[/quote]
,
,
.
lies on 
such that 
.
passing trough 
at 
.
.
such that
,
.
and 
the midpoints of 
and 
.
and 
meet 
and 
at 
and 
,
,
bisects both angles 
,
,
.
be the circumcircle of triangle 
,
be the tangent line of 
passing 
.
be the points each on side 
such that 
.
meets 
.
meets 
passing 
.
and tangent to line 
.
![\[
\text{Find all prime numbers } (p, q) \text{ such that } p^2 + 3pq + q^2 \text{ is a fifth power of an integer.}
\]](http://latex.artofproblemsolving.com/c/f/8/cf855cb087102024a50d41f49c67e7fd1ef22767.png)
be an integer, and let 
be positive integers. Show that there exist positive integers 
satisfying the following three conditions:
for 
the remainders of 
are pairwise different; and
denotes the integer part of real number 
,
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 
.
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 
is equilateral with sides 
where 
,![$[CMN]- \text{the~red~area}$](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 
lunes. This can be shown by connecting intersection points to the mid point of the corresponding semicircle. The area of a lune can be found as the area of a 
degree sector minus the area of an equilateral triangle of side length 
.
