Re: Problem
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If the planet is radius r, lamppost height h, and number of lampposts n:
note: the angle between the lampposts (from the center of the planet)
is 2*PI/n
the angle between the lampposts and the position of the fringe of
their light is PI/n (half of the above.
The line representing the last lightbeam to barely touch the
earth is tangent to the surface, thus creating a right triangle between
the top of the lamppost, the center of the planet and the point at the
fringe of the light created by the lamppost. the radius of the earth is
given, and the angle between the lamppost position and the light fringe
position is Pi/n. The height can be derrived as h=(r/cos(Pi/n)-r) or
h=r*(1/cos(Pi/n)-1). The sumation of the heights will simply equal
H=n*r*(1/cos(Pi/n)-r). take the derrivative of H and set it equal to 0
to find the number of lampposts needed.
I think.
-jesse
Adam Trilling wrote:
>
> I don't know how on-topic this is, but I saw this problem in a magazine
> and I figured someone here had to know enough math to figure it out.
>
> Supose you have a perfectly spherical planet and you want to put
> lightposts around the equator so that the entire equator is illuminated
> and the total height of lightposts is minimized. You must use at least 3
> lightposts. Thus you could use very few tall lightposts or many small
> ones. CAN SOMEBODY HELP ME FIGURE THIS OUT???
>
> Adam Trilling
> Hanegev Online Services Chairperson
> adamt@shadow.net
>
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