TITLE: Can one always map a given triangle into a triangle with chosen angles by means of a parallel projection?
QUESTION [11 upvotes]: This is something that seems to be true from experience by playing with shadows from the sun:
If one cuts a paper triangle, he can turn it in a way to make its shadow be a triangle of any given angles (of course, not exceeding internal sum of 180°), for example: If one draws a right triangle with internal angles 90°, 60° and 30°,  one can turn it in a way that the shadow triangle can be of angles 60°, 60°, 60° (equilateral) or 120°, 30° and 30° (isosceles), or any other triangle possibilities.
Is it true? How can we prove it? (if possible, showing not only that it's possible, but how to obtain the specific mapping that does it)

REPLY [4 votes]: Here's an explicit solution via linear algebra.
Place both triangles on the plane with one vertex at the origin. Then we need to map the two edges going out from the chosen vertex of one triangle, call them $u_1$ and $v_1$, to the corresponding edges of the other triangle, $u_2$ and $v_2$. Both pairs of vertices are linearly independent, so there is a $2\times2$ matrix $M$ that maps $u_1$ to $u_2$ and $v_1$ to $v_2$.
This matrix has a singular value decomposition $M=U\Sigma V^T$ where $U$ and $V$ are orthogonal and $\Sigma$ is diagonal. $U$ and $V^T$ correspond to rotations and reflections in the plane, so the distortion of the triangle is encoded entirely in $\Sigma$.
Let the diagonal entries of $\Sigma$ be $\sigma_1$ and $\sigma_2$ with $\sigma_1\ge\sigma_2$. How to obtain a rigid 3D transformation whose projection to the plane is similar to $\Sigma$? Consider the matrix $\tilde\Sigma = \frac1{\sigma_1}\Sigma$ whose diagonal entries are $1$ and $\sigma_2/\sigma_1$. This is the projection of a 3D rotation about the $x$-axis by an angle of $\theta=\cos^{-1}(\sigma_2/\sigma_1)$.
So:

Rotate or flip the first triangle in the $xy$-plane according to $V^T$.
Rotate it about the $x$-axis by an angle $\theta$, taking it out of the $xy$-plane.
Rotate or flip it again according to $U$.

The shadow of the result is a $1/\sigma_1$-scale copy of the second triangle.