AoPS Academy
[/quote]
, do not answer with "
is a solution" only. Either you post any kind of proof or at least something unexpected (like "
is the smallest solution). Someone that does not see that is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
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. for which it is always possible to color crosses in two colors such that any balanced set has crosses of both colors.
be an acute triangle with 
, and let 
be the projection from 
onto 
. Let 
be a point on the extension of 
past such that 
. Let 
be on the perpendicular bisector of 
such that and 
are on the same side of and![\[\frac12\angle ALE=1.4\angle ABE+3.4\angle ACE-558^\circ\]](http://latex.artofproblemsolving.com/c/6/3/c63eeb053f1fa8a29bef93516814ae121382219e.png)
Let the reflection of across 
and 
be 
and 
, respectively. Let 
and 
such that 
. Let 
and 
intersect the circumcircles of 
and 
at 
and 
, respectively. Let 
and 
intersect 
and 
at 
and 
. Let 
intersect at 
. Prove that 
.
![\[
C(p-1, k-1) \equiv (-1)^{k-1} \pmod{p}
\]](http://latex.artofproblemsolving.com/4/0/7/4072235afbaa1f2d8d4c9ca044615a113a4306ce.png)
for 
, where 
is the binomial coefficient 
choose 
.
satisfy that: 
and 
. Find the maximum: 
be a convex quadrilateral. Suppose that the circles with diameters and 
intersect at points 
and . Let 
and 
. Prove that the points , , and are concyclic. and are not the diagnols)
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 
-gon 
we have 
. Prove that 
for all reals 
.
be positive integers, where 
. It is given that there exist a positive integer 
such that
. Prove that the above equation is true for all positive integers 
. (Here 
is the sum of digits of 
taken in base 
).
, we have 
, 
, and 
centimeter. Find the length of diagonal .
, we have , , and 
centimeter. Find the length of diagonal .
, we have , , and centimeter. Find the length of diagonal .
, the point can be any point on some circle.
, we have , 
, and centimeter. Find the length of diagonal ., the point can be any point on some circle.
; any point onto the bigger arc 
has 
, that is, 
is cyclic, wherefrom 
.
so
But the LHS has the same sign as that of
so that 
.
such that 
is isosceles with angle 
=50° Then notice that 
is cyclic and 
=90° and then get that is the center of the circumscribed circle. From here it is obvious that is the radius 
(the case 
is analogous). Let the perpendicular bisector of 
intersect at (this is due to in 
). Since 
is isosceles, the perpendicular bisector will pass through . We now have 
and 
, so 
. Also, 
and if we let 
, we will get 
, but 
too, so 
is cyclic. Now we have 
, so 
is isosceles and 
. Now in the case 
we directly get that is a deltoid, so 
, while 
and so .
such that 
where 
the circumcircle of 
, we have 
hence 
are cyclic but 
li e on the 
implies 