[/quote]
,
denotes a polynomial of degree 
such that 
for 
,
.
with 
,
to the sides 
are 
,
is the orthocenter. 
is the midpoint of segment 
.
and 
intersect at 
.
from 
and 
intersect at 
.
are colinear
and 
with real coefficients satisfying
be three positive reals such that 
.![\[ \frac { x^5-x^2 }{x^5+y^2+z^2} + \frac {y^5-y^2}{x^2+y^5+z^2} + \frac {z^5-z^2}{x^2+y^2+z^5} \geq 0 . \]](http://latex.artofproblemsolving.com/a/1/4/a14adf0f1e35df57c734c7df701d5a67da22dc3c.png)
is 
respectively. A point 
satisfies that 
and 
are both tangent to 
and tangent to 
intesects 
.
is an arbitrary point on 
,
is the second intesection point of 
and 
.
are concyclic.
so that 
can be a perfect square
be an isosceles triangle with 
.
be a point on 
.
be a point inside the triangle such that 
and![\[
CL \cdot BD = BL \cdot CD.
\]](http://latex.artofproblemsolving.com/1/6/7/16771838c95f86e92f79b8d049e46ab473e6287d.png)
Prove that the circumcenter of triangle 
lies on line 
.
such that
for all real numbers 
such that 
.
such that for all 
,![\[f(x+yf(x))+y = xy + f(x+y).\]](http://latex.artofproblemsolving.com/c/f/3/cf3d20a041c27244e90876119b4568b7a3e13c03.png)
be an integer, let 
be a set of 
,
such that ![\[ A_i \cap A_j \not= \emptyset, A_i \cap A_k \not= \emptyset, A_j \cap A_k \not= \emptyset, \;\textrm{imply}\ \;A_i \cap A_j \cap A_k \not= \emptyset \ .\]](http://latex.artofproblemsolving.com/c/1/1/c113f5cdce00dfec6714c382a8e1bc74f188f025.png)
Show that 
.
such that 
or 
,![\[
1 \leq a_1 < a_2 < \cdots < a_{(p-1)/2} < p
\]](http://latex.artofproblemsolving.com/6/8/e/68e3d73edec43b2cc480832adf23bbbb5f4c7bee.png)
such that![\[
\prod_{\substack{1 \leq i < j \leq (p-1)/2}} (a_i + a_j)^2 \equiv 1 \pmod{p}.
\]](http://latex.artofproblemsolving.com/2/9/e/29edf6ab8cee2a526e7753a140c5a291e987856a.png)
denote the maximum number of 
-
board such that each 
such that![\[
f(N) \leq cN^2
\]](http://latex.artofproblemsolving.com/7/4/8/748109c3d63203f9a67a167434e10a909f38e83f.png)
for all positive integers 
.
be a graph colored using 
colors. We say that a vertex is forced if it has neighbors in all the other 
colors.
-
colors such that at least 
of the colors have a forced vertex of that color.
vertices of the same color may be adjacent.
Y by cubres
Exactly 
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
![\[
\frac{1}{3}n^{\alpha} \leq f(n) \leq 3n^{\alpha}
\]](//latex.artofproblemsolving.com/0/a/0/0a06d4ff52479c8cb785c6450cea8748bf7769b6.png)
for any 
.
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 
.This post has been edited 1 time. Last edited by Assassino9931, Apr 9, 2025, 8:54 AM
from bottom to top and left to right, respectively.
and 
as 
.
be the largest integer such that 
.
,
in ascending order. Next, consider the columns congruent to 
,
cells such that the distance between any two black cells is at least 
new black squares. Note that the distance between any two black cells is at least 
new black cells. Since 
,
.
.
white cells such that the black cells satisfy the required condition.
black squares on an 
grid, there exists two squares whose distance is at most 
.
,
.
(these may lie outside the 
squares. Since the pairwise distance between black squares is at least 
,
.
Solving the inequality leads to a contradiction.
.![$$\sum_{m=2}^{n}\sqrt{\dfrac{2}{m+1}}\cdot (n+1) + n\leq \left[2(n+1)\right]^{3/2} + n<3n^{3/2}$$](http://latex.artofproblemsolving.com/c/0/c/c0ca85f7cfe1db7159b2cc4379a5c127aacaa6e0.png)
This completes the proof.
,
for 
.
additional cells. We also have 
cells from the 
columns and ![\[
n(m - 1) + mn + n = 2mn < 3n^{3/2}.
\]](http://latex.artofproblemsolving.com/9/a/e/9aefa63d61b39e763258bcfdf78b07a3458e1f29.png)
.
where 
(they are in the table, since 
,
)
,
within the starting black cell). These sets for different points do not intersect (they may only share boundary points).
black cells from the boundary of the initial black cell to the boundary of the corresponding area (if that number is fractional, then at least one of the newly added cells and on the boundary between two areas and not belonging fully to either of the two neighboring regions). This means that:![\[
f(n) \geq \frac{n}{2}(m - 1) \geq \frac{n}{2}(\sqrt{n - 1} - 2)
\]](http://latex.artofproblemsolving.com/3/a/5/3a5990385ddd8dd1ccd1a075e7da10b0a6742f06.png)
![\[
\frac{1}{2}(\sqrt{n - 1} - 2) \geq \frac{1}{3}\sqrt{n}
\]](http://latex.artofproblemsolving.com/2/1/1/21159c9efd1f45974f1266a2ce1f191533ef2421.png)
![\[
3\sqrt{n - 1} \geq 2\sqrt{n} + 6
\]](http://latex.artofproblemsolving.com/9/6/5/9651e6016594997925413ed005e80681bb0acf8d.png)
![\[
9n - 9 \geq 4n + 24\sqrt{n} + 36
\]](http://latex.artofproblemsolving.com/0/2/4/024ab2f23a94a9d3eae0197a04a7d04d7b0a3bd6.png)
![\[
5n - 45 \geq 24\sqrt{n}
\]](http://latex.artofproblemsolving.com/5/f/1/5f1c2b0fb034af8a8ef34228573def2c5403bc8a.png)
,
.
(here 
)
and 
)
and color in black a centered cell (unique if 
starting from the black cell and going out the 
square it belongs to; hence the total number of new squares is at least 
,
for something like 
.
instead, as it is anyway one of my favourite constants... Fortunately, this affected only one student and by not much (there were a few others in this trap, but they either quickly figured out an optimization or just bashed more efficiently smaller cases and this did not affect their overall performance in the day). All others with very good partials had combinatorial issues for the lower bound (like trying a spanning tree, which does not work easily/at all) or had very lazy constructions which required an upper bound constant much bigger than 
.
with 
vertices (cells) and edges that connect each two adjacent cells (sharing a side). We want to choose 
An easy calculation shows that it's enough to take 
.
be the set of 
be a connected subgraph of 
such that 
.
,
from 
,
,
and each edge of 
.
.
as to the white vertices and to the vertices in 
be a white vertex of 
be the neighbor of 
whose length is at least 
,
be all the leaves that 
.
where we assume 
.
since all 
are black. Summing up these inequalities, we get:
Now, we delete the vertex 
.
black vertices, the corresponding deleted edges have total length at least 
.
as 
.
