In the formulas...

r = the radius of the circumscribed circle

i = the radius of the respective inscribed circle

s = the sagitta of the respective circular segment

**Method A** (Area of circular segments minus area of inscribed circles)

1. Total area of the 3 circular segments = 481.49203......respective segment areas = r²ArcCos((r - s)/r) - (r - s)√(2rs - s²)

2. Total area of the 3 inscribed circles R, M & L = 175.92919......respective circle areas = πi²

481.49203

- 175.92919

------------

305.56284 = shaded region area

=======

**Method B** (Area of circumscribed circle minus areas of inscribed objects)

1. Area of circumscribed circle = 756.47691......πr²

2. Total area of the 3 inscribed circles R, M & L = 175.92919......respective circle areas = πi²

3. Area of triangle ABC, using Heron's Formula on the 3 circular segment chord lengths AB, AC & BC = 274.98488......respective chord lengths = 2√(2sr - s²)

756.47691

- 175.92919

- 274.98488

------------

305.56284 = shaded region area

=======

*Edit: I first had Method B as my only method (same as the one in my post #5), but then discovered Method A.*

I have zero idea of how to solve for 'r' in that, other than in Mathematica, or online in W|A.

W|A's 'Wolfram Language code' for finding 'r' is:

Solve[2 r ArcSin[Sqrt[8 r - 16]/r] + 2 r ArcSin[Sqrt[16 r - 64]/r] + 2 r ArcSin[Sqrt[24 r - 144]/r] == 2 Pi r, r]

*Edit: I had to click on the 'Approximate forms' button in W|A to get their answer, but was given 2 values to choose from. I chose the one (r≈15.5175) that resulted in the area of the shaded region being a positive value, the other yielding negative.*

My wife thinks I 'cheated' by using a computer to generate the answer.

Hi Bob;

Your wife would have had an argument on that topic with bobbym, who said this about that! (which mathsyperson later commented on)

Anyway, I did as you did, and resorted to 'cheating' by using W|A to solve a horrible equation.

The equation was for the sum of the lengths of the arcs in the M, R and L circular sectors = 2πr...and W|A said that r≈15.5175.

]]>As that formula can be proved by basic Euclidean methods, there must be a way to get to a result using such methods. So I'll keep working on this. It may take a while.

Bob

]]>OD = r-8 so

which simplifies toSo

similarly

If a,b and c are the lengths of the sides of a triangle then there's a formula connecting these with the circum-radius:

I put this system of four equations into Wolfram Alpha and it came up with r = 15.5175.

Bob

]]>I can't find a proper solution, but drew it up in Geogebra to get some sort of answer.

I don't know how accurate that is.

]]>Whilst being sorry for you that you haven't done it, you have made me happy that I'm not the only one stuck on this.

Just getting a diagram on 'Sketchpad' took a lot of trial and improvement and now I've realised I'll have to start again with a scaled down diagram so I can put in some new construction lines. Hhhmmm. Quite a problem.

Bob

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