# Internal Distances and Angles in Terms of Barycentric Coordinates and Side Lengths

## weblog/

Suppose we are given a triangle $ijk$ with known metric --- so that $\ell_i ∈ \mathbb{R}^+$ is the length of the edge opposite corner $i$ --- and a point $p$ within this triangle identified by barycentric coordinates $[w_i,w_j,w_k]$. We are interested in deriving expressions for the distance $d_i$ from $p$ to the corner $i$ and the internal angle $φ_i$ subtended by the side $jk$ at $p$.

Momentarily embed the triangle in $\mathbb{R}^2$; the choice of translation and rotation are irrelevant. We can write the point-to-point squared distance expression in terms of barycentric coordinates: $$d_i^2 = ‖ \mathbf{p}_i - w_i \mathbf{p}_i - w_j \mathbf{p}_j - w_k \mathbf{p}_k ‖².$$ Because the barycentric coordinates sum to one ($w_i+w_j+w_k = 1$), we can simplify this to: \begin{align} d_i^2 &= ‖ \mathbf{p}_i - (1-w_j-w_k) \mathbf{p}_i - w_j \mathbf{p}_j - w_k \mathbf{p}_k ‖², \\ &= ‖ w_j (\mathbf{p}_i-\mathbf{p}_j) + w_k (\mathbf{p}_i-\mathbf{p}_k) ‖². \end{align} Expanding the square produces an expression in terms of scalar lengths and angles: $$d_i^2 = w_j² \ell_k² + w_k² \ell_j² + 2 w_j w_k \ell_j \ell_k \cos θ_i,$$ where $θ_i$ is the corner angle at $i$.

The Law of Cosines states that: $$2 \ell_j \ell_k \cos θ_i = -\ell_i² +\ell_j² + \ell_k².$$ Substituting this above, we have an expression for $d_i²$ that only involves the side lengths and barycentric coordinates:

$$d_i^2 = w_j² \ell_k² + w_k² \ell_j² - w_j w_k (\ell_i² - \ell_j² - \ell_k²).$$

Consider the sub-triangle $jpk$, we can express the cosine of the subtended angle $φ_i$ in terms of the analogously derived lengths $d_j,d_k$ via the Law of Cosines:

$$\cos φ_i = \frac{-\ell_i²+d_j²+d_k²}{2 d_j d_k}.$$

As far as I can tell, there is no way to avoid the square root arising from this expression.