

Title: Area of the overlap Post by BMAD on May 23^{rd}, 2014, 3:13pm 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? 

Title: Re: Area of the overlap Post by BMAD on May 31^{st}, 2014, 9:07pm Bump 

Title: Re: Area of the overlap Post by dudiobugtron on Jun 1^{st}, 2014, 4:05am Here's an outline of how I would start to approach it. I don't actually have time to investigate it further now, though: [hide]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 halfsegments. 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.[/hide] 

Title: Re: Area of the overlap Post by rloginunix on Jun 1^{st}, 2014, 8:47am 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. 

Title: Re: Area of the overlap Post by BMAD on Jun 1^{st}, 2014, 8:52am I moved the unanswered question back to the top of the forum by bumping it up 

Title: Re: Area of the overlap Post by Grimbal on Jun 1^{st}, 2014, 2:59pm Without using calculus: [hide] pi/3  sqrt(3) + 1 [/hide] [hide]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.[/hide] 

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