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Topic: Area of the overlap (Read 1404 times) |
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BMAD
Junior Member
Posts: 57
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Area of the overlap
« on: May 23rd, 2014, 3:13pm » |
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Suppose you have a unit circle. You cut the circle into four equal parts but cutting vertically and horizontally through the diameter (perpendicular through the origin). Putting one of the four pieces on each of the following coordinates, (0,0) , (0,1) , (1,1) , (1,0) and fitting them together so that the radii produce a square causes the interior to have overlapping arcs. There is a piece, in the innermost center, where all four arcs overlap. Without using calculus, what is the area of the overlapped center?
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BMAD
Junior Member
Posts: 57
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Re: Area of the overlap
« Reply #1 on: May 31st, 2014, 9:07pm » |
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Bump
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dudiobugtron
Uberpuzzler
Posts: 735
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Re: Area of the overlap
« Reply #2 on: Jun 1st, 2014, 4:05am » |
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Here's an outline of how I would start to approach it. I don't actually have time to investigate it further now, though:
The corners of the middle shape are all findable using pythagoras, since they are on a right triangle with base = 0.5 and hypotenuse = 1.
(so they are each sqrt(3)/2 away from their opposite side).
The trick for finding the areas of the overlapping bits is to consider them as being made up of two half-segments. To work out the area of a segment, work out the area of the sector it is part of, then subtract the inner triangle.
You can find the angles using trig since you know all of the sides.
Once you know those areas, it should be relatively easy to work out what to subtract.
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rloginunix
Uberpuzzler
Posts: 1029
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Re: Area of the overlap
« Reply #3 on: Jun 1st, 2014, 8:47am » |
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I apologize if I am missing some piece of common knowledge but I don't understand the meaning of "Bump".
Could anyone please explain? What does it mean?
Thanks in advance.
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BMAD
Junior Member
Posts: 57
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Re: Area of the overlap
« Reply #4 on: Jun 1st, 2014, 8:52am » |
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I moved the unanswered question back to the top of the forum by bumping it up
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Grimbal
wu::riddles Moderator
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Re: Area of the overlap
« Reply #5 on: Jun 1st, 2014, 2:59pm » |
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Without using calculus:
pi/3 - sqrt(3) + 1
The idea is to cut the pizza in 12 and keeping only the middle 1/3 of each quarter. It does all the arcs. You have some extra surfact to remove, but these are 8 identical triangles, easy to compute.
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