[/quote]
,
be a diameter of a circle 
,
be the tangent at 
.
be a fixed, positive real, and consider all pairs of points 
and 
lying on 
.
and 
intersect 
and 
,
pass through a common point.
such that
for all 
satisfying 
and 
.
,
,
,
with circumcenter 
.
and 
intersect the altitude from 
at points 
and 
respectively. 
is the circumcenter of triangle 
and 
is the reflection of 
is the second intersection of circumcircles of triangles 
and 
.
are collinear.
intersects the altitudes of 
.
lies on a median of triangle 
of natural numbers satisfying![\[(a+14\sqrt{3})(b-14c\sqrt{3})=2024,\]](http://latex.artofproblemsolving.com/b/c/c/bcc656bbea625f4cf5798440b497110b5d30a64a.png)
determine the one with the maximal value of 
.
Prove that
Prove that
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)
.
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 
