[/quote]
,
,
by letting 
and 
for 
.
.
such that, for some positive integer 
are all prime.
a
.
cities in a country, where 
is an integer. Some pairs of cities are connected by direct (two-way) flights. For two cities 
A 
between 
,
,
and 
for every 
;
A 
between 
A 
between 
be the total number of pairs of cities in the country that are connected by direct flights. In terms of 
such that 
for all real number 
and 
be a sequence of perfect squares such that 
can be obtained by concatenating a digit to the right of 
.
. Prove that
be 
for 
such that 
st for every two subsets 
of 
, there exists a polynomial 
with integer coefficients st either 
or 
where the equations are considered mod n.
is the set of reminders of 
where 
mod n.
be the set of positive real numbers. Find all functions 
such that, for all 
,
Y by centslordm, HWenslawski, megarnie
An architect is building a structure that will place vertical pillars at the vertices of regular hexagon 
, which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of the pillars at 
, 
, and 
are 
and 
meters, respectively. What is the height, in meters, of the pillar at 
?
, which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of the pillars at 
, 
, and 
are 
and 
meters, respectively. What is the height, in meters, of the pillar at 
?
be the heights at the vertices. Note that 
,
.
,
.
is a line, 
corresponds to the height at the center, and similarly 
and 
do too, therefore they must be equal.
and 
and so the height at the center is 
and 
.
units. Since the side length of the hexagon doesn’t matter, let it be 
.
.
,
,
,
is the side length of 
Therefore, the equation of the plane is
Therefore, if 
is the requested distance, we have
The answer is 
.
:diablo:
be the center of the regular hexagon 
,
.
.
.
,
.
and 
.
,
.![\[A=(0,0,0), B=(1,0,0), C=\left(\frac{3}{2},\frac{\sqrt{3}}{2},0\right), D=(1,\sqrt{3},0), E=(0,\sqrt{3},0), F=\left(-\frac{1}{2},\frac{\sqrt{3}}{2},0\right)\]](http://latex.artofproblemsolving.com/3/a/5/3a5ae90aa6d232815fbac539a6e5e7a78eb831d3.png)
![\[(0,0,12), (1,0,9),\left(\frac{3}{2},\frac{\sqrt{3}}{2},10\right), (1,\sqrt{3},d'), (0,\sqrt{3},e), \left(-\frac{1}{2},\frac{\sqrt{3}}{2},f\right)\]](http://latex.artofproblemsolving.com/0/d/f/0df5333b36567d9324489b0993f764acb249d044.png)
All these points must be on the same plane.
.![\[12c+d=0, a+9c+d=0, \frac{3}{2}a+\frac{\sqrt{3}}{2}b+10c+d=0\]](http://latex.artofproblemsolving.com/c/1/a/c1abefcddd13f019d45163c05e744eeba49f1c55.png)
.![\[14.5c+d+\frac{\sqrt{3}}{2}b=0\implies \frac{\sqrt{3}}{2}b=-\frac{5}{2}c\implies \sqrt{3}b=-5c.\]](http://latex.artofproblemsolving.com/8/f/2/8f2ffd339137d89078a2c1607d9293b894cf01a1.png)
.
is 
Also, linearity gives 
since 
are collinear. Thus 
,
,
.
.
,
is the desired height. Note that 
must be coplanar.
.
.
by plugging into point 
by plugging into point 
.
.
,
equal to 
instead of 
