How to test whether two triangles can be made to coincide by rotation and translation only

Question

I am interested in conditions that determine if two triangles can be made to coincide with each other by rotation and translation only (without mirroring). So, given two triangles, one should be able to conclude if any of those two are "a match".

Any tips are welcome :)

Answer 1

In a coordinate setting, one way to check that the orientations of two triangles match up is to use the determinant formula for area: $$ S_{ABC} = \frac12 \det \begin{bmatrix}x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1\end{bmatrix} $$ Usually, we take the absolute value of this formula to find the area of a triangle. Without the absolute value, however, it gives the signed area: a positive number when $A,B,C$ are in counterclockwise order around the triangle and a negative number when $A,B,C$ are in clockwise order.

So to test if $\triangle ABC$ and $\triangle DEF$ are congruent-without-reflections given the coordinates $(x_A,y_A), \dots, (x_F, y_F)$ of all six points, we can test if they're congruent by checking side lengths as usual: check if the multisets $$ \{(x_A-x_B)^2 + (y_A-y_B)^2, (x_A-x_C)^2 + (y_A-y_C)^2, (x_B-x_C)^2 + (y_B-y_C)^2\} $$ and $$ \{(x_D-x_E)^2 + (y_D-y_E)^2, (x_D-x_F)^2 + (y_D-y_F)^2, (x_E-x_F)^2 + (y_E-y_F)^2\} $$ are the same. If they are, this tells us which points are supposed to match up to each other: for example, if $AB = EF$, $BC = DE$, and $AC = DF$, this tells us that $A$ matches up with $F$, $B$ matches up with $E$, and $C$ matches up with $D$.

To do the final check and verify that $\triangle ABC$ and $\triangle FED$ have the same orientation, check if $S_{ABC}$ and $S_{FED}$ have the same sign: if not, then a reflection is necessary.

Answer 2

What you are looking for is the congruence of triangles. However, this needs to be tweaked a bit so that forbidding reflections is accounted for.

I am just spitballing here,

$\triangle ABC$ "coincides" with $\triangle PQR$ whenever starting at some $\angle A_1$ of $\triangle ABC$ and then go around clockwise and form a sequence of $A_n =A_1,A_2,A_3,A_1,A_2,\dots$ and we do the same with $\triangle PQR$ getting $P_n=P_1,P_2,P_3,P_1,P_2,\dots$ then $A_n$ and $P_n$ are subsequences of each other.

And then same holds for lengths of the lines of the triangles.

I believe if you could satisfy those criteria then your criteria for coincides would be met.

Hope this helped. Cheers!

NB: If $A_n$ and $P_n$ are subsequences of each other this doesn't mean that $A_n = P_n$