be the set of positive integers. A function 
satisfies the equation ![\[\underbrace{f(f(\ldots f}_{f(n)\text{ times}}(n)\ldots))=\frac{n^2}{f(f(n))}\]](http://latex.artofproblemsolving.com/2/e/7/2e73ff2f75b103e9d6a4004a42e4ed7f4ee64a67.png)
for all positive integers 
.
.
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 
and 
?
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 Williamgolly, Inconsistent, megarnie, Adventure10
As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 
so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles?
![[asy]
size(140);
fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4));
fill(arc((2,0),1,180,0)--(2,0)--cycle,white);
fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white);
fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white);
fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white);
fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white);
fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white);
draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0));
draw(arc((2,0),1,180,0)--(2,0)--cycle);
draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle);
draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle);
draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle);
draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle);
draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle);
label("$2$",(3.5,3sqrt(3)/2),NE);
[/asy]](//latex.artofproblemsolving.com/c/c/4/cc4a8eea36390b30289f00cdba179c25b9ee3543.png)
so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region—inside the hexagon but outside all of the semicircles?![[asy]
size(140);
fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4));
fill(arc((2,0),1,180,0)--(2,0)--cycle,white);
fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white);
fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white);
fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white);
fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white);
fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white);
draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0));
draw(arc((2,0),1,180,0)--(2,0)--cycle);
draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle);
draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle);
draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle);
draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle);
draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle);
label("$2$",(3.5,3sqrt(3)/2),NE);
[/asy]](http://latex.artofproblemsolving.com/c/c/4/cc4a8eea36390b30289f00cdba179c25b9ee3543.png)
This post has been edited 3 times. Last edited by djmathman, Feb 7, 2020, 5:40 AM
anyone confirm?
(
Examine only 
of the original diagram (ignore the green parts and only focus on the light-blue parts) and label the points as shown 
is the midpoint of 
)
where 
are the midpoints of sides 
and ![$[CMN]- \text{the~red~area}$](http://latex.artofproblemsolving.com/8/f/2/8f2810be11b356cd8b7ecd1380a790bb0b87c1b9.png)
.![$[CMN]=\frac14[ABC]=\frac14\left(\frac12\cdot2\cdot2\cdot\sin\left(\frac{\pi}3\right)\right)=\frac{\sqrt3}4$](http://latex.artofproblemsolving.com/c/a/7/ca747b6d1a9e75171f7033ac844cb675dd9830d2.png)
.
.
and the equation of the line passing through 
.
and 
,
.![$[CMN]- \text{the~red~area} =\frac{\sqrt3}4-\left(\frac{\pi}6-\frac{\sqrt3}4\right)=\frac{\sqrt3}4-\frac{\pi}6$](http://latex.artofproblemsolving.com/d/3/e/d3e9fde235f902f3c909dcef9d5d433035aa2782.png)
and the answer to the problem is 
.
minus 
degree sector minus the area of an equilateral triangle of side length 
.
