[/quote]
,
crosses are placed. A set of 2025 cells we will call balanced if no two of them are in the same row or column. It is known that any balanced set has at least 
crosses.
,
be the projection from 
onto 
.
be a point on the extension of 
past 
.
be on the perpendicular bisector of 
such that 
are on the same side of ![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of 
and 
be 
and 
,
and 
such that 
.
and 
intersect the circumcircles of 
and 
at 
and 
,
and 
intersect 
and 
at 
and 
.
intersect 
.
.
such that 
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
,
is the binomial coefficient 
choose 
.
satisfy that: 
and 
.
be a convex quadrilateral. Suppose that the circles with diameters 
intersect at points 
and 
and 
.
be positive integers and 
be an odd prime such that:![\[
a^p \equiv 1 \pmod{p^n}.
\]](http://latex.artofproblemsolving.com/3/4/8/34815b8211ded2f37b0653525dc4b436356e6eee.png)
Prove that:![\[
a \equiv 1 \pmod{p^{n-1}}.
\]](http://latex.artofproblemsolving.com/0/7/a/07a4ed2ab3c94f41028080f7789cfb6e2973955c.png)
be a function such that for every regular 
-
we have 
.
for all reals 
.
be positive integers, where 
.
.
.
is the sum of digits of 
taken in base 
)
where 
Y by ehuseyinyigit
Let 
be circumcenter of a non-isosceles triangle 
and 
be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and 
is the circumcenter of , 
. Prove that 
bisects 
be circumcenter of a non-isosceles triangle 
and 
be a point in the interior of 
. Let 
be foots of perpendicular lines from to 
. Suppose that 
is cyclic and 
is the circumcenter of , 
. Prove that 
bisects 
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This post has been edited 2 times. Last edited by m4thbl3nd3r, Mar 28, 2025, 5:00 PM
antipode of 
can be easly MMP'd as 
boh have degree one by simply checking where your cases are 
and 
now which means that 
colinear is degree 4 and the cases are 
and 
so you are done.
which is the Miquel point of 
.
so 
collinear and passes through the 
.
Note that 
and
.
Since 
we are done.
)
be diameter of 
,
be midpoint of 
be foot of perpendicular line from 
.
is collinear and 
(Because 
)
.
and 
.
and we're done
