Let $A$, $B$, and $C$ be three different points such that compositions of the following three rotations $$ R(C, 256^\circ) \circ R(B, 244^\circ) \circ R(A, 220\circ) $$ has a fixed point. Determine the measures of angles of $\triangle ABC$. Here $R(X, a)$ means rotation about $X$ through an angle $a$.
How to solve it? How Victor Prasolov's (Problems in Plane and Solid Geometry V-1) problem 18.40 is applicable here? If possible suggest some books where this type of discussion can be found. Thank you in advance.
Edit: Let angles α, β, γ be such that 0 < α, β, γ < π and α+β +γ = π. We know If the composition of rotations R(C,2γ) ◦ R(B, 2β) ◦ R(A, 2α) has fixed points then it is an identity transformation, then the angles of triangle ABC are equal to α, β, γ. Here if we take α=64 °, β=61°, γ=55° then α+β +γ = π the condition satisfied and the composition of rotations becomes R(C,4γ) ◦ R(B, 4β) ◦ R(A, 4α). My confusion is how this composition also gives the angles of a triangle α, β, γ.
Here is a way to solve the issue using complex numbers.
A rotation with angle $a$ around point $c$ is given by
$$r(z)-c=e^{i a}(z-c) \iff r(z)=c+e^{i a}(z-c)$$
Here if we take $A$ at the origin, we have to compose (with $\alpha =220°, \beta=244°, \gamma= 256°$) :
$$\begin{cases}R_1 : z \rightarrow z_1=e^{i\alpha }z \\ R_2 : z_1 \to z_2=b+e^{i \beta}(z_1-b)\\ R_3 : z_2 \to z_3=c+e^{i \gamma}(z_2-c)\\ \end{cases}$$
Composing them, the image of $z$ is :
$$z'=R_3(R_2(R_1(Z)))=c+e^{i\gamma}(b+e^{i\beta}(e^{i \alpha} z - b))$$
$$z'=c+e^{i\gamma}b+\underbrace{e^{i(\gamma+\beta+\alpha)}}_{= 1}z - e^{i(\gamma+\beta)} b$$
(the central exponential is equal to $1$ because the sum of the three angles equals $720° = 2 \times 360°$). Otherwise said :
$$z'=z+t \ \ \text{where} \ \ t:=c+(-e^{i(\gamma+\beta)}) + e^{i\gamma})b\tag{1}$$
We must resist here to the temptation to conclude that the composition of the 3 rotations is a translation.
Indeed, as it is said in the question that a certain fixed point $Z$ exists, (1) gives:
$$Z=Z+t \ \iff \ t=0$$
yielding the following constraint :
$$c= \underbrace{\left(e^{i(\gamma+\beta)} - e^{i\gamma}\right)}_m \ \ b\tag{1}$$
We can, again WLOG, take $b=1$, giving :
$$c=e^{i(\gamma+\beta)} - e^{i\gamma}$$
yielding finaly a triangle with angles :
$$\hat{A}=arg(c)=108°, \ \hat{B}=arg(-b/(c-b))=46.624°, \ \hat{C}=25.376°$$
which are not the same as the values you have found.