# Proof: Rounded Corners on Path Segment

Given a path segment by three points $$A$$, $$B$$ and $$C$$ as well as a the radius $$r$$ we want to have at point $$B$$. Now the vectors the vectors from $$B$$ to $$A$$ and $$C$$ respectively are $$\mathbf{a}=A-B$$ and $$\mathbf{b}=C-B$$. The bisector of these two vectors is $$\mathbf{v}=\hat{\mathbf{a}}+\hat{\mathbf{b}}$$.

The mid-point of the circle $$M$$ is given by

$\begin{array}{rl} M&= B + \hat{\mathbf{v}}\frac{r}{\sin\frac{\theta}{2}}\\ &= B + \hat{\mathbf{v}}\frac{r}{\frac{\sqrt{1-\hat{\mathbf{a}}\cdot\hat{\mathbf{b}}}}{\sqrt{2}}}\\ &= B + \hat{\mathbf{v}}\frac{\sqrt{2}r}{\sqrt{1-\hat{\mathbf{a}}\cdot\hat{\mathbf{b}}}}\\ &= B + \frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{\sqrt{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|^2}}\cdot\frac{\sqrt{2}r}{\sqrt{1-\hat{\mathbf{a}}\cdot\hat{\mathbf{b}}}}\\ &= B + (\hat{\mathbf{a}}+\hat{\mathbf{b}})\frac{\sqrt{2}r}{\sqrt{(\hat{\mathbf{a}}+\hat{\mathbf{b}})\cdot(\hat{\mathbf{a}}+\hat{\mathbf{b}})\cdot(1-\hat{\mathbf{a}}\cdot\hat{\mathbf{b}})}}\\ &= B + (\hat{\mathbf{a}}+\hat{\mathbf{b}})\frac{r}{\sqrt{1-(\hat{\mathbf{a}}\cdot\hat{\mathbf{b}})^2}}\\ \end{array}$

Now the intersection points $$X$$ and $$Y$$ on $$BA$$ and $$BC$$ respectively can be found with

$X = B + \hat{\mathbf{a}}w$

$Y = B + \hat{\mathbf{b}}w$

where the weight $$w$$ is

$\begin{array}{rl} w &= \frac{r}{\tan\frac{\theta}{2}}\\ &= \frac{r}{\tan\left(\frac{1}{2}\cos^{-1}(\hat{\mathbf{a}}\cdot\hat{\mathbf{b}})\right)}\\ &= \frac{r}{\frac{\sqrt{1-\hat{\mathbf{a}}\cdot\hat{\mathbf{b}}}}{\sqrt{1+\hat{\mathbf{a}}\cdot\hat{\mathbf{b}}}}}\\ &= \frac{r}{\sqrt{\frac{2}{1+\hat{\mathbf{a}}\cdot\hat{\mathbf{b}}}-1}}\\ &= r\cdot \sqrt{\frac{2}{1-\hat{\mathbf{a}}\cdot\hat{\mathbf{b}}}-1} \end{array}$