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.
satisfying![\[\begin{aligned}
\begin{cases}
a+b+c+d=0,\\
a^2+b^2+c^2+d^2=12,\\
abcd=-3.\\
\end{cases}
\end{aligned}\]](http://latex.artofproblemsolving.com/8/b/1/8b1923e0303c0aa79332ee4540ee1885a9cf3f08.png)
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 
).
satisfy 
; in other words, let be an isosceles triangle with base . Let be a point inside the triangle such that 
. Denote by the midpoint of the segment . Show that 
.
moves on a circle tangent in 
to 
respectively. Let be the second intersection of 
with this circle. 
is a harmonic cyclic quadrilateral, so, if 
is the intersection between the tangent at to the circle and , we find 
, meaning that is the symmedian of 
, which is precisely what we want.
satisfy ; in other words, let be an isosceles triangle with base . Let be a point inside the triangle such that . Denote by the midpoint of the segment . Show that .
, 
, 
, 
, 
,
, 
.
The theorem of Sinus in the triangle 
. The trigonometrical form of the Menelaus' theorem in for the inner point 
.
, 
, 
results: 
, i.e. the ray 
is in the triangle 
, meaning
be a triangle and be the midpoint of . Suppose is the intersection of the tangents of the circumcircle at and 
. Then 
.
and 
are the tangents of 
at and .
). 
is isosceles 
(U) is also tangent to CA at A, AB is polar of C WRT the circle (U). CP meets (U) at P and again at Q. If P' is a reflection of P in 
and CP' meets (U) again at Q'. PP'Q'Q is isosceles trapezod symmetrical WRT CM, the diagonals PQ', P'Q meet on CM and on the polar AB of C, 
. Let 
be on 
such 
is parallogram. Then 
. Hence 
are isogonal points wrt 
. Hence 
is isogonal line . This imply 
.
be a triangle and 
its circumcircle. Let the tangent to at and intersect at . Then coincides with the - symmedian of 
. The proof can be found here. and are tangents to the circumcircle of 
at points and respectively. The tangents and intersect at . So 
coincides with the - symmedian of . Since 
is the median of , 
. or 
. So 
.
is the symmedian.
. Then the problem is equivalent to 
, aka. 
is a symmedian in 
. Notice that and are the tangents to 
so we are done.
is the humpty point, and so 
is the midpoint of . Furthermore, we have that 
and 
. Hence, we get that 
where this follows from similar triangles 
and 
.
meet circumcircle of at S. Since then is tangent to and since 
then 
is symmedian of 
so 
and so 
.
. The angle condition implies that and are tangent to the circumcircle of , then we know that 
is the -symmedian of , which implies that 
thus we have that 
, which clearly gives us that 
be a symmedian in triangle 
, so it suffices to show that lies along the symmedian. But this is evident by the tangent intersection definition.
satisfying 
such that 
or on the line 
.
be the point of intersection of the circumcircle of the triangle 
with line 
(Fig. 1 and 2). Since 
,
lie on a circle. Also![\[(1)\qquad\angle BPC=\angle BQC=\angle AQC.\]](http://latex.artofproblemsolving.com/a/9/8/a988a64bff438e45b57a05249ab94c426c3398f4.png)
![\[\angle APM+\angle BPC=\angle APM+\angle AQC=180^\circ.\]](http://latex.artofproblemsolving.com/2/5/9/259d348a3417b67c0faa0548f45e32447269a7a7.png)
![\[\angle APM+\angle BPC=\angle AQM+\angle AQC=180^\circ.\]](http://latex.artofproblemsolving.com/9/c/f/9cfac1d50178272a49af4608a3d80936c833cb06.png)
. Since 
, so by analogy with the above, we obtain the equality 
.![\[\angle APM+\angle BPC=180^\circ,\]](http://latex.artofproblemsolving.com/9/6/1/9618056d4f538976f67525c7e20375d9c00e9972.png)
.
.
.
,
and 
Hence, 
is the 
Notice that 
,
,
and 
.
So
Therefore 
,
,
is cyclic and 