Blue or Yellow plus Red

Since the yellow area and blue area are equal, the area of the yellow equilateral triangle is half that of the unit equilateral triangle, and the area of one of the blue triangle is $\dfrac 16$ that of the unit equilateral triangle or $A = \dfrac 16 \cdot \dfrac {\sqrt 3}4 = \dfrac 1{8\sqrt 3}$ .

Since the yellow equilateral triangle is half that of the unit equilateral triangle, its side length is $\dfrac 1{\sqrt 2}$ . Let us consider one of the blue triangles $\triangle BDE$ . Then $DE = \dfrac 1{\sqrt 2}$ . Let $BE = a$ ; then $AD = a$ and $DB = 1-a$ . The semiperimeter $s = \dfrac {\frac 1{\sqrt 2} + a + 1 - a}2 = \dfrac {\frac 1{\sqrt 2} + 1}2$ .

The radius of a red circle

$r = \frac As = \frac 1{4\sqrt 3\left(\frac 1{\sqrt 2}+1\right)} = \frac {\sqrt 2 \cdot \sqrt 3(\sqrt 2-1)}{4\cdot 3 (\sqrt 2+ 1)(\sqrt 2 - 1)} = \frac {2\sqrt 3 - \sqrt 6}{12}$

The required answer is $a+b+c+d = 2 + 3 + 6 + 12 = \boxed{23}$ .

Yellow area : yellow-blue area = 1 : 2
Yellow side : yellow-blue side = 1 : √2
Therefore, FD = (1 / √2) × AC = 1 / √2

AD = FC, so
One blue perimeter
= FD + DA + AF
= FD + CF + AF
= FD + (CF + FA)
= FD + CA
= 1/√2 + 1

One blue area
= (1 / 3) × (1 / 2) × ABC area
= (1 / 6) × (1² × sin 60° / 2)
= √3 / 24
= (1 / 2) × (One blue perimeter) × R
= (1 + √2)R / 2√2

R = 2√6 / 24(1 + √2)
= (√2 – 1) / 2√6
= (√12 – √6) / 12
= (2√3 – √6) / 12

Answer = 2 + 3 + 6 + 12 = 23