[/quote]
,
is such that 
for any positive integers 
.
such that 
for all positive integers 
.
.
satisfy that 
.
and 
.
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 
.
and circumcircle of 
and 
be the midpoint of 
and 
be the midpoint of 
.
,
and 
,
in the array can be changed into either 
or 
so that the row-sums and column-sums remain unchanged. (Note that 
be 
given points in the plane, and let 
be a real number. Alice and Bob play the following game. Firstly, Alice constructs a connected graph with vertices at the points 
a non-negative real number 
,
,
.
such that 
and 
are connected by an edge for every 
.
is not fixed and the first selected vertex always has to be 
.![\[
\frac{1}{k+1} \sum_{j=0}^{k} r_{i_j} \geq r;
\]](http://latex.artofproblemsolving.com/1/7/9/1795db8e4a509dd465f6ff462093ae75b04de2b4.png)
otherwise, Alice wins. Depending on 
,
for which Bobby has a winning strategy.
,
,
,
,
,
be positive integers and let 
.
divides 
and 
.
is the centre of the excircle opposite the vertex 
This excircle is tangent to the side 
,
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 
Y by MathAllTheWay, mathematicsy, Quidditch, ys311, Adventure10, ItsBesi, Rounak_iitr, DEKT
Let 
be a triangle with circumcentre 
. The points 
lie in the interiors of the sides 
respectively, such that 
is perpendicular to 
and 
is perpendicular to 
. (By interior we mean, for example, that the point 
lies on the line 
and is between 
and 
on that line.)
Let
be the circumcentre of triangle 
. Prove that the lines 
and are perpendicular.
Netherlands (Merlijn Staps)
be a triangle with circumcentre 
. The points 
lie in the interiors of the sides 
respectively, such that 
is perpendicular to 
and 
is perpendicular to 
. (By interior we mean, for example, that the point 
lies on the line 
and is between 
and 
on that line.)Let
be the circumcentre of triangle 
. Prove that the lines 
and are perpendicular.Netherlands (Merlijn Staps)
and 
and 
and 
are cyclic. 
.
be the angle bisector of 
,
the point on the same side of 
is cyclic. Clearly 
.
not including 
'
.
is cyclic, 
.
.
and 
.
and 
.
,
.
,
is cyclic. So 
,
is a line segment.
.
is antiparallel to 
,
.
cuts 
and 
.
is cyclic isosceles trapezoid inscribed in circle 
,
intersecting at 
.
in this case. Since 
,
is tangent to 
in this case. Since 
,
is tangent to 
be intersection of circles 
other than 
cuts 
is A-median of 
.
respectively ![$\color[rgb]{0.75,0,0} \angle CZB = \pi - (\angle ZBC + \angle BCZ)= \pi - (\angle ZAB + \angle CAZ) = \pi - \angle CAB$](http://latex.artofproblemsolving.com/9/4/4/944b8dcfcb7b3739c1a890d37125a69521798c6b.png)
.
be foot of the A-symmedian of 
in this case.
are tangents of 
at 
is on A-median 
of 
are antiparallels to 
WRT 
,
are cyclic ![$\color[rgb]{0.75,0,0} \angle CZ'B = \pi - (\angle Z'BC + \angle BCZ') = \pi - (\angle EBP + \angle PCF) =$](http://latex.artofproblemsolving.com/e/1/e/e1e9d01cccef02aed0f746f0940fbf6d2f833491.png)
![$\color[rgb]{0.75,0,0} \pi - (\angle CAP + \angle PAB) = \pi - \angle CAB$](http://latex.artofproblemsolving.com/7/9/7/79760c1f4e18cd1340822a49f7444ff22dda63f2.png)
.
are identical, circles 
form a pencil intersecting at 
and their centers 
collinear.
,
and points 
are identical.
is arbitrary, isosceles 
form centrally similar files with similarity centers 
of trapezoids 
are fixed 
are always collinear and points 
are always collinear.
.
for arbitrary 
.
be 
.
,
are concyclic, implying 
concyclic, implying 
.
,
.
is 
.
.
be the midpoint of 
wrt 
.
is the radical axis of all circles 
,
Therefore, 
are concyclic. Similarly, 
are concyclic. Then after inversion with pole 
Denote 
,
Prove that 
and 
are orthogonal.
meet at 
.
Meanwhile, Pascal's Theorem applied to cyclic hexagon 
yields 
But as 
,
Hence, 
and the desired result follows. 
be the projection of 
meet at 
By Pascal's Theorem applied to cyclic hexagon 
,
Therefore, 
,
Then the homothety 
implies that the locus of 
be the intersection of 
with 
with 
Because the points 
respectively, and the quadrilateral 
is cyclic 
we have ![\[\angle EDF=\angle YDX=\pi-\angle BOC=\pi-2\angle BAC=\pi-\angle FKE.\]](http://latex.artofproblemsolving.com/5/1/6/51620d72382179ca991a75cf7df9175f1ea96539.png)
Therefore, the quadrilateral 
is cyclic, because 
we deduce that 
Because 
we can see that ![\[\angle CDE=\angle CDY=\frac{\pi}{2}-\angle YCD=\frac{\pi}{2}-\angle XBD=\angle XDB=\angle FDB.\]](http://latex.artofproblemsolving.com/7/d/1/7d19f8a517bdc9f06874481d2b35bb6b668b7008.png)
Finally, we conclude that 
as claim.
condition and 
are kind of a joke; they just mean that the point 
,
.
be the centres of the circles 
is the perpendicular bisector of the segment 
,
and 
But now it is obvious. (Here, 
denotes the power)
we must have 
gives 
we obtain 
as desired 
on 
are isogonal conjugate . 
, similar with 
on 
, then 
.
so 
, then 
and 
, which implies 
.
.
.
,
.
,
.
we have 
as desired.
and 
are cyclic as 
and 
.
which finishes by perpendicularity lemma.
,
,
,
are 
as they are not on the unit circle.
cyclic
indeed
![\[
\angle FDB=90^{\circ}-\angle OBC=\angle A.
\]](http://latex.artofproblemsolving.com/7/0/3/70383ebe7679b987e5f3ae688c81fac3c2a22967.png)
Similarly, 
as well. Therefore,![\[
\angle FKE=2\angle A=\angle FDB+\angle EDC=180^{\circ}-\angle FDE,
\]](http://latex.artofproblemsolving.com/a/d/5/ad5a2b6bc82649c4f8de086c268bd3be8b78f107.png)
so quadrilateral 
,
,
,
,
and ![\[2\measuredangle EDC = 2\measuredangle OCD = 2\measuredangle OCB = \measuredangle COB = 2\measuredangle CAB \]](http://latex.artofproblemsolving.com/4/d/f/4df4ae1bdf128f4cb456546dea53694df91b4a48.png)
Thus, 
implying that ![\[\measuredangle EDC = \measuredangle CAB = \measuredangle BDF\]](http://latex.artofproblemsolving.com/e/a/e/eae36a7a7478e5f948e08f344cce6b0d475163bd.png)
Now, we have the next claim.
and is in fact the midpoint of arc ![\[ \measuredangle EKF = 2\measuredangle EAF = 2\measuredangle CAB =\measuredangle EDF \]](http://latex.artofproblemsolving.com/d/0/4/d047ace2d648cb3272594d59301dd376e4b9ce4b.png)
Thus, 
-
as well, we can conclude that 
.
.
,
.
is cyclic.
,
,
as desired.
,
and 
are antiparallel in 
so 
is cyclic. Thus, 
,
bisects 
.
we see that 
too, and 
.
is cyclic. Let 
be the orthocenter of 
due to 
.
.
has center 
.
,
,
,
and 
be 
is cyclic by opposite right angles. Then 
.
so 
.
so 
as desired.
.
are concyclic 
are cyclic
![\[ \angle DEF = 90 - \angle OCE = 90 - \angle OBF = \angle DFE. \]](http://latex.artofproblemsolving.com/4/a/a/4aabed5c70ffa36cfd3ada2739f1f8daf9f8b3ca.png)
Thus, 
is isosceles. But, 
,
are cyclic quadrilaterals. From this, we have 
.
so points 
lie on a circle. For obvious reasons 
being cyclic and 
