Although I haven’t checked, d = 40.5 could be right. However I think there’s a problem with the solution of 22.5. With that, the centre of the smaller circle is inside the larger circle, i.e. the overlap is more than half of the area of the smaller circle. The problem may occur either because at some stage you are adding areas instead of subtracting areas (or vice versa), or because the segment of the smaller circle (in my earlier analysis) is more than 180 degrees, and the formula is assuming less than 180 degrees.
The 22.5 number is a correct solution to the equation, but it’s obviously not a correct solution to the problem; the equation won’t work, as you said. I somehow got the wrong number when I substituted in 40.6, but it does work, and I believe it’s the correct answer
I don’t see how anyone could get a quadratic equation out of the expression for the area of a lens.
I get 18.335814 and -18.33581, but I think your mistake, to which ultrafilter alluded, was using the quadratic for a 4th degree function. Either solution yields a shared area one unit from the center of the smaller circle. Someone who knows more than I do can try to determine if this is accurate.
I mean, since the Venn circles are just abstractions of sets of people, who cares about the geometric area of the overlap? Did I miss something?
Spectre:
Bolding mine.
Two pages of simplifying formulas and I gave up for the mean time - it was driving me crazy. My theory is that you can use the area of the triangle created between the intersection of the circles, those two points, and the center of the circle and then subtract the other circle’s sector’s area. Do this twice and you are good to go. The problem lies in the fact that I would need to find the cooridinates for the center of the circle. Much pulling of the hair ensues.
(btw - I really think it is cool how I can expect to see the same people in a math thread. You guys rock!)
We have a formula already; it’s in the Mathworld link presented in post #3. Just to put it in the thread, it’s A = r[sup]2[/sup]arccos((d[sup]2[/sup] + r[sup]2[/sup] - R[sup]2[/sup])/2dr) + R[sup]2[/sup]arccos((d[sup]2[/sup] + R[sup]2[/sup] - r[sup]2[/sup])/2dR) - sqrt((d + R + r)(-d + R + r)(d - R + r)(d + R - r))/2. If that can be solved algebraically, it requires more smarts than I have, so I’m thinking the thing to do from here on out is to plug in values for d (we have R = 27 and r = 19) and see what gets us closest to A = 424. So far, we know that with d = 34, A = 250, and that d must be no greater than r + R.
Ah, sorry about that. But, I think I may have my redemtion thanks to a graphing calculator. When y=422, x=42.577. Is that right?
(42.2224,424). I remembered it wrong, sorry!
I dunno. The closer d is to R + r, the smaller the area should be. I could’ve calculated the area for d = 34 incorrectly.
If arccos is the same thing as Cos[sup]-1[/sup], I can solve that on my handy dandy ti-83. Otherwise, it might not be quite as easy. Or it might.
Ok, it has been a long week and I am tired so excuse me. d=41.1009 using my graphing calculator. Before I was thinking of the wrong thing and I also forgot all things under the square root. But, the intersection of y= that equation where d=x and the equation of y=424 leaves us with (41.009, 424) or d=41.1009.
It is but it would take forever. Ti-89 all the way, baby!
Now that you mention it, I’m getting a value of 10.266 giving a result of 4.4E6 and 10.267 giving -1.3E7, so I’m guessing there’s a 0 in there. The Table feature on the 83 won’t go further than the thousandth place, I think.
Hoping that’s correct and that someone can use that value.
I should mention there are other values, according to the equation I got when I simplified things, between 22 and 23, 28 and 29, 29 and 30, at 35 exactly, 34.8, etc. Should I be getting so many zeroes?
Where do you see them as zeros? The table? Going to “find zeros”? Holes in the graph? I graphed the equation from the link and then I graphed 424 as Y2. Mine was a piece-wise function but the other parts were for negative y values. I didn’t bother changing it to a dot graph because I figured it would be a solid relationship. Perhaps I was mistaken?
I didn’t use formula provided in the Math-World site; I instead plugged things into a spreadsheet, solving for the area of the two circular segments for assumed values of a (the length of the common chord, per the Math-World site’s nomenclature), and I get:
d = 28.48349
a = 34.6344
d1 = 8.17359 (for the circle with area=1152)
d2 = 20.3099 (for the circle with area=2238)
Both. I first substituted (as I gather many of us did) the values in for the variables, then simplified. I can post the equation I used if desired. I then moved 424 to the X side of the equation so that I would be able to find zeroes of the equation instead of having to look for places where the value was 424. Incidentally, the value I get between 28 and 29 is between 28.63 and 28.64. I’m working in degrees; should I be working in radians instead?
You folks have a lot more patience than I do. Thanks again! I appreciate the help.