[/quote]
,
satisfying![\[f(x + yf(x)) + f(xy) = f(x) + f(2019y),\]](http://latex.artofproblemsolving.com/8/0/f/80fee0440e9fc4031fa4dcff96b9605011ef0e68.png)
for all real numbers 
and 
.
,
rows in which the majority of cells are blue, and exactly 
such that![\[
2^a \cdot 3^b + 1 = 5^c.
\]](http://latex.artofproblemsolving.com/0/a/c/0ac1f22162eacf698847bceb33f9a337bd53c44b.png)
be an equilateral triangle with circumcenter 
.
and 
be two points on segments 
and 
,
.
is the reflection of 
,
and 
are parallel.
from the set of positive integers to the set of positive integers such that, for all positive integers 
and 
,![\[ a, f(b) \text{ and } f(b + f(a) - 1).\]](http://latex.artofproblemsolving.com/4/a/b/4ab97da27e05d0f9e7762f74d0d0bd2dc8b9842f.png)
be an acute-angled triangle with 
.
be the circumcircle of 
be the midpoint of the arc 
of 
.
meets 
at 
and meets 
.
at 
.
by 
.
.
meets line 
.
be a cyclic pentagon inscribed in a circle of centre 
which has angles 
.
and 
meet at a point belonging to the diameter 
.
and the segment 
intersects the circumcircle of 
in a point 
.
be the midpoint of 
is parallel to 
.
be the set of all nonnegative integers. Find all the functions 
satisfying the relation![\[ f(f(f(n))) = f(n+1 ) +1 \]](http://latex.artofproblemsolving.com/8/b/9/8b953428105df2c2344963cbaf3eb81d8201e0df.png)
.
from the nonnegative integers to the integers satisfying 
and ![\[ f(mn) = f(m) + f(n) + f(m)f(n) \]](http://latex.artofproblemsolving.com/e/f/f/eff10a32cffaec742403ebfc46ff09bc8ef7b7c5.png)
for all nonnegative integers 
and 
.
,
,
![\[ \frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|
\]](http://latex.artofproblemsolving.com/3/b/a/3ba7733caf51cc70a090920ca21f8b47aad9b1ec.png)
and 
.
satisfying
for all integers 
.
,
,
means 
means 
applied 
Y by
Niek has 
square cards that are yellow on one side and red on the other. He puts down the cards to form a 
-square. Some of the cards show their yellow side and some show their red side. For a colour pattern he calculates the monochromaticity as follows. For every pair of adjacent cards that share a side he counts 
or 
according to the following rule: if the adjacent cards show the same colour, and if the adjacent cards show different colours. Adding this all together gives the monochromaticity (which might be negative). For example, if he lays down the cards as below, there are 
pairs of adjacent cards showing the same colour, and 
such pairs showing different colours.
![[asy]
unitsize(1 cm);
int i;
fill(shift((0,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((1,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((2,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((3,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((0,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((1,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red);
fill(shift((2,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((3,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((0,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((1,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((2,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((3,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red);
fill(shift((0,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red);
fill(shift((1,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((2,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((3,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red);
for (i = 0; i <= 4; ++i) {
draw((i,0)--(i,4));
draw((0,i)--(4,i));
}
[/asy]](//latex.artofproblemsolving.com/f/e/6/fe6f80da7b7e4a7e62297a87af023d8da327bdce.png)
The monochromaticity of this pattern is thus
. Niek investigates all possible colour patterns and makes a list of all possible numbers that appear at least once as a value of the monochromaticity. That is, Niek makes a list with all numbers such that there exists a colour pattern that has this number as its monochromaticity.
(a) What are the three largest numbers on his list?
(Explain your answer. If your answer is, for example,
, and 
, then you have to show that these numbers do in fact appear on the list by giving a colouring for each of these numbers, and furthermore prove that the numbers 
, 
, 
, 
and all numbers bigger than do not appear.)
(b) What are the three smallest (most negative) numbers on his list?
(c) What is the smallest positive number (so, greater than
) on his list?
square cards that are yellow on one side and red on the other. He puts down the cards to form a 
-square. Some of the cards show their yellow side and some show their red side. For a colour pattern he calculates the monochromaticity as follows. For every pair of adjacent cards that share a side he counts 
or 
according to the following rule: if the adjacent cards show the same colour, and if the adjacent cards show different colours. Adding this all together gives the monochromaticity (which might be negative). For example, if he lays down the cards as below, there are 
pairs of adjacent cards showing the same colour, and 
such pairs showing different colours.![[asy]
unitsize(1 cm);
int i;
fill(shift((0,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((1,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((2,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((3,0))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((0,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((1,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red);
fill(shift((2,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((3,1))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((0,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((1,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((2,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((3,2))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red);
fill(shift((0,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red);
fill(shift((1,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((2,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), yellow);
fill(shift((3,3))*((0,0)--(1,0)--(1,1)--(0,1)--cycle), red);
for (i = 0; i <= 4; ++i) {
draw((i,0)--(i,4));
draw((0,i)--(4,i));
}
[/asy]](http://latex.artofproblemsolving.com/f/e/6/fe6f80da7b7e4a7e62297a87af023d8da327bdce.png)
The monochromaticity of this pattern is thus
. Niek investigates all possible colour patterns and makes a list of all possible numbers that appear at least once as a value of the monochromaticity. That is, Niek makes a list with all numbers such that there exists a colour pattern that has this number as its monochromaticity.(a) What are the three largest numbers on his list?
(Explain your answer. If your answer is, for example,
, and 
, then you have to show that these numbers do in fact appear on the list by giving a colouring for each of these numbers, and furthermore prove that the numbers 
, 
, 
, 
and all numbers bigger than do not appear.)(b) What are the three smallest (most negative) numbers on his list?
(c) What is the smallest positive number (so, greater than
) on his list?This post has been edited 4 times. Last edited by nsato, Feb 13, 2023, 7:39 PM
.
,
,
.
.
is not possible. If it were, it would mean that there are 
+1s and 
-1. However, since the pair of squares with the border as 
,
.
,
