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# Halving half a disk

Posted 10 years ago
 I want (for starters) to cut half a disk (which I think is the same thing as a "semicircular lamina") into two pieces of the same area, with a cut parallel to the straight ("diametrical") side of the half disk. (In all the following, radius is 1.) At first I applied the formula for the area enclosed by a chord at the Mathworld "chord" page, area = ArcCos-r Sqrt[1-r squared], where r is the distance from the center of the circle to the chord, to try to get an area of Pi/4. I couldn't solve it, and my Mathematica couldn't solve it, but by trial-and-error I got an r of around 0.4040 (I don't need a mathematical solution; a numerical approximation will do). Then I saw there's a Mathworld "semicircle" page, where the "geometric centroid of the semicircular lamina" (which I think is another way of describing what I'm looking for) is given as 4/(3Pi), which is about 0.4244. So either one or both pages is wrong, or what is of course much more likely, I'm missing something.I'm calling it a night so I won't be rejoining this discussion until tomorrow.
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Posted 10 years ago
 Thank you. This is useful, as is the answer I got when I posted under another discussion title, http://mrob.com/pub/ries/index.html
Posted 10 years ago
 GoogleInverse Symbolic Calculatorbut it does not include this particular constant.http://oldweb.cecm.sfu.ca/projects/ISC/ISCmain.html
Posted 10 years ago
 I have a general question I'll post also under a new discussion title, but in my case it happens to be related to the above question.I have a book, A Dictionary of Real Numbers, by Borwein and Borwein. If you're not familiar with it, it's a list of numbers that are the result of a limited number of operations on a limited number of numbers. The numbers printed are eight digits after the decimal point, truncated, not rounded. I took the number I got by trial-and error, 0.403972, and looked up 403972, but it's not in the book. But if for example in another problem I laboriously come up with -2.403441, I look in the book and find 40344144, and the book shows it came from 3Sqrt/4-2Sqrt. Of course this isn't a mathematical proof that the number I calculated is equal to 3Sqrt/4-2Sqrt, but it does suggest a direction I might want to go if I do try to prove it.My question is, is there anything like this online?
Posted 10 years ago
 This info is great. Thank you both very much.
Posted 10 years ago
 Suppose the lamina is not exactly half disk. We can compute this extended problem easily with Mathematica. Particulary, we are looking for the line P3_P4 that cuts the lamina into two pieces of same area. So the area above p3_p4 is basically that  the area of Disk P3_P0_P4 minus the area of the blue triangle. The entire area of the lamina is that the area of Disk P0_P1_P3_P4_P2 plus the area of the red triangle. Call the obtuse angle of the red triangle: "phi";  the obtuse angle of the blue triangle : "theta", we have r=Sqrt[(1/2)^2+y0^2];\[Phi]=2*ArcTan[1/(2*y0)];s=(2\[Pi]-\[Phi])/2*r^2+1/2*r^2*Sin[\[Phi]];res={r,\[Phi],\[Theta]/.FindRoot[r^2/2*\[Theta]-Sin[\[Theta]]/2*r^2==s/2,{\[Theta],\[Pi]/2}]}where r is radius of the lamina, P0 = {0,y0} the center and the coordinate of p1 and p2 be {+/- 1/2 ,0} respectively . The above code is all you need to resolve this problem. I can merge the result and some extra graphics code to make a dynamic presentation: where y0 is the distance between P0 and chord P1_P2 and the ratio in the vertical axis is defined as 1 minus (height of blue triangle divides the radius of the circle). Certainly , when the center moves very far this ratio should be one, which means the diameter cuts the laminar into two identical parts approximately. Copyable code:  Manipulate[  {r, \[Phi], \[Theta]} = f[y0];  p0 = {0, y0};  p1 = {-1/2, 0};  p2 = {1/2, 0};  p3 = {-r*Sin[\[Theta]/2], y0 + r*Cos[\[Theta]/2]};  p4 = {r*Sin[\[Theta]/2], y0 + r*Cos[\[Theta]/2]};  tracker = {y0, 1 - Cos[f[y0][]/2]};  g = Graphics[   {    Circle[{0, y0},      r, {-((\[Pi] - \[Phi])/2), (3*\[Pi] - \[Phi])/2}],    {Red, Polygon[{p1, p2, {0, y0}}]},    {Blue, PointSize[0.03], Point[{p1, p2}]},    {Blue, PointSize[0.03], Point[{p3, p4}]},    {Blue, Line[{p0, p3, p4, p0}]},    Text["p0", p0 + {0, 0.1}],    Text["p1", p1 - {0, 0.1}],    Text["p2", p2 - {0, 0.1}],    Text["p3", p3 - {0.1, 0}],    Text["p4", p4 + {0.1, 0}]    }]; Plot[1 - Cos[f[x][]/2], {x, 0.001, 1.3},  PlotRange -> Full,  AxesLabel -> {"y0", "ratio"}, LabelStyle -> Directive[13, Italic],  Epilog -> {    Inset[Style[g, Magnification -> 1.3], {0.9, 0.75 + y0/35}],    {PointSize[0.02], Point[tracker]},    {Dashed, Line[{{0, tracker[]}, tracker, {tracker[], 0}}]}    }, ImageSize -> {500, 400}  ] , {y0, 0.001, 1.3}, Initialization :> (   f[y0_] := Module[{r, \[Phi], s, res},     r = Sqrt[(1/2)^2 + y0^2];     \[Phi] = 2*ArcTan[1/(2*y0)];     s = (2 \[Pi] - \[Phi])/2*r^2 + 1/2*r^2*Sin[\[Phi]];     res = {r, \[Phi], \[Theta] /.         FindRoot[         r^2/2*\[Theta] - Sin[\[Theta]]/2*r^2 == s/          2, {\[Theta], \[Pi]/2}]}     ])]
Posted 10 years ago
 The equation can be solved numerically to any desired precision with FindRoot, for example In:= FindRoot[ArcCos[r] - r Sqrt[1 - r^2] - Pi/4, {r, 0}, WorkingPrecision -> 20]Out= {r -> 0.40397275329951720932} As for the discrepancy, this is because the centroid does not have the sought area-bisection property. To see the difference, consider for example a simple isosceles triangle. It is easy to check that the centroid is situated at 1/3 of the height, however the half-area line is at 1-1/Sqrt of the height, which is lower than the centroid. 