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Making a perfect venn diagram

Posted: Sun Jan 11, 2009 3:56 am UTC
by v1ND
This problem has been bugging me over the last 2 days. My math-fu is lacking so hopefully someone can help me out.

Image

In a nutshell, what is the ratio (red line:blue line) of the perfect venn diagram (ie each portion has an equal area).

The green area would have to be equal to half of either of circles. It doesn't strike me as a very difficult problem but I can't seem to find any equations for lunes/lenses/vesica piscis.

In other related questions:

What about if the area of the green is equal to the total of the yellow.

Is it possible to do this with three circles and have 7 equal areas?
If not, what about 6 equal portions ignoring the centre.

Re: Making a perfect venn diagram

Posted: Sun Jan 11, 2009 4:29 am UTC
by skeptical scientist
Spoiler:
Let's assume that we have unit circles, and the length of the red line is r. Then the green area must be pi/2, but we can also find it in terms of r using calculus: [math]A=4\int_{1-r}^1 \sqrt{1-x^2} \, dx[/math][math]=4\int_{\arcsin(1-r)}^{\frac{\pi}{2}} \sqrt{1-(\sin t)^2} \, \cos t \, dt[/math][math]=4\int_{\arcsin(1-r)}^{\frac{\pi}{2}} \cos^2 t \, dt[/math][math]=4\int_{\arcsin(1-r)}^{\frac{\pi}{2}} \frac{1}{2}+\frac{1}{2}\cos(2t) \, dt[/math][math]= \left [ 2t+\sin(2t) \right ]_{\arcsin(1-r)}^{\frac{\pi}{2}}[/math][math]= \left [ 2t+2\sin t \cos t \right ]_{\arcsin(1-r)}^{\frac{\pi}{2}}[/math][math]= \pi-2\arcsin(1-r)-2(1-r)\sqrt{2r-r^2}[/math]
So we want to find r such that [imath]2\arcsin(1-r)+2(1-r)\sqrt{2r-r^2}=\frac{\pi}{2}[/imath]. I'm pretty sure this is intractable analytically, but it's pretty easy to solve numerically.

Re: Making a perfect venn diagram

Posted: Sun Jan 11, 2009 4:30 am UTC
by gnuoym
OK, I don't know how to do the markups to make the pretty math symbols, but I think I can explain it without.

Spoiler:
Let the left circle be the unit circle centered at the origin and the right circle be a unit circle with center (a,0).
These circles intersect at x=a/2.

Consider now only the semicircle above the x-axis with equations y = sqrt(1-x^2) and y = sqrt(1-(x-a)^2).

The integral of each circle across its domain is pi/2; so half that is pi/4.

Find a such that the difference of the integrals of the left circle from [a/2,1] and the right circle on [a/2,a+1] is pi/4.

The ratio is then a:1.

I'll post again after I figure out the answer and edit to add the equations.


Darn, beaten to it

Re: Making a perfect venn diagram

Posted: Sun Jan 11, 2009 5:54 am UTC
by t0rajir0u
Not to be preachy, but this is a classic example of a problem that people think can't be solved without calculus that requires only one observation to solve with elementary methods. Which is this:
Spoiler:
Draw lines from the centers of both circles to the intersection points. The area of the middle section is twice the difference between the area of a sector and the area of a triangle. Both areas can be found by elementary methods. See the solution here.

Re: Making a perfect venn diagram

Posted: Sun Jan 11, 2009 7:01 am UTC
by skeptical scientist
Is there something wrong with calculus? I never said I needed it, it's just the first thing I thought of (perhaps because I'm teaching trig substitution right now).

Re: Making a perfect venn diagram

Posted: Sun Jan 11, 2009 7:36 pm UTC
by v1ND
Is the same feat possible with 3 circles?

Re: Making a perfect venn diagram

Posted: Sun Jan 11, 2009 9:42 pm UTC
by t0rajir0u
Yes. The center of a Venn diagram with three circles is the union of three half-lunes (which are the difference of a sector and a triangle) and a triangle, and once you've got the area of the center you've got the area of the other three intersections.

Re: Making a perfect venn diagram

Posted: Thu Nov 30, 2017 11:22 am UTC
by Quilkind
You just need to find the area of the center. Damn, old college memories are kicking in x.x