Revenge of the impossible Sangaku

I wish we could flag a problem as "brilliant", because this one definitely is.

The workings get rather messy, so the following is just a sketch of how I found each of $P,B,Y,G,R$ . For clarity, let the common side have length $s=20$ (and note that this is also the radius of the circumscribed circle). Working from the top...

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$\color{#CEBB00}{\text{Yellow circle}}$

This circle's diameter is the height of the circle segment cut off by the hexagon; that is $2Y=s \left(1-\cos 30^{\circ} \right)$

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$\color{#D61F06}{\text{Red circle}}$

The diameter of this circle is the height of the hexagon minus the height of the pentagon; so $2R=2s \cos 30^{\circ} - \frac12 s \tan 72^{\circ}$

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$\color{#20A900}{\text{Green circle}}$

The first tricky one. If we extend the side of the square that is tangent to this circle, we form a triangle $\Delta$ with part of the top two sides of the pentagon. The green circle is the incircle of $\Delta$ .

$\Delta$ is clearly isosceles, with angles $36^{\circ},36^{\circ},108^{\circ}$ . Its height is the height of the pentagon minus the height of the square. This is enough to uniquely define $\Delta$ , and so we can find its inradius.

After some ugly algebra, and using standard trigonometric values for angles in a pentagon, we find $G=\frac{-2 + \sqrt{5 + 2 \sqrt5}}{2 \sqrt5} s$

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$\color{#3D99F6}{\text{Blue circle(s)}}$

We use a similar trick for this one. The blue circle is the incircle of a $30-60-90^{\circ}$ triangle, whose longer leg is $s$ .

A bit more algebra, and $B=\frac12 s \left(1-\frac{1}{\sqrt3} \right)$

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$\color{magenta}{\text{Purple circle}}$

This one is just the incircle of an equilateral triangle with side $s$ , so $P=\frac{\sqrt3}{6} s$ .

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Putting all these together, and following the required form of the solution, we find $P+B+Y-G-R=5\sqrt{5+2\sqrt5}- 2\sqrt5 \sqrt{5+2\sqrt5} + 2\cdot 5 \cdot 2 + 2\cdot 2\sqrt5 - 5 \cdot 3 \sqrt3$ , that is $a=5$ , $b=2$ and $c=3$ so that $a+b+c=\boxed{10}$ .