Is there an elegant solution?

Question

I want to solve the equation

3arcsin(5/(2R)) + arcsin(6/(2R)) + arcsin(8/(2R)) = pi

for R, where R is a positive real number. Of course R > 4 for the last term to be defined. I know that the solution is 5, and that this solution is unique [the derivative is negative, so the left side is monotonic]. However, I got this solution essentially from a numeric solver. Is there a nice way to find this solution?

I've thought of plugging both sides into the sine function, but then it turns into a long pseudo-polynomial problem, which doesn't seem helpful.


If it helps, this equation comes up in the following geometry problem. A pentagon is inscribed in a circle and has edge lengths 5, 5, 5, 6, and 8. What is the radius of the circle? An alternative computation of the radius is also acceptable.

Indica · Accepted Answer

x=R²  satisfies the quartic f(x) = (3906.25−649.5x+31.25x²)² − 576x²(144-25x+x²) = 0

x=25 is the only real root

But f(x) = 0.0625(x−25)²(6409x²−98650x+390625) so it's a double root

edit  :

Alternatively if 3θ is angle between sides of length 8 & 6

100 − 96cos(3θ) = 25{sin(3θ)/sin(θ)}²

Putting x=cos(θ) leads to 400x⁴+ 384x³ – 200x² – 288x – 75 = 0

which factors as (4x²−3)(100x²+96x+25) = 0

The only real solutions are x=±½√3 whence θ=π/6

Is there an elegant solution?

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