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.
+1 w
of the vertices of the regular polygon 
we write the number 
and in the remaining ones we write the number 
Let 
be the number written on the vertex 
A vertex is good if ![\[x_i+x_{i+1}+\cdots+x_j>0\quad\text{and}\quad x_i+x_{i-1}+\cdots+x_k>0,\]](http://latex.artofproblemsolving.com/b/f/4/bf4cf984af7f310c1eb90b6b37126bfca9181305.png)
for any integers 
and 
such that 
Note that the indices are taken modulo 
Determine the greatest possible value of such that, regardless of numbering, there always exists a good vertex.
satisfying ![$$(ab+bc+ca)\left[\frac{1}{(a+b)^{2}}+\frac{1}{(b+c)^{2}}+\frac{1}{(c+a)^{2}}\right]\ge \frac{9}{4}+k\cdot\frac{a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b)}{(a+b+c)^{3}}$$](http://latex.artofproblemsolving.com/f/6/e/f6ed10f7fff1cc94edd8f451e75718a0916a8bfa.png)
holds for all 
be a scalene triangle, 
be the median through 
, and 
be the incircle. Let touch 
at point 
and segment 
meet for the second time at point 
. Let 
be the triangle formed by lines and and the tangent to at . Prove that the incircle of triangle is tangent to the circumcircle of triangle .
be real numbers such that 
Prove that
Where 
Where 
are are the lengths of the sides of a 
gon such that
then![$$(n-2)\left[\sum_{i=1}^{n}{\frac{a_i^2}{(1-a_i)^2}}-\frac n{(n-1)^2}\right]\geq(2n-1)\left(\sum_{i=1}^{n}{\frac{a_i}{1-a_i}}-\frac n{n-1}\right)^2.$$](http://latex.artofproblemsolving.com/9/3/e/93e5529518807b92cd9e0382f5fb07a293128b04.png)
be an acute triangle. Tangents on the circumscribed circle of triangle at points 
and 
intersect at point . Let 
and 
be a foot of the altitudes from onto 
and 
and let 
be the midpoint of . Prove: is the orthocenter of the triangle 
.
cuts 
in half.
be a positive integer and let 
be an irrational number. Show that for every integer 
, there exists a positive integer 
such that 
.
and be positive integers with 
. In a group of people, each one or always, what is the parity of people who speak to is a subset of size 
of the set of people. The answer can only
” or “
”. and is it possible to determine which people speak the truth and
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.
and 
.
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 