.
,
.
lies on 
such that 
.
passing trough 
at 
.
.
such that
be an integer. Rowan and Colin play a game on an 
grid of squares, where each square is colored either red or blue. Rowan is allowed to permute the rows of the grid and Colin is allowed to permute the columns. A grid coloring is orderly if: [list] [*]no matter how Rowan permutes the rows of the coloring, Colin can then permute the columns to restore the original grid coloring; and [*]no matter how Colin permutes the columns of the coloring, Rowan can then permute the rows to restore the original grid coloring. [/list] In terms of 
,
is selected inside acute 
so that 
and 
.
is chosen on ray 
so that 
.
be the midpoint of 
.
is tangent to the circumcircle of triangle 
.
that satisfy![\[
f(x^2-y)+2yf(x)=f(f(x))+f(y)
\]](http://latex.artofproblemsolving.com/f/b/b/fbb693770cd90d8a144026608074122e54da5802.png)
for all 
.
and 
be positive integers. The cells of an 
grid are colored amber and bronze such that there are at least 
amber cells and at least 
bronze cells. Prove that it is possible to choose 
chosen cells lie in the same row or column.
and 
are erected outside an acute triangle 
Suppose that ![\[\angle BC_1C+\angle CA_1A+\angle AB_1B=180^{\circ}.\]](http://latex.artofproblemsolving.com/3/3/7/3378f0c4409d14bc0afddf2616992c13f624be3b.png)
Prove that lines 
and 
are concurrent.
-
of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let 
?
of identical dominoes is a maximal grid-aligned configuration on the board if 
dominoes where each domino covers exactly two neighboring squares and the dominoes don't overlap: 
be the number of distinct maximal grid-aligned configurations obtainable from 
Y by Mango247
For every positive integer 
prove the inquality
(The integer part![$[x]$](//latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
of the number 
is the largest integer not exceeding . The fractional part of the number is a number 
such that ![$[x]+\{x\}=x$](//latex.artofproblemsolving.com/0/3/a/03ae0917760023ab705aee0ca548af5adc9ba945.png)
.)
A. Golovanov
prove the inquality
(The integer part
![$[x]$](http://latex.artofproblemsolving.com/b/c/e/bceb7b14e55d33a8bca29b7863ad3cdae95afce4.png)
of the number 
is the largest integer not exceeding . The fractional part of the number is a number 
such that ![$[x]+\{x\}=x$](http://latex.artofproblemsolving.com/0/3/a/03ae0917760023ab705aee0ca548af5adc9ba945.png)
.)A. Golovanov
This post has been edited 6 times. Last edited by Hopeooooo, Jul 25, 2024, 5:09 PM
Reason: 1
Reason: 1
where 
is the fractional part of 
.
.
.![\[
|2\sqrt{m}-2k-1|>\frac{1}{4(\sqrt{m}+1)}\Leftrightarrow \frac{|4m-(2k+1)^2|}{2\sqrt{m}+2k+1}>\frac{1}{4(\sqrt{m}+1)}.
\]](http://latex.artofproblemsolving.com/1/e/0/1e09c9abb86cd903b051995f222370c8fd8de90e.png)
Note that 
,![\[
\frac{|4m-(2k+1)^2|}{2\sqrt{m}+2k+1}>\frac{1}{4(\sqrt{m}+1)}\ge \frac{2}{2\sqrt{m}+2k+1}\ge \frac{2}{4\sqrt{m}+1}>\frac{1}{4\sqrt{m}+4}
\]](http://latex.artofproblemsolving.com/6/8/1/68110c4cbf0e70a166d6d3e86b0d0b2b67d905d8.png)
for trivial reasons.
.
,
.![\[
\frac{1}{2\sqrt{k^2+k}+2k+1}>\frac{1}{4\sqrt{k^2+k}+1}\Leftrightarrow 2\sqrt{k^2+k}>2k,
\]](http://latex.artofproblemsolving.com/5/4/0/54096d53aec6ebbfd77f9412c2e38a786183f4d0.png)
which is clearly the case.
,![\[ \sqrt{m} - \frac{1}{2} \ge \frac{1}{2} > \frac{1}{16} \ge \frac{1}{8(\sqrt{m} + 1)} \]](http://latex.artofproblemsolving.com/6/c/8/6c85769fed2b1b165b05647f2d02b43cb052da0e.png)
