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Draw $IG$ and consider $\triangle GHI$ . Let $O$ be the center of the circle with a radius of $1$ . Since $\angle GHI = 22.5°$ , central angle $\angle GOI = 45°$ . Draw the perpendicular bisector of $GI$ at $J$ , and draw $HK \perp OJ$ at $K$ , and let $x = \angle HOI$ .

Then $\angle IOJ = 22.5°$ , so that from $\triangle IOJ$ we have $IJ = \sin 22.5°$ and $OJ = \cos 22.5°$ , and from $\triangle HOK$ we have $OK = \cos (x + 22.5°)$ . Therefore, $JK = \cos 22.5° - \cos (x + 22.5°)$ and $GI = 2 \sin 22.5°$ , so that $\triangle GHI$ has an area of $A_{\triangle GHI} = \frac{1}{2} \cdot GI \cdot JK = \sin 22.5° (\cos 22.5° - \cos (x + 22.5°))$ .

Since inscribed angles $\angle ABC = \angle BCD = \angle CDE = \angle DEF = \angle EFG = \angle FGH = 22.5°$ , central angles $\angle AOC = \angle BOD = \angle COE = \angle DOF = \angle EOG = \angle FOH = 45°$ . Then by a similar method from above, $A_{\triangle EFG} = \sin 22.5° (\cos 22.5° - \cos (x + 22.5° + 90°))$ , $A_{\triangle CDE} = \sin 22.5° (\cos 22.5° - \cos (x + 22.5° + 180°))$ , and $A_{\triangle ABC} = \sin 22.5° (\cos 22.5° - \cos (x + 22.5° + 270°))$ .

That means the sum of the areas of the red triangles is $A = \sin 22.5° (\cos 22.5° - \cos (x + 22.5°)) + \sin 22.5° (\cos 22.5° - \cos (x + 22.5° + 90°))$ $+$ $\sin 22.5° (\cos 22.5° - \cos (x + 22.5° + 180°)) + \sin 22.5° (\cos 22.5° - \cos (x + 22.5° + 270°))$ $=$ $4 \sin 22.5° \cos 22.5°$ , which is equivalent to the sum of the areas of congruent triangles $\triangle AOC$ , $\triangle COE$ , $\triangle EOG$ , and $\triangle GOI$ , which means the red area is half the circle, which means the two areas are equal .

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The solution above is for the specific case of inscribed angles of $\frac{90°}{4} = 22.5°$ where $n = 4$ . The same method can be used for a more general solution for inscribed angles of $\frac{90°}{n}$ where $n$ is a positive integer greater than $1$ , to find that one set of same-colored triangles have an area sum of $\displaystyle \sum_{k=0}^{n - 1} \sin(\frac{90°}{n})(\cos(\frac{90°}{n}) - \cos(\frac{90°}{n} + x + k\frac{180°}{n})) = n \sin (\frac{90°}{n}) \cos (\frac{90°}{n})$ , which is equivalent to the area of half a regular polygon with $2n$ sides, to conclude that the two areas are equal for any integer $n > 1$ .