[/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 
,
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 positive reals such that 
.
Y by khan.academy, cubres, Mathdreams
Consider a triangle 
and some point 
on 
. The perpendicular from to 
intersects the circumcircle of 
at 
and the perpendicular from to 
intersects the circumcircle of 
at 
. Show that the line 
does not depend on the choice of .
(Shining Sun, USA)
and some point 
on 
. The perpendicular from to 
intersects the circumcircle of 
at 
and the perpendicular from to 
intersects the circumcircle of 
at 
. Show that the line 
does not depend on the choice of .(Shining Sun, USA)
This post has been edited 6 times. Last edited by Bata325, Apr 13, 2025, 2:54 AM
Reason: italics
Reason: italics
denote the circumcenter of 
.
.
denote the foot of the altitude from ![\[\angle{BAO} = \frac{180^{\circ} - \angle{AOB}}{2} = \frac{180^{\circ} - 2\angle{C}}{2} = 90^{\circ} - \angle{C} = \angle{HXC} = 180^{\circ} - \angle{BXH} = \angle{BAQ}.\]](http://latex.artofproblemsolving.com/e/2/d/e2d9375826c14c5e7197102e8c1d14f2dad41068.png)
Hence, the points 
are collinear. Similarly 
are all collinear.
hence 
are the centres of the circles 
respectively
and 
such that 
and 
and 
are the perpendicular bisectors of the radial axes 
and 
hence 
and 
so 
and 
and the required claim is proved
![\[\angle{BAO} = \frac{180^{\circ} - \angle{AOB}}{2} = \frac{180^{\circ} - \angle{C}}{2} = 90^{\circ} - \angle{C} = \angle{HXC} = 180^{\circ} - \angle{BXH} = \angle{BAQ}.\]](http://latex.artofproblemsolving.com/7/d/9/7d972f3079ea48f9b57a8f171777f9dabada44e0.png)
Hence, the points 
hence 
where ![\[\measuredangle BAQ = \measuredangle CXQ = 90 + \measuredangle BCA\]](http://latex.artofproblemsolving.com/9/4/2/9429375d4f0c28c6a6939e228bc971645d77e8ec.png)
which implies that ![\[\measuredangle CAQ = \measuredangle CXP = 90 + \measuredangle CBA\]](http://latex.artofproblemsolving.com/0/0/a/00ab53c14582f60a8519bd3daef06f3048629879.png)
which implies that 
