[/quote]
,
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 
be an isosceles triangle 
)
.
passes through 
.
and 
,
be the midpoint of 
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 
be positive reals such that 
.
Y by Adventure10
On a circle with diameter 
, let 
be an arbitrary point distinct from 
and 
. Points 
are the midpoints of chords 
, and points 
are the midpoints of smaller arcs restricted by these chords. Lines 
and 
meet at point 
, and lines 
and 
meet at point 
. Prove that lines 
and 
concur.
, let 
be an arbitrary point distinct from 
and 
. Points 
are the midpoints of chords 
, and points 
are the midpoints of smaller arcs restricted by these chords. Lines 
and 
meet at point 
, and lines 
and 
meet at point 
. Prove that lines 
and 
concur.
as the reference triangle. We have,
Since 
Plugging this into the circumcircle equation gives,
In a similar fashion we have
Trivially,
Also,
Therefore, the equation for line 
The equation for line 
is
The equation for line 
is
By the concurrency lemma, it suffices to show that
However, the determinant in question turns out to be
by the Pythagorean Theorem.
and 
.
