Find closest point in triangle given barycentric coordinates outside Given a non-degenerate triangle ABC and an arbitrary point P in 3D space, I can project P onto the plane defined by ABC and check whether the triangle contains it as described here. I end up with barycentric coordinates of the point.
In case the point does not lie inside the triangle, i.e. any of alpha, beta or gamma are not in range [0..1], I now want to find the point inside the triangle which is closest to the projection of P onto the plane defined by ABC. How would I do that?
 A: You know that 
$$
\begin{align}
P = \alpha A + \beta B + \gamma C &= \beta(B-A) + \gamma(C-A) + A\\
&= \alpha (A-B) + \gamma(C-B) + B \\
&= \alpha (A-C) + \beta(B-C) + C.
\end{align}
$$
We can classify the outside of the triangle in six regions:
1) $0\leq\beta$, $0\leq\gamma$, and $1\leq\beta+\gamma$, in which case the point is closest to $BC$. (In which case would it be closest to the $B$ or $C$?)
2) $0\leq\alpha$, $0\leq\gamma$, and $1\leq\alpha+\gamma$, in which case the point is closest to $AC$.
3) $0\leq\alpha$, $0\leq\beta$, and $1\leq\alpha+\beta$, in which case the point is closest to $AB$.
I'll let you work the other three cases which involve mixed signs, for example $1<\gamma$, $\beta<0$, for which the vertex $C$ is the closest point.
A: As @PaulDuBois commented, the above answer (@hjhjhj57) is incorrect for obtuse angles.
In fact, we can classify the outside of the triangle in three regions:
1) $\alpha\geq 0$ and $\beta<0$, in which case $P$ is closest to either $AB$ or $AC$,
2) $\beta\geq 0$ and $\gamma<0$, in which case $P$ is closest to either $AB$ or $BC$,
3) $\gamma\geq 0$ and $\alpha<0$, in which case $P$ is closest to either $AC$ or $BC$.
The three regions are equivalent under substitution of variables and I will work out case 1:
If $\gamma<0$, and $(P-A)(B-A)>0$, $i.e.$ the projection of $P$ onto $AB$ where vector multiplication is the dot product, then $P$ is closest to $AB$ (verify that this can only be the case if $A$ is obtuse). Then
\begin{eqnarray}
  \beta'  &=& \text{min}(1, \frac{(P-A)(B-A)}{(B-A)(B-A)} ) \\
  \gamma' &=& 0
\end{eqnarray}
Otherwise $P$ is closest to $AC$ with
\begin{eqnarray}
  \beta'  &=& 0 \\
  \gamma' &=& \text{min}(1, \text{max}(0, \frac{(P-A)(C-A)}{(C-A)(C-A)} ) )
\end{eqnarray}
And lastly
\begin{eqnarray}
  \alpha' &=& 1 - \beta' - \gamma' \\
  P'      &=& \alpha'A + \beta'B + \gamma'C
\end{eqnarray}
