into 
or 
can sometimes be quite complex. That's why I’ve written a program to assist with this process.
such that for all positive real numbers 
and 
,![\[
\frac{x + f(y)}{x f(y)} = f\left( \frac{1}{y} + f\left( \frac{1}{x} \right) \right)
\]](http://latex.artofproblemsolving.com/7/d/f/7df25755bc33bef970fffb392fb5b6b789390881.png)
cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from 
to 
cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?
.
be a triangle of orthocenter 
.
and the circumcircle of triangle 
intersect a second time in 
.
passes through the midpoint of segment 
.
.
Let 
.
Let 
.
Let 
with edges 
.
three perpendiculars to the three lateral faces.
and radius 
.
.
exist that satisfy 
for all real 
and the domain of f is strictly within the set 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.
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 
.
