Isometry between two identical shape in $\mathbb{R}^2$ I know that it could be a silly question, but if I have two identical convex compact shapes  in $\mathbb{R}^2$ (e.g. two identical equilateral triangles with side length equal to $1$), what is the isometry between them? I know they are isometric up to a rigid motion, but I'd like an explicit formula of such isometry. 
 A: Isometry is for small quadrilateral differential lengths as part of larger surfaces. . In the case of equilateral triangles it is a congruence or identity.
A: Every isometry of $\mathbb R^2$ is a composition of a rotation, a reflection, and a translation.  Every isometry either preserves orientation (so that clockwise loops map to clockwise loops) or reverses it (so that clockwise maps to counterclockwise). In the former case, you just have a composition of a translation and a rotation:
\begin{array}{rccc}
x & \mapsto & ax-by & \mapsto & ax-by + c \\
y & \mapsto & bx+ay & \mapsto & bx+ay + d \\[6pt]
& \uparrow & & \uparrow \\
& \text{rotation} & & \text{translation} \\
& (\text{where } \\
& a^2+b^2 \\
& =1.)
\end{array}
In the above, if you find $a$, then you have $b=\pm\sqrt{1-a^2}$ so you've got $b$ except for $\text{“}\pm\text{''}$.
Find two particular points whose images you know: thus $(x_0,y_0)$ maps to $(x_1,y_1)$ and $(x_2,y_2)$ maps to $(x_3,y_3)$, and plug them in to the the system of equations above (including $a^2+b^2=1$), and solve for $a,b,c,d$.
