Tetra: added steps to proof of sign of denominator lemma

Research/Edge Intersection/Tetrahedra/mccoid2020tetrahedra.tex

@@ -183,7 +183,7 @@ This allows us to consider the intersection of these shapes with the face of $Y$
 Suppose $\spi{i} \neq \spi{j}$ for some $\gamma$.
 Then there is an intersection between the edge of $X$ lying between the $i$--th and $j$--th vertices and the plane $P_\gamma$.
 This intersection lies in the plane $P_\gamma$ and so its value of $p_\gamma$ is zero.
-There remains two coordinates needed to ascertain its position in $\bbr^3$.
+There remain two coordinates needed to ascertain its position in $\bbr^3$.
 
 We parametrize the plane $P_\gamma$ with the coordinates $q_\gamma$ and $r_\gamma$.
 These are chosen such that the face of $Y$ lies between the lines $q_\gamma=0$, $r_\gamma=0$ and $q_\gamma + r_\gamma=1$.
@@ -259,10 +259,7 @@ For example, if $\sign(q_\gamma^{ij}) \neq \sign(q_\gamma^{ik})$ then the line $
 
 \subsection{Intersections between faces of $X$ and edges of $Y$}
 
-Intersections between edges of $Y$ and faces of $X$ may be treated as a 2D intersection problem between $G$, the intersection of $X$ with a given plane of $Y$, and the face of $Y$ that lies in that plane.
-This may be done in the same manner as (nb: self-cite).
-
-Take, as an example, the intersection between the $(ijk)$--th plane of $X$ with the line $q_\gamma=0$ for some $\gamma$.
+Continuing the example from above, suppose there is an intersection between the $(ijk)$--th plane of $X$ with the line $q_\gamma=0$ for some $\gamma$.
 Suppose there is an edge of $G$ between its $(ij)$--th and $(ik)$--th vertices.
 Then the intersection along $q_\gamma=0$ is
 \begin{equation*}
@@ -339,7 +336,7 @@ Cramer's rule (nb: cite?) gives the solution as
 \frac{c}{d} = \frac{ \begin{vmatrix} x_i & y_i & 1 \\ x_j & y_j & 1 \\ x_k & y_k & 1 \end{vmatrix} }{\detijk}.
 \end{equation*}
 The values of $t_{y,z}^{ijk}$, $t_{x,z}^{ijk}$ and $t_{x,y}^{ijk}$ are then the inverses of these fractions.
-The values of $1-t_{y,z}^{ijk}$, $1-t_{x,z}^{ijk}$ and $1-t_{x,y}^{ijk}$ are trivial to simplify using known properties of the determinant.
+The values of $1-t_{y,z}^{ijk}$, $1-t_{x,z}^{ijk}$ and $1-t_{x,y}^{ijk}$ can be simplified using known properties of the determinant.
 
 The value of $t_{x,xyz}^{ijk}$ is
 \begin{align*}
@@ -382,16 +379,18 @@ If the line coincides with $q_\gamma+r_\gamma=1$ then
 
 \begin{proof}
 The intersection along $q_\gamma=0$ has already been given and involves division by $q_\gamma^{ij} - q_\gamma^{ik}$.
-Since this intersection exists only if $\sign(q_\gamma^{ij}) \neq \sign(q_\gamma^{ik})$ this denominator has the same sign as $q_\gamma^{ij}$, which itself has the same sign as
-\begin{equation*}
-\begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix} \pxi{i} .
-\end{equation*}
 The value of this denominator is
 \begin{align*}
 q_\gamma^{ij} - q_\gamma^{ik} = & \frac{\begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix}}{\pxi{i} - \pxi{j}} - \frac{\begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{k} & \qxi{k}\end{vmatrix}}{\pxi{i} - \pxi{k}} \\
-\vdots \\
+% = & \frac{(\pxi{i} - \pxi{k}) \begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix} - (\pxi{i} - \pxi{j}) \begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{k} & \qxi{k} \end{vmatrix} }{ (\pxi{i} - \pxi{j})(\pxi{i} - \pxi{k}) } \\
+= & \frac{\pxi{i} \begin{vmatrix} 0 & \pxi{i} & \qxi{i} \\ 1 & \pxi{j} & \qxi{j} \\ 1 & \pxi{k} & \qxi{k} \end{vmatrix} - \begin{vmatrix} 0 & \pxi{i} & \qxi{i} \\ \pxi{j} & \pxi{j} & \qxi{j} \\ \pxi{k} & \pxi{k} & \qxi{k} \end{vmatrix} }{ (\pxi{i} - \pxi{j})(\pxi{i} - \pxi{k}) } \\
+= & \frac{\pxi{i} \begin{vmatrix} 0 & \pxi{i} & \qxi{i} \\ 1 & \pxi{j} & \qxi{j} \\ 1 & \pxi{k} & \qxi{k} \end{vmatrix} - \begin{vmatrix} -\pxi{i} & \pxi{i} & \qxi{i} \\ 0 & \pxi{j} & \qxi{j} \\ 0 & \pxi{k} & \qxi{k} \end{vmatrix} }{ (\pxi{i} - \pxi{j})(\pxi{i} - \pxi{k}) } \\
 = & \frac{\pxi{i} \begin{vmatrix} 1 & \pxi{i} & \qxi{i} \\ 1 & \pxi{j} & \qxi{j} \\ 1 & \pxi{k} & \qxi{k} \end{vmatrix}}{\left ( \pxi{i} - \pxi{j} \right ) \left ( \pxi{i} - \pxi{k} \right )} .
 \end{align*}
+Since this intersection exists only if $\sign(q_\gamma^{ij}) \neq \sign(q_\gamma^{ik})$ this denominator has the same sign as $q_\gamma^{ij}$, which has the sign of
+\begin{equation*}
+\begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix} \pxi{i} .
+\end{equation*}
 By comparing these signs it is clear that
 \begin{equation*}
 \sign \left ( \begin{vmatrix} 1 & \pxi{i} & \qxi{i} \\ 1 & \pxi{j} & \qxi{j} \\ 1 & \pxi{k} & \qxi{k} \end{vmatrix} \right ) = \sign \left ( \begin{vmatrix} \pxi{i} & \qxi{i} \\ \pxi{j} & \qxi{j} \end{vmatrix} \right ) .