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The centers of four circles are at the vertices of a square of sidelength 100m. Each circle has the radius of 100m. Which is the area of their intersection?

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Are the centers of the circles in the picture each one unit apart? –  Tim Ratigan Dec 6 '13 at 15:31
    
are you familiar with integration –  trafalgar_law Dec 6 '13 at 15:44
    
The centers are on the vertex of a square. –  haia Dec 6 '13 at 16:25

4 Answers 4

I think this question becomes much easier if we can know the four intersection points. And we can do this through observing the equilateral triangles if we join lines between centers of the circles.

If my calculation right, C $(\frac{\sqrt3}{2}r,\frac{1}{2}r)$, B $(\frac{1}{2}r,\frac{\sqrt3}{2}r)$, and you can work out the leftmost and downmost ones because of symmetry about the center $(\frac{1}{2}r,\frac{1}{2}r)$ but not even necessary.

One fourth of the area (suppose the upperright quarter) can be represented as:

$$\frac{S}{4} = \pi r^2\times\frac{30}{360}-\frac{(\frac{\sqrt3}{2}r-\frac{1}{2}r)\frac{1}{2}r}{2}\times 2$$

Simplified:

$$S=\frac{\pi}{3}r^2-(\sqrt3-1)r^2$$

Graph and the hilarious font :) enter image description here Look! This is integration-free!

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Yes, but how did you know that one forth of the area is πr2×30360−(3√2r−12r)12r2×2 ???? –  haia Dec 6 '13 at 16:54
    
@haia See my graph above, one quarter = 1/12 of circle - two identical triangule. –  Ray Dec 6 '13 at 16:59
    
How do you kwow it is 1/12, i cannot see it, maybe it is just an aproximation –  haia Dec 6 '13 at 17:20
    
Sorry now I see it. –  haia Dec 6 '13 at 17:26
    
@haia Ha! Just about to explain :-) –  Ray Dec 6 '13 at 17:28

The intersection point at the east is where $$x^2 + y^2 = 1$$ and $y = \frac{1}{2}$, so $x = \frac{\sqrt{3}}{2}$.

So by symmetry the quantity you desire is $$ 400 \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \sqrt{1-x^2} - \frac{1}{2} dx $$ which Wolfram Alpha says is $$ \frac{100\pi}{3} - 100(\sqrt{3} - 1). $$

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Thank you for adding more answer! –  haia Dec 7 '13 at 13:59

lets define some points: (Assuming a radius of 1)

O = (0,0)
A = (0.5, 0.5)
B = (cos(30°), sin(30°))
C = (cos(60°), sin(60°))

The rounded "triangle" ABC is a quarter of the area you are looking for.

The rounded triangle OBC is a sector with radius 1 and angle 30°, so its area is 1/12 of a disk, $\pi/12$.

From this sector we substract the two triangles OBA and OAC which are congruent and have a base of cos(30°) - 0.5 and a height of 0.5, so their total area is 0.5*(cos(30°)-0.5)

the area of ABC is then $$\pi/12 - (\sqrt{3}/2 - 1/2)/2$$ and the total area four times as much: $$\pi/3 - \sqrt{3} + 1$$

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Assuming that the radiuses are 100 and centers rest on two other circles as in the diagram.

If I were to place a square within the black portion with the same corners, the solution would be the area of that square plus 4 curved pieces.

The curved pieces area can be found by noting that is consists of one 12th the difference in area between a circle and regular dodecagon. That area is $c=\frac{10000 \pi-30000}{12}$.

The side of the square is the side length of the dodecagon so $s=50\sqrt{2}(\sqrt{3}-1)$

So the area is $$s^2+4c = 5000(\sqrt{3}-1)^2+10000\frac{\pi-3}{3}\approx 3151$$

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Note, no calculus and no trig! –  kaine Dec 6 '13 at 16:15
    
Thank you! cool –  haia Dec 7 '13 at 14:00

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