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
cells of an 
square grid are colored black, and the remaining cells are white. The cost of such a coloring is the minimum number of white cells that need to be recolored black so that from any black cell 
, one can reach any other black cell 
through a sequence 
of black cells where each consecutive pair 
are adjacent (sharing a common side) for every 
. Let 
denote the maximum possible cost over all initial colorings with exactly black cells. Determine a constant 
such that![\[
\frac{1}{3}n^{\alpha} \leq f(n) \leq 3n^{\alpha}
\]](http://latex.artofproblemsolving.com/0/a/0/0a06d4ff52479c8cb785c6450cea8748bf7769b6.png)
for any 
.
. Let 
be an interior point such that 
, 
and 
. Find 
.
be real numbers satisfying 
. Find the maximum value of![\[P = \dfrac{a(b+c)}{a^2+2bc+3} + \dfrac{b(a+c) }{b^2+2ca +3 } + \dfrac{c(a+b)}{c^2+2ab+3}.\]](http://latex.artofproblemsolving.com/7/b/2/7b2ec0a87ca05d3cb29da888a4dfa73e10f238f8.png)
with orthocenter 
, isogonal conjugate 
lies on the McCay cubic. Let 
be the pedal triangle of wrt and let 
. Let 
be the second intersection of 
with 
and let 
be the Poncelet point of 
. Then the lines 
are concurrent.
and let 
be the second intersection of with 
be the pedal triangle of 
wrt . Then 
.
be the circumcevian triangle of wrt . Then 
and 
are homothetic.
. On the other hand, from the Lemma 1, we see that 
. Therefore from the Lemma 2, we see that 
and 
, respectively, are collinear and 
is the insimilicenter of 
and , so 
.
of is tangent to at . Hence by applying Monge-D'Alembert's theorem to 
, we see that 
are collinear.
, 
. Prove that
such that 
and there exists an integer 
such that 
is a multiple of 
.
is chosen on the side 
of a rectangle 
. From the vertex 
, the perpendicular 
is dropped to the segment 
. The segments and divide the rectangle into three parts such that each of them has the inscribed circle (see figure). Prove that if the circles tangent to are equal then the third circle is also equal to them.
be a non-constant function from the set of positive integers into the set of positive integer, such that 
divides 
for all distinct positive integers 
, 
. Prove that there exist infinitely many primes 
such that divides 
for some positive integer 
.
of degree 
has exactly 
real roots, counted with multiplicity. We know that the coefficient of 
is 
, all the roots are less than or equal to , and 
. What values can 
take?
lies inside triangle 
such that 
and 
. Point 
is the midpoint of segment 
. Point 
lies on segment 
with 
. Prove that 
.
be a triangle such that 
. Let 
be points on 
respectively such that 
. Let 
meet the circumcircle of triangle at point 
. Let be a point on 
. Prove that 
if and only if lies on the circumcircle of triangle 
.
must be between and , here I found a nice solution, but if it is not, then I am getting a contradiciton! 
must be between and , here I found a nice solution, but if it is not, then I am getting a contradiciton! is outside the segment , but in my drawing lies on the segment 
, while liues on the ray 
, with 
between and 
. I think it might depend on this, but I'm too tired to analyze the cases. I'll just go ahead and write what my drawing is showing me 
. is the second intersection of with the circle 
, and let be the midpoint of the arc of which does not contain . By Ptolemy's Theorem applied to the quadrilateral 
, we get 
, and, since 
and 
, we get 
. However, the triangles 
are similar, and 
, while 
lie on the circumcircles of 
respectively, so the figures 
are similar. This gives 
. If we compare this with 
we get exactly what we want: 
.
on satisfying 
. This can be seen as follows: the locus of the points with 
is an ellipse with foci , passing through , the second intersection point between and , as was shown above. In order to prove that is the only intersection point between this ellipse and , we would have to show that is tangent to the ellipse in or, in other words, that it is the external angle bisector of 
. However, it's clear from the construction above that 
is the angle bisector of , so the converse of what we have shown in the paragraph above holds iff 
, which is equivalent to 
, for example.
be a triangle such that . Let be points on respectively such that . Let meet the circumcircle of triangle at point . Let be a point on . Prove that if and only if lies on the circumcircle of triangle .
. Let be a point on 
such 
which follows due to angle chasing 
(
). Let 
. A similar procedure shows 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/f/b/d/fbdfd7a97b3889e9ec36ea30918b51ad46108d05.png)
?
Does EV theorem work?
be real numbers . Prove that
![$$-\frac{1}{18}\leq \frac{ab+a+b+3}{(a^2+3)(b^2+3)}\leq \frac{2+\sqrt[3] 2+\sqrt[3] {4}}{12} $$](http://latex.artofproblemsolving.com/9/9/9/9997397eb38cd4bceec2ff8c85951c78e2797d0f.png)
and 
.![\[ab+bc+ac\leqslant 27\]](http://latex.artofproblemsolving.com/c/b/5/cb5e2a1efd462abd2354f4136f5fa64408f3138c.png)
,