spiral similarity in (*) sends 
In other words, we wish to show that the spiral similarity rotates things by 
Indeed, it rotates 
to 
and the angle between these two lines is 
Thus 
as claimed.
and 
we see that 
and 
Since 
Therefore, 
and 
are all cyclic suffices as this gives that 
and 
is 
and thus gives that 
to 
,
is similar to 
,
,
,
in the sensible way.
are cyclic. (Proof: power of point on 
)
.
by angle chasing. Hence:
.
as desired.
,
.
are bisectors.
is cyclic.![\[ II_1 \cdot IB = II_2 \cdot IC \iff \left(\frac{BI}{CI}\right)^2 = \frac{1 - \cos C}{1 - \cos B} \]](http://latex.artofproblemsolving.com/3/b/7/3b7254a0edc38d219ca5662588a8e0259facf97a.png)
which is obvious. 
and ![\[ \measuredangle I_1I_2 = \measuredangle BI_1 + \measuredangle I_2C - \measuredangle BC = \measuredangle BI + \measuredangle CI - BC \]](http://latex.artofproblemsolving.com/5/b/b/5bb3c599b6ab83705a78be1a52ac1a24ca7862d1.png)
is parallel to the tnagent at 
which is perpendicular to 
