is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
and 
at 
and 
,
and 
meet at 
,
and 
meet at 
Let 
be the point of intersection of the lines 
and 
be the point of intersection of the lines 
and 
Prove that 
is the circle that is tangent to the line segment 
,
.
be a positive integer. Consider the following equation: ![\[ \{x\}+\{2x\}+ \dots + \{nx\} = \lfloor x \rfloor + \lfloor 2x \rfloor + \dots + \lfloor 2nx \rfloor\]](http://latex.artofproblemsolving.com/8/2/9/8299b3ceb1650418c0a373536cb1d2931900efd9.png)
a) For 
,
.
real solutions.
and 
intersect at two points 
and 
.
be the line tangent to these circles at 
and 
,
be the line parallel to 
on 
on 
and 
meet at 
;
and 
;
and 
.
.
is such that 
for any positive integers 
.
such that 
for all positive integers 
.
.
satisfy that 
.
with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of 
be real numbers . Prove that : There exist positive integer 
such that 
Where 
)
passes through 
,
be the midpoint of 
.
,
and 
Y by Infinityfun, MathLuis
Let 
be the incenter of triangle 
, and 
be the perpendicular bisector of 
. Suppose that 
is on the circumcircle of triangle , and line 
and intersect at point 
. Point 
is on such that 
.Suppose that line 
and the midsegment of that is parallel to 
intersect at 
. Show that 
(Note: In a triangle, a line connecting two midpoints is called a midsegment.)
be the incenter of triangle 
, and 
be the perpendicular bisector of 
. Suppose that 
is on the circumcircle of triangle , and line 
and intersect at point 
. Point 
is on such that 
.Suppose that line 
and the midsegment of that is parallel to 
intersect at 
. Show that 
(Note: In a triangle, a line connecting two midpoints is called a midsegment.)
and Let 
and let 
with 
being closer to 
m
be the reflection of 
now 
so 
are collinear and 
so 
is cyclic 
now 
so 
is cyclic so as 
since 
so inverting with center 
we have 
since 
so let 
then since 
lies on 
we have 
now if 
at 
so similarly taking 
we conclude that the fixed line is 
.
is obviously projective so 
.
.
cases. Let 
and 
,
.
.
and 
so we need to show 
.
Calculating 
.
.
And now it is easily seen that we are done in this case.
.
where 
and 
are intersections of 
with 
and 
.
Notice that 
Which is equivalent with 
But 
And again as before its easily seen that we are done. And now the hardest case 
(we symmetricaly get 
Via angle chasing we can see 
then its not hard to see 
Next just calculate 
Now its trivial to check that we are done. Just noticed sythetic for 
then 
and claim follows easily.
(where 
,
.
be a line passing through 
is the Newton's line of quadrangle 
.
are harmonic quadrilateral, so 
,
is perpendicular to 
,
,
and 
and 
be the reflection of 
be the circumcenter of 
.
.
,
.
be the reflection of 
.
and 
in 
,
.![\[AX' \cdot AL = AN \cdot AI \Longleftrightarrow \frac{AX'}{AN} = \frac{AI}{AL}\]](http://latex.artofproblemsolving.com/6/8/7/68750ab1dedb95f573f65c3ce7ea5268914400a2.png)
Just notice that![\[\frac{AN}{NX'} = \frac{AN}{NX} = \frac{AI}{IF} = \frac{AC}{CF} = \frac{AL}{LB} = \frac{AL}{LI}\]](http://latex.artofproblemsolving.com/b/b/c/bbc45b34e4021d66216ff1abdcbfc3c29bb6898e.png)
so 
,
,
so we are done. ![[asy]
defaultpen(fontsize(10pt));
size(12cm);
pen mydash = linetype(new real[] {5,5});
pair A = dir(120);
pair B = dir(225);
pair C = dir(315);
pair I = incenter(A,B,C);
pair F = extension(A,I,B,C);
pair O = circumcenter(A,B,C);
pair L = 2*foot(O,A,I)-A;
pair X = midpoint(A--F);
pair N = midpoint(A--I);
pair X1 = 2*N-X;
pair S = midpoint(A--C);
pair T = midpoint(A--B);
pair P = dir(238);
pair D = 2*foot(O,B,I)-B;
pair E = 2*foot(O,C,I)-C;
pair Q = extension(A,P,D,E);
pair M = extension(Q,I,S,T);
pair R = extension(P,rotate(90,P)*I,D,E);
pair M1 = 2*foot(M,D,E)-M;
real s = 0.9;
pair K1 = s*unit(M1-X1) + extension(M1,X1,A,O);
pair K2 = s*unit(X1-M1) + extension(M1,X1,A,O);
draw(K1--K2, royalblue+1);
draw(A--B--C--cycle, black+1);
draw(B--D, mydash);
draw(C--E, mydash);
draw(R--D);
draw(A--L);
draw(A--P);
draw(P--R);
draw(T--S);
draw(Q--I);
draw(R--M);
draw(circumcircle(A,B,C));
draw(circumcircle(P,I,N));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$I$", I, dir(315));
dot("$F$", F, dir(225));
dot("$L$", L, dir(270));
dot("$X$", X, dir(45));
dot("$N$", N, dir(60));
dot("$X'$", X1, dir(60));
dot(S);
dot(T);
dot("$P$", P, dir(270));
dot(D);
dot(E);
dot("$Q$", Q, dir(55));
dot("$M$", M, dir(225));
dot("$R$", R, dir(180));
dot("$M'$", M1, 2*dir(180));
label("$k$", K2, dir(60));
[/asy]](http://latex.artofproblemsolving.com/f/7/3/f73975566fbd5366f0b9d054be93ce1afe023a5c.png)
be four different points on the plane such that no three of them are collinear and 
.
s
passes through 
,
.
.
Hence, 
is symmedian, this finishes the first part. Also,
Thus; 
is cyclic. Finally,
intersects 
.
and 
.
Also, 
is tangent to 
.
is tangent to 
.
.
.
.
is a symmedian of 
.![$[AK]$](http://latex.artofproblemsolving.com/3/2/a/32a180e681de5ca6aeffe6bfa17cf23f043b9361.png)
.
.
.
be the center of 
.
.
.
,
so it's enough to prove that 
because 
.
the reflexion of 
so it's enough to prove that 
.
and 
then see that 
is almost similar to 
then let 
then we can guess that 
but with angles: 
But 
so 
(well-known) and this is sufficient to prove that 
, we don't know if this point it's on 
.
the excenters relative to 
,
in center of 
and 
then 
and 
.
This implies that 
is a parallelogram and we need to prove that 
are concyclic. By homothety with center 
with ratio 
we need to prove that the midpoints of 
,
,
it's on the circumcircle of 
and it's well-known that the others points also are on this circle. 
then 
later see that the conjugate isogonal of 
it's the 
it's a median this implies that 
and 
.
so 
is harmonic and 
then 
this implies that 
(well-known) so 
and 
.
cyclic as this would prove 
cyclic giving 
.
are reflections about 
is parallel to the tangent to 
.
and 
because it lies on the perpendicular bisectors of 
respectively. Then 
cyclic.
and 
.
so if 
then the lines 
and 
,
so 
giving 
.
as needed, so 
and similarly 
are cyclic.
where the first equality comes from shooting lemma, so by power of a point, 
again at 
implies 
satisfies 
,
and 
;
,
.
be the reflection of 
.
intersects 
is cyclic, and 
,
is tangent to 
as well, thus 
is cyclic by Power of a Point. This implies that 
,
.
. Consequently, it is well-known that 
by alternate segment theorem, which provides the result.
,
.
by alternate segment theorem, and symmedian property, which suffices.
.
.
is cyclic. Now, if 
again at 
by right angle and bisector. Projecting this onto 
is harmonic, and as 
are antipodes of each other. Consequently, 
,
inversion.
.
and 
at 
are concyclic.
,
iff 
s
.
.
and circumcenter 
.
.
meets the perpendicular to 
intersects the ninepoint circle at 
is a parallelogram.
is a parallelogram. Since the ninepoint circle is the circle shifted from 
with length 
be the reflection of 
.
are concyclic, by reflection over 
are concyclic. 
and 
thus, 
.
are concyclic. These imply that 
are concyclic and since 
we get 
as desired
