You have an isosceles triangle with points labelled ABC
(source)
"standing" vertically, so B is the apex at the top. Now some rotation then translation in the plane is applied, after which A,B,C are now at the locations A',B',C' respectively. You know the coordinate pairs of A',B',C' and which one of those is B', but you don't know yet which is A' and which is C'.
Figure out which of the two remaining coordinates is A' and C', purely in terms of coordinates, with trig level math, if possible.
I'm not sure if this is relevant, but this is for a computer program so a solution must exclude visual observation.
Thank you to the folks at freenode.net #math for helping me to formulate my question into the proper terms. A lead on an answer was given that I might investigate Matrix Rotation:
That math is beyond the length of my sleeve at the moment, but I'm betting that this might be able to be solved with easier math?
One thought I had was that the slopes a and c will always be negative and positive in relation to a median line from B to the mid point of A and C but, the only way I know to calculate slope is in relation to the coordinate axis’s.
Many Thanks in advance for help with my quandary.
One tool to distinguish orientation is the planar cross product, defined for two vectors as $$(a,b)×(c,d)=ad-bc$$ This is also the magnitude of the true cross product of $(a,b,0)$ and $(c,d,0)$, hence the name. It distinguishes orientation in that rotating the two vectors does not change the product, but reflecting the vectors across any line through the origin inverts its sign.
With all points being considered as position vectors, let $n=(A-B)×(C-B)$. Now let $P_1$ and $P_2$ be the two points which are $A'$ and $C'$ but in unknown order; compute $(P_1-B')×(P_2-B')$. If this has the same sign as $n$, $P_1=A'$ and $P_3=C'$; otherwise it is the other way round. No matrices are required at all, and this works for any triangle in any orientation, as long as only rotations and translations are applied.