[/quote]
,
.
be reals such that 
and 
Find the value of 
players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least 
next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
and 
Find the value of 
be a triangle and 
,
,
be the midpoints of arcs 
,
,
on the circumcircle. Line 
passes through the feet of the perpendiculars from 
to 
and 
.
passes through the feet of the perpendiculars from 
.
denote the intersection of lines 
and 
similarly. Prove that triangle 
and 
are similar to each other.
such that : 
and 
table filled with natural numbers, we say it is a divisor table if:
-
,
-
,
for every 
.
is given. Determine the smallest natural number 
,
.
.
,![\[\frac{a^3}{(b-c)^2+bc}+\frac{b^3}{(c-a)^2+ca}+\frac{c^3}{(a-b)^2+ab}\ge a+b+c.\]](http://latex.artofproblemsolving.com/1/c/8/1c80f369f5e5daf2155c9620dff0463ed8d284a0.png)
and internal bisector 
.
again at 
and 
cuts 
.
lie on 
.
at 
respectively. Let circumcircle of 
cuts 
.
.
,
is a permutation of the set 
.
-![\[
\left\lfloor \frac{k_i}{j+1} \right\rfloor.
\]](http://latex.artofproblemsolving.com/7/5/8/7583af9655a43eb3f6167f955148984a468d4e70.png)
If 
,
.
in circumcircle of 
.
is a point on the line segment of 
.
.
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.
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)
,
.
)
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 
.
