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 
).
passes through points and and intersects the sides and of triangle at points and 
, respectively. The circumcircles of triangles and 
meet at distinct points 
and . Prove that 
.
passes through points and and intersects the sides and of triangle at points and , respectively. The circumcircles of triangles and meet at distinct points and . Prove that .
denote the image of . An inversion 
around causes 
to be collinear, and 
to be collinear. Then, also sends the circumcircle of 
to the circumcircle 
of 
. As such, 
lies on the midpoint of the tangents from to .
lies on the polar of , hence 
. Inverting back implies the result. 
![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -9.627272727272732, xmax = 11.245454545454548, ymin = -6.2381818181818245, ymax = 7.652727272727277; /* image dimensions */
pen ffxfqq = rgb(1,0.4980392156862745,0); pen yqqqqq = rgb(0.5019607843137255,0,0);
/* draw figures */
draw(circle((-1.1727272727272753,0.16181818181818255), 3.4075754207256947), linewidth(1) + ffxfqq);
draw((-3.2454545454545487,5.052727272727277)--(-4.136876511792053,-1.51907132570026), linewidth(1) + red);
draw((-4.136876511792053,-1.51907132570026)--(1.1909090909090891,-2.292727272727274), linewidth(1) + red);
draw((1.1909090909090891,-2.292727272727274)--(-3.2454545454545487,5.052727272727277), linewidth(1) + red);
draw(circle((-0.9928108389069386,1.4008136157456776), 4.290789790551068), linewidth(1) + ffxfqq);
draw(circle((-3.4253709792748843,3.8137318387997823), 1.2519902589284826), linewidth(1) + red);
draw((-4.662867079350029,3.6237765526121537)--(-1.1727272727272753,0.16181818181818255), linewidth(1) + yqqqqq);
draw((-4.662867079350029,3.6237765526121537)--(-3.2454545454545487,5.052727272727277), linewidth(1) + yqqqqq);
/* dots and labels */
dot((-1.1727272727272753,0.16181818181818255),dotstyle);
label("$O$", (-1.1,0.3436363636363624), NE * labelscalefactor);
dot((1.1909090909090891,-2.292727272727274),dotstyle);
label("$A$", (1.3,-2.7290909090909126), NE * labelscalefactor);
dot((-4.136876511792053,-1.51907132570026),dotstyle);
label("$C$", (-4.063636363636367,-1.3290909090909115), NE * labelscalefactor);
dot((-3.2454545454545487,5.052727272727277),dotstyle);
label("$B$", (-3.172727272727276,5.234545454545457), NE * labelscalefactor);
dot((-2.2450261848480517,3.3962803150017655),linewidth(4pt) + dotstyle);
label("$K$", (-2.1727272727272755,3.543636363636365), NE * labelscalefactor);
dot((-3.5820063289067603,2.57157847640949),linewidth(4pt) + dotstyle);
label("$N$", (-3.9909090909090943,2.8709090909090915), NE * labelscalefactor);
dot((-4.662867079350029,3.6237765526121537),linewidth(4pt) + dotstyle);
label("$M$", (-5.318181818181822,3.6890909090909103), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]](http://latex.artofproblemsolving.com/2/b/3/2b31342bc35be2d6d6015784950642be638b62ce.png)
such that 
where 
