TITLE: How can one distribute six dots within a semicircle in order to minimise the distance between any single point and one of the six dots?
QUESTION [5 upvotes]: I am a biologist studying flight behaviour in the Manx Shearwater. For a project I am doing I am looking at the influence of wind on flight behaviour. I know my birds are within a semi-circle of radius 50km from their nest sites, but I do not know their exact positions. But knowing their position or somewhere close by is important to be able to estimate the wind vectors they are being exposed to. 
I am able to acquire six locations of modelled wind data from the Met Office. To make the most of this I want to choose six locations that would enable at least one of these locations to at least be representative of any possible position a bird is at within this semi-circle. So I imagine there is an optimal distribution of the 6 locations within this semi-circle that minimises the maximum distance a bird could be from any one location. I have a possible way of working out this distribution below and it would be very much appreciated if anyone could comment on the suitability of this method or come up with any other methods that would enable a solution to the problem. Thank you. 
Let $S$  be the unit semicircle in the plane. 
We want to find points $x_1  , x_2  , x_3  , x_4  , x_5  , x_6$   in $S$  so as to minimise $\max${$\min${$d(x,x_1),\ldots ,d(x,x_6 )$}:$x∈ S$} .

REPLY [2 votes]: Notes 
The explicit form of the lower bound in this answer may be wrong. In the updated numerical results from Mark, it can produce a configuration with a $r$ smaller than the lower bound here. It looks like the constraint on one of the points (the most likely candidate is point $F$) is redundant. We are essentially back to square one....  C'est la vie....

Please consider this as a supplement to Mark H's answer.
In Mark's answer, the centers are roughly symmetrical in the horizontal direction. 
If one compute the Voronoi diagram associated with these centers, one obtain a figure
looks roughly what is shown below.
$\hspace0.75in$ 
The points $A, B, C, D$ are those $4$ centers in the first quadrant.
The orange lines are the boundary of the Voronoi cells. The points
$E, F, G, H, I$ are the intersections of the boundaries of these cells and the semi-circle. Let $O = (0,0)$ and $X = (1,0)$. 
The key observation is the distances (those illustrated in pink) among these points
are all roughly equal. If the algorithm in Mark's answer converge to some configuration of centers, that configuration should be a local minimum of the maximum-minimum distance functional. This means these distances should be equal to each other exactly. i.e.
$$|AX| = |AE| = |AF| = |AG| = |BE| = |BF| = |BH|\\
 = |CF| = |CG| = |CH| = |CI| = |DH| = |DI|$$
To find such a configuration, we first relax the constraint that $D$ lies on $y$-axis. We assume $A$ is located near what's in Mark's answer. 
Let $r$ be the common values of above $13$ distances. Let $A = (1-u,v)$ and $B = (0,w)$. The condition $|AX| = r$ leads to $r^2 = u^2 + v^2$.
We then proceed  to express the positions of $E, F, G, C, H, I$ (in that order) and finally $D$ in terms of these 3 variables $u,v,w$. 
Aside from the formula of $I$ and $D$, they are not that horrible.

$|AE| = r$ implies $E = (1-2u,0)$.
$|BE| = r$ leads to
$$w^2 + (1-2u)^2 - u^2 - v^2 = 0\tag{*1}$$
$|AE| = |BE| = |AF| = |BF|$ implies $F = A + B - E = (u,v+w)$.
$|AX| = |GX|$ and $|OX| = |OG|$ implies
$$G = \verb/Refl/(A,X) = 
\left(\frac{2(1-u)^2}{v^2+(1-u)^2}-1,\frac{2(1-u)v}{v^2+(1-u)^2}\right)$$
where $\displaystyle\;\verb/Refl/(\vec{U},\vec{V}) = 2\frac{\vec{V}\cdot\vec{U}}{|\vec{U}|^2}\vec{U} - \vec{V}$ maps point $V$ to its mirror image with respect to $OU$.
$|AF| = |AG| = |CF| = |CG|$ implies
$$C = F + G - A = \left(\frac{2(1-u)^2}{v^2+(1-u)^2}+2u-2,w+\frac{2(1-u)v}{v^2+(1-u)^2}\right)$$
$|BF| = |BH| = |CF| = |CH|$ implies
$$H = B + C - F = 
\left(\frac{2(1-u)^2}{v^2+(1-u)^2}+u-2,w+ \frac{2(1-u)v}{v^2+(1-u)^2}-v\right)$$
$|CG| = |CI|$ and $|OG| = |OI|$ implies $ I = \verb/Refl/(C,G) = $ a horrible mess!
$|CH| = |CI| = |DH| = |DI|$ implies $D = H + I - C = $ another horrible mess!

If we put back the constraint that $D$ lies on the $y$-axis, we obtain
$$((u-1)v^2+u^3-u^2-u+1)w^2 + (4u^2-8u+4)vw\\
+ (4u^3-4u^2-4u+4)v^2 + 4u^5-12u^4+12u^3-4u^2\\
= 0\tag{*2}$$
We can eliminate $w$ by computing the resultant between the two polynomials in $(*1)$ and $(*2)$.
The resultant has the form $(u-1)^2f(u,v)$ where $f(u,v)$ is a polynomial of degree $8$ in $u, v$.
We can simplify this expression a little bit by a change of variable.
Let $(u,v) = (rs, r\sqrt{1-s^2})$,
the condition becomes
$$\begin{align}
g(r,s) \stackrel{def}{=} &\;\; f(rs,r\sqrt{1-s^2})\\
= &\;\;16r^6s^4+(-16r^7-32r^5-16r^3)s^3+(24r^6+48r^4+24r^2)s^2\\
&\;\; + (8r^7-24r^5-40r^3-8r)s+r^8-12r^6+22r^4+4r^2+1\\
= &\;\; 0
\end{align}
$$
To complete our task, we need to find the point along the curve $g(r,s) = 0$
with minimal $r$. At that point, the tangent vector of the curve will be pointing in the $s$-direction. This means $\frac{\partial g}{\partial s}(r,s) = 0$ at that particular point.
We can eliminate $s$ by computing the resultant between $g(r,s)$ and $\frac{\partial g}{\partial s}(r,s)$. The resultant is reasonably simple
$$-1048576 r^{18} (r^2-1)^{12} (27r^8 - 324 r^6 - 270 r^4-36 r^2 + 11)$$
Eliminating the impossible values $0$ and $1$ for $r$, the local extremium
of  maximum-minimum distance functional should be a root of the quartic equation:
$$27r^8 - 324 r^6 - 270 r^4-36 r^2 + 11 = 0$$
Solving this equation and compare the numerical values we find from Mark's answer, the $r$ we seek equals to
$$r = \sqrt{ 3 + 2\sqrt{3} - 2\sqrt{5 + \sqrt{27} - \frac{1}{\sqrt{27}}}} \approx 0.3719887399641683$$
and at the least, this is a local minimum for the maximum-minimum distance functional. With this, we can numerically back out the parameters $u,v,w$
$$\begin{cases}
u &\approx 0.3259601005065833\\
v &\approx 0.1792362562035586\\
w &\approx 0.1312100460994327
\end{cases}$$
and the location of the centers
$$\left\{\begin{array}{lcll}
x_5 \leftrightarrow A & \approx & (0.6740398994934167,& 0.1792362562035586),\\
x_1 \leftrightarrow B & \approx & (0,& 0.1312100460994327),\\
x_3 \leftrightarrow C & \approx & (0.5198397097356646, & 0.6279149145866445),\\
x_2 \leftrightarrow D & \approx & (0,& 0.7661472706667444)
\end{array}\right.$$