Wikipedia:Reference desk/Archives/Mathematics/2022 December 21

= December 21 =

Calculate distance between centers of circles given area of intersections
I need help checking my work. I'm working on a representation of multiple regression effect sizes using a Venn Diagram for educational policy research. I see that the area of a geometric lens has a closed form solution. Given a Venn diagram made from circles $$A$$, $$B$$, and $$C$$ with centers $$X$$, $$Y$$, and $$Z$$, that $$r=R=\sqrt{1/\pi}$$, and $$A_{A\cap B}=0.23$$, then $$d_{XY}\approx 0.429$$. Can someone validate this? The Venn diagram is represented at this link here. Schyler ( exquirito veritatem bonumque ) 04:07, 21 December 2022 (UTC)


 * Circle $$C$$ has no role. I cannot replicate these numbers. When each circle passes through the centre of the other circle, so the circle centres are $$r\approx 0.56419$$ apart, a rhombus of two equilateral triangles fits within the lens. The area of this rhombus is $$\tfrac{\sqrt 3}{2\pi}\approx 0.27566,$$ so when $$d_{XY}=0.4290.27566,$$ contradicting $$A_{A\cap B}\approx 0.23\,.$$ My calculations give me that distance $$d_{XY}=0.429$$ gets you $$A_{A\cap B}\approx 0.52785\,.$$ Conversely, to get lens area $$A_{A\cap B}=0.23\,,$$ I find we need $$d_{XY}\approx 0.73938\,.$$ --Lambiam 07:38, 21 December 2022 (UTC)
 * Yes, circle $$C$$ has no role. Let $$\theta$$ be the angle at X (or Y) between the lines to the junctions of ABD and of CEFG.  The lens has area $$A =  R^2\left(\theta - \sin \theta \right)$$, where $$R=\sqrt{1/\pi}$$, and the diagram says $$A_{A\cap B}$$ (D+G) has area 0.03, so $$\theta - \sin \theta = 0.03 \pi$$.  That gives $$\theta \approx 0.8235$$ (in radians).  The length from X (or Y) to the midpoint of XY is $$R \cos(\theta / 2)$$, and twice this gives $$d_{XY}\approx 1.0341$$.  My trig is rusty and I may have made some errors, but I think the principle and the order of magnitude are right.  If $$A_{A\cap B}$$ were 0.23, we'd get $$\theta \approx 1.1377$$ and $$d_{XY}\approx 0.9507$$. Certes (talk) 13:31, 21 December 2022 (UTC)
 * For $$\theta=0.8235,$$ we have $$(\theta-\sin \theta)/\pi\approx 0.02864,$$ not quite $$0.03$$ but at least in the ballpark. But in the second case, you missed a factor $$\pi$$: for $$\theta=1.1377,$$ we have $$\theta-\sin \theta\approx 0.23$$ while $$(\theta-\sin \theta)/\pi\approx 0.07322.$$ --Lambiam 15:52, 21 December 2022 (UTC)
 * Here are the calculations for lens area $$0.23,$$ step by step:
 * $$\begin{array}{cccc}

\theta &=& 1.71254 \\ \sin \theta &=& \sin 1.71254 &\approx& 0.98997 \\ \theta-\sin \theta &\approx& 1.71254-0.98997 &\approx& 0.72257 \\ (\theta-\sin \theta)/\pi &\approx& 0.72257/3.14159 &\approx& 0.23000 \\ \cos(\theta/2) &\approx& \cos 0.85627 &\approx& 0.65526 \\ 2R\cos(\theta/2) &\approx& 2 \times 0.56419 \times 0.65526&\approx& 0.73938 \\ \end{array}$$
 * --Lambiam 16:03, 21 December 2022 (UTC)
 * Oops. Thanks, that looks more credible. Certes (talk) 18:40, 21 December 2022 (UTC)

Okay, well this is interesting. Yes, in the example, circle C has no role here. Someone said "When each circle passes through the centre of the other circle," but I do not think that is probable. Here were my steps:

Circles $$A$$, $$B$$, and $$C$$ have centers $$X$$, $$Y$$, and $$Z$$ and radii $$r_A=r_B=r_C=\sqrt{1/\pi}$$. The centers form $$\triangle XYZ$$ and intersect such that $$Area_{A\cap B}=0.03$$, $$Area_{B\cap C}=0.11$$, $$Area_{A\cap C}=0.23$$

$$d_1=\overline{\rm XY};d_2=\overline{\rm XZ};d_3=\overline{\rm YZ}$$

$$Area_{A\cap C}=r^2_A\cos^{-1}(\frac{d^2_1+r^2_A-r_C^2}{2d_1r_A})+r_C^2\cos^{-1}(\frac{d^2_1+r_C^2-r^2_A}{2d_1r_C})-2\Delta$$

$$\Delta=\frac{1}{4}\sqrt{(-d_1+r_A+r_B)(d_1-r_A+r_B)(d_1+r_A-r_B)(d_1+r_A+r_B)}$$

Simplifying:

$$\cancel{0.23=2(\frac{1}\pi \cos^{-1}(\frac{d^2_1}{2d_1\sqrt{\frac{1}{\pi}}}))-\frac{1}{2}\sqrt{(-d_1+2\frac{1}{\pi})(d_1+2\frac{1}{\pi})}}$$

oh I see this was my problem, I think... I distributed the 2 onto $$\frac{1}{4}\Delta$$

$$\cancel{0.23=2(\frac{1}\pi \cos^{-1}(\frac{d^2_1}{2d_1\sqrt{\frac{1}{\pi}}}))-\frac{1}{4}\sqrt{(-d_1+2\frac{1}{\pi})(d_1+2\frac{1}{\pi})}}$$

Update: well, here i am checking my own work again. i found another error. i deleted some values of d within $$\Delta$$. I should know better than to ask for help right away... i can do this... but the doubt is strong in this one

$$\cancel{0.23=2(\frac{1}{\pi} \cos^{-1}(\frac{d^2_1}{2d_1\sqrt{\frac{1}{\pi}}}))-\frac{1}{4}\sqrt{(-d_1+2\frac{1}{\pi})(d_1+2\frac{1}{\pi})}}$$

no that's wrong too, $$r=\sqrt{\frac{1}{\pi}}$$... i got it mixed up in $$\Delta$$ again...
 * You can shift two equal-sized circles such that each passes through the other's centre; I used this special case merely to easily establish bounds that were violated by a purported solution.
 * The function $$f(\theta)=(\theta-\sin \theta)/\pi$$ is transcendental, and one cannot hope to solve the equation $$f(\theta)=x$$ by a combination of algebraic and trigonometric manipulations for rational values of $$x,$$ except when $$x$$ is an integer, in which case the equation is solved by $$\theta=x\pi.$$ --Lambiam 09:01, 22 December 2022 (UTC)