Suppose you have a labelled triangle with known side lengths, and you take one image of this triangle using a known pinhole camera (i.e. the focal length is known), from a point with known coordinates, and also the orientation of the camera is known. How can you find the coordinates of the three vertices of the triangle? And are these coordinates unique, or is it possible that a single image will corresponds to more than one possible set of coordinates for the three vertices ?
My Attempt: Since the triangle vertices are labelled we have a known correspondence between the vertices and their images. If the vertices are $P_A, P_B, P_C$, and their images are $Q_A, Q_B, Q_C$, then
$P_A = P_0 + t_1 R Q_A $
$P_B = P_0 + t_2 R Q_B $
$P_C = P_0 + t_3 R Q_C $
note that since the focal length of the pinhole camera is known, then $Q_A, Q_B, Q_C$ are known vectors. Also, since the position of the pinhole and the orientation of the camera is known, then vector $P_0$ and matrix $R$ are also known. The unknowns are the scalars $t_1, t_2, t_3$ and the vectors $P_A, P_B, P_C $. This is where knowledge of the side lengths comes into play. The sides are $a = \| P_B P_C \| , b = \| P_A P_C \|$ , and $c = \| P_A P_B \|$. From the expressions above, and using the fact the $R$ is a rotation matrix, so it does not change of the length, then
$a^2 = (t_3 Q_C - t_2 Q_B)^T (t_3 Q_C - t_2 Q_B ) $
$b^2 = (t_3 Q_C - t_1 Q_A )^T (t_3 Q_C - t_1 Q_A) $
$ c^2 = (t_2 Q_B - t_1 Q_A)^T (t_2 Q_B - t_1 Q_A ) $
which are three quadratic equations in the three unknowns $t_1, t_2, t_3$ , that can be solved simultaneously by numerical methods.
Once $t_1,t_2, t_3$ are found, then we're done, because now we can compute the coordinates $P_A, P_B, P_C $.