Another sequel to the Odd Sangaku Problem by Michael Huang.

Label the diagram as follows, add segments $HF$ and $FK$ , and let $x = DE = EF = FG = GD$ .

By the Pythagorean Theorem on $\triangle ABC$ , $AC = \sqrt{27^2 + 36^2} = 45$ .

All the triangles are similar to each other by AA similarity, and since the purple and red circles have the same radii, $\triangle EBK \cong \triangle FGI$ , so $EB = x$ .

Since $\triangle ABC \sim \triangle ADE$ , $AE = ED \cdot \cfrac{AC}{BC} = x \cdot \cfrac{45}{36} = \cfrac{5}{4}x$ .

Then $AB = AE + EB = \cfrac{5}{4}x + x = 27$ , which solves to $x = 12$ .

Since $\triangle EHF \sim \triangle ABC$ , $EH = EF \cdot \cfrac{AB}{AC} = 12 \cdot \cfrac{27}{45} = \cfrac{36}{5}$ .

Then $IL = HB = EB - EH = 12 - \cfrac{36}{5} = \cfrac{24}{5}$ .

Since $\triangle ILC \sim \triangle ABC$ , $LC = IL \cdot \cfrac{BC}{AB} = \cfrac{24}{5} \cdot \cfrac{36}{27} = \cfrac{32}{5}$ and $IC = IL \cdot \cfrac{AC}{AB} = \cfrac{24}{5} \cdot \cfrac{45}{27} = 8$ .

Therefore, the radius of the pink circle is $r = \cfrac{1}{2}(IL + LC - IC) = \cfrac{1}{2}\bigg(\cfrac{24}{5} + \cfrac{32}{5} - 8\bigg) = \cfrac{8}{5}$ , and its square is $r^2 = \cfrac{8^2}{5^2} = \cfrac{64}{25} = \boxed{2.56}$ .