Commit 13caa5bb by Conor McCoid

Tetra: added steps to proof of sign of denominator lemma

parent 3db4a39a
 ... ... @@ -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$t_{x,xyz}^{ijk}is \begin{align*} ... ... @@ -382,16 +379,18 @@ If the line coincides withq_\gamma+r_\gamma=1$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 ) . ... ...
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