A geometry problem for my coding project...

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Ryan Burnside (Posted 2009) [#1]

I'm making some art in the style of Chuck Close. This artist makes pictures using concentric circles of color. The colors blur in the viewer's eyes from far away forming crude pixels. I want to make an image converter that outputs in a similar style. So I need to have all the circle shapes have the same area.

Gabriel (Posted 2009) [#2]

Misread the question, sorry. I thought you were going out, not in.

Ryan Burnside (Posted 2009) [#3]

I mean that all 3 colored sections have equal areas. blue=teal=green Each area will be one of the RGB colormode components so the area needs to be equal so they blend in the eye. Think of these as giant circular computer pixels. :)

Warpy (Posted 2009) [#4]

Measured from the centre, say the green circle has radius r1, teal has radius r2, and blue r3. The area of the teal ring is the area of the middle circle minus the area of the centre circle. pi*(r1)^2 = pi*(r2^2-r1^2) Cancel pi and rearrange: 2*(r1^2) = r2^2 r2 = sqrt(2)*r1 And the same idea applies for r3 = sqrt(2)*r2.

Gabriel (Posted 2009) [#5]

.

PGF (Posted 2009) [#6]

Warpy was right for r2 but made an incorrect leap for r3. In fact the n'th circle has radius rn = sqrt(n)*r1. So starting at r1, r2 = sqrt(2)*r1, r3 = sqrt(3) * r1, ... r17 = sqrt(17) * r1 ... Or since you are giving D which is 2 * r3, r1 = D/(2*sqrt(3))

Ryan Burnside (Posted 2009) [#7]

Thanks again guys. :)

Warpy (Posted 2009) [#8]

Yep, I was wrong. In my defence, I wrote that at 3am when woken up by something in my eye...