Circular Coverage
Consider a disk with a radius of one unit. We need to cover the disk’s surface with some smaller disks so that none of the larger disk can be seen. These smaller disks have a diameter of one unit. How many (the least number) will it take to completely cover the surface?
Labels: mathemagic
posted by Ragknot at
12:18 PM
20 Comments:
Let's start with 8.
The large disk has a circumference of 2*pi = 6.28..., so seven are needed to cover the perimeter.
Inscribe a regular 7-sided polygon (heptagon?) inside the circle and place the smaller disks with their centres just in from each corner so that the perimeter of the larger disk is just covered all round.
Would a small disk in the centre of the larger one cover what's left? I dunno - I'm just doing this in my head without compass, calculations or software.
I would say eight also. Since the diameter of the small disc in 1/2 the diameter of the large disc, it would take four to cover from the edge to the center. These would overlap at the center leaving no area uncovered there, but there would be four area, fairly small, uncovered at the peripheries. This would take four more discs to cover.
If Large circle Radius = R
and smaller circle radius =0.5R
Then to minimize number of circles, max arc length of larger circle must be covered by each smaller circle. Envision one smaller circle taken from the center of the larger circle (with vertical and horizontal axes in the centre of the large circle) and brought up the vertical axis, until it it intersects the arc of the larger circle. The maximum arc coverage is achieved when the distance of the two intersections of the arc are R apart (on opposite sides of the smaller circle).
Using cos x= a/h
we see that a=0.5 R (portion on right side of the vertical axis)
h= R (distance from centre to edge of larger circle)
cos x = 0.5
thus X =60 degrees
90-X =30 degrees * 2 sides of circle= 60 degrees
360/60 degrees= 6 circles to cover edge
Now to fill Remaining area in centre.
We know the angle of X is 60 degrees, forming a 1,2,sqrt(3)ratio triangle
distance from centre of large circle to centre of small circles is sqrt(3)*R/2 indicating some overlap
from centre of each edge circle to intersection with centre small circle edge to centre to centre of edge and centre circle we can form
a 0.5R,0.5R,sqrt(3)*R/2 triangle
if angle from centre circle to small circle intersection <=30 then adequate coverage by one circle is maintained
solving for angle from centre circle to intersection we use
cos c= (a^2+b^2-c^2)/2ab
cos c= [(0.5*R)^2+(R*sqrt(3)/2)^2-(0.5*R)^2]/[2*0.5*R*sqrt(3)*R/2]
cos c= (3/4)/(sqrt(3)/2)=sqrt(3)/2
c= 30 degrees
indicates adequate coverage
ANSWER:
7 circles total required...
Cam
I got the answer 4, however i've gotta be up early for work in the morning and don't really have the time to write up my explanation, i'll try and get on at lunch to write it up for you guys.
Also i am extremely curious about the answer for super women. Is there actually an answer?
Cam,
You almost got it.
One glaring error though.
Come see me at my blog.
I always prefer to do these puzzles off the top of my head (which is why I get so many wrong), however on this occasion I yielded to temptation and got some coins out to try it out in practice.
To me it looks as if Cam's six small circles don't quite cover the perimeter (very small gaps outside), and the space left in the middle requires three more to cover, not one.
My previous answer of seven round the perimeter works there, and one plugs the hole in the middle. So, I'll stick with EIGHT as my answer!
My latest post just disappeared in cyberspace, so here it is again.
I abandoned my usual theoretical stance and tried this with coins over a circle on paper.
Maybe my props weren't completely accurate, but it looked to me if Cam's six small circles didn't quite cover the perimeter (small gaps outside) and three more were needed to cover the inside.
My previous answer of seven outside and one in the middle worked a treat.
So I'll stick with EIGHT as my answer.
My penultimate post, which I thought had disappeared, turned up just as I did it again. So, my apologies for repeating myself. EIGHT is still my answer!
Ragknot,
I stand corrected.....
ANSWER:
7 DISKS are required.
BTW in my proof it should state: if angle from centre disk to small disk intersection >=30 then adequate coverage by one disk is maintained
<30 would imply inadequate coverage of the channel. Since it is exactly 30 I didn't catch the error first time.
Cam
Of course, if I can chop the disks into infitesmally small pieces then, A_bigcircle/A_smallcircle =4, and only 4 small pulverized disks are required. ;)
Cam
my god!!!!!
i was going to say that the only logical thing i found is that it is illogical to me. :P
The solution?
See this
http://ragknot.blogspot.com/
Nice one, Ragknot!
Thanks Wiz,
I enjoyed it also.
And in case someone didn't catch on, Cam got it correct early, but I said he had almost got it...
Actually he said 7 circles and it should have been 7 disks.
I didn't want others to give up early. I was expecting someone to say 7 had to be correct.
I posted pictures on my blog.
Hey Cam, why don't you solve the minumum size of the small disks that it would take for 5 to cover a larger disk of unit size? Logic would say it has to be larger than 1/2, since it takes 7 to cover 1/2 size. Then post it as a ToM.
Warning,
Five is not a TRIVAL!
Ragknot,
Did you have an answer for 5 ?
I have a construction, and an optimal answer for the construction, but I can't be certain the construction itself is optimal. (construction differs from 7 disk solution in that disks are no longer placed to optimally cover perimeter)
Cam
Hi Cam, I don't think what you say is possible. If a given perimeter coin is not where Ragknot showed, it must be covering less than a sixth of the perimeter, so you'd need more that 7 coins on the perimeter.
I'm pretty sure that Ragknot's solution is the only 7 coin solution and that it can't be done with less than 7 coins.
... I meant more than 6 coins on the perimeter.
Chris,
Agreed that 7 disks is the minimum # if smaller disks R is 0.5 R of bigger disk. I already solved that.
The question at hand that Ragknot asked following that answer was, what was the minumum size of the small disks that it would take for 5 to cover a larger disk of unit size?
I have an answer, for a construction, which is likely optimal, but I'm not 100% on it.
If someone wants to see the solution I came up with I could post it, but I don't think I'll post it as a Tom unless someone can independently confirm the answer (preferably with proof that the construction is optimal).
Cam
Cam
Hi Cam. Sorry I hadn't spotted the new problem set by Ragknot.
With regards to the followup question Ragknot posed of finding minimum R for 5 smaller disks, the minimum R, that I came up with, for the smaller disks is ~0.618*R of the larger disk.
Can anyone confirm ?
Cam
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