A Chomby was walking down a path one day when Jhudora appeared in front of it. "Solve this puzzle, and you may pass," she said.
Suppose you have two identical circles that overlap such that the center point of one circle is on the edge of the other circle. If the overlapping area is exactly 20000 square centimetres, what is the radius of the circles? Please round to the nearest whole centimetre, and please submit only a number as your answer
NVM didn't get a whole number, I set my calculator to round for me xD
Wed 18 Nov, 2009 05:00 pm
I was able to find enough bits and pieces for this type of puzzle on the Internet, but I am almost certainly wrong. I got an answer where I rounded up.
Wed 18 Nov, 2009 05:09 pm
Solution is a bit hard to describe... the two overlapping circles will create an equilateral triangle between them (all three sides of the triangle are of the length 'radius').
As all angles in an equilateral triangle are 60degrees, you know the sections of the circles that overlap are both 120degrees (or 1/3rd of circle surface).
We can now express both the surface area of these triangles in the circle and the area of the section of the circle in their radius. If we were to calculate twice the area of such a circle-section (once for each circle) we would have added in the surface of the two triangles within twice, so we'll have to subtract these again.
Surface area of the overlap in r is then: 2 * (1/3 * pi * r²) - 2 * ( root(0.75)*0.5 *r²). This will have to equal to 20000, working back we can conclude that r = 127.4, rounded down as the LC requests makes for 127!