how can i styling overlapping part of two circles with filling of hatching?

卫博 shared this question 5 months ago
Answered

hi,

thanks for the great tool to engage math teaching.

How can i styling overlapping section of two circles with filling of hatching?

Currently, i can only see different color lightness or hue changed bit. But i want to style it with hatching filling. How can i implement that? Is there any workaround solution for it?

thanks.

Comments (14)

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Hi,


- for the top semi circle, you can use a circularsector that can be filled

- for other domains, you can


  • create 2 circular arcs
  • create a free M point on the sequence of 2 such circular arcs
  • create MM=M just to create a locus MM from M
  • you can fill or hatch the domain of the locus with the style

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arc: CircularArc(centre,point,point)

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hi, Michel, currently i have understood your approach of locus for this problem. It is perfect and beautiful and can work for many situations. Great! thanks~

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I like this way

Files: foro.ggb
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hi, mathmagic,

If the overlapped region is a semi-circle, can we use your approach of inequality?

I have not found a valid leftSide(semiCircle) expression.

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using the circle c with the diametral line (not vertical) e.g. f(x)=2x+3, just do the intersect of the domains (inequation of the semi-plane and the inequation of the disk related by && (that yields ^)) like that :


(y>=f(x))&&(LeftSide(c)<=RightSide(c))


LeftSide(c)<=RightSide(c) <=> CM^2<= r^2 <=> CM<= r with C the center and r the ray

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for this case you can use semicircle trough 2 points also (it is valid for vertical diameters)

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@mathmagic, assumming k=semicircle through 2 points, but following expression is not valid:

LeftSide(k)<RightSide(k)

Is there some workaround solution?

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@Michel, thank you so much for your effort helping me! Can you give a solution based on inequality approach for this final product in the attachment?

thanks

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With intersections of domains

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thanks, michel, i caught it

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thank you very much. You helped me out one day~. I like @mathmagic way. It is simple and just work.

I am a new hand, solution from Michel and alfabeta2 works well, but i can not understand well. I will do more study and understand these solutions. thanks again~

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my solution using a custom tool:

https://ggbm.at/XmvnxSZT

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some usefull figures without inequality

Files: foro.ggb
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