euclideangeometry (10apr20)

git-svn-id: svn://tug.org/texlive/trunk@54659 c570f23f-e606-0410-a88d-b1316a301751

4 files changed, 169 insertions, 13 deletions

diff --git a/Master/texmf-dist/doc/latex/euclideangeometry/README.txt b/Master/texmf-dist/doc/latex/euclideangeometry/README.txt
index 6402a6e9ff3..b0ea7a4bf8e 100644
--- a/Master/texmf-dist/doc/latex/euclideangeometry/README.txt
+++ b/Master/texmf-dist/doc/latex/euclideangeometry/README.txt
@@ -10,7 +10,7 @@
 %%   License information appended
 %% 
 File README.txt for package euclideangeometry
-        [2020-02-12 v.0.1.5 Extension package for curve2e]
+        [2020-03-30 v.0.1.7 Extension package for curve2e]
 
 The package bundle euclideangeometry is composed of the following files
 
diff --git a/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry-man.pdf b/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry-man.pdf
index d36d79326ed..c7ab75ca885 100644
--- a/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry-man.pdf
+++ b/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry-man.pdf
diff --git a/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry-man.tex b/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry-man.tex
index 4c09bb60256..3df15a96c43 100644
--- a/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry-man.tex
+++ b/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry-man.tex
@@ -884,7 +884,7 @@ z^2      &= m^2\eu^{\iu 2\phi}\\
 \cs{IntersecionOfLines}\parg{point1}\parg{dir1} and\parg{point2}\parg{dir2}
 \qquad to\meta{vector}
 \end{ttsyntax}
- were each line is identified with its \meta{point} and its direction
+ were each line is identified with its \meta{point} and its \emph{direction}
  \meta{dir}; the intersection coordinates go to the output \meta{vector}.
 
 \item A second command\cs{IntersectionOfSegments} does almost the same
@@ -898,7 +898,59 @@ z^2      &= m^2\eu^{\iu 2\phi}\\
  Again the intersection point coordinates go to the output \meta{vector}.
  The first segment is between points 11 and 12, and, similarly, the second
  segment is between points 21 and 22.
-
+\item Another “intersection” command is \cs{IntersectionsOfLine} to 
+ determine the two intersection of a line with a circle. The syntax is:
+ \begin{ttsyntax}
+\cs{IntersectionsOfLine}\parg{point}\parg{dir}WithCircle\parg{center}\marg{radius}to\meta{int1} and\meta{int2}
+\end{ttsyntax}
+ where \meta{point} and \meta{dir} identify a line with a given direction 
+ and passing through a given point; the circle is identified with its 
+ \meta{center} and \meta{radius}; the intersection points \meta{int1} and 
+ \meta{int2} are the coordinates of the intersection points; If the line 
+ and circle do not intersect, a warning message is issued, shown in the 
+ console and written to the \file{.log} file; the intersection points are 
+ assigned the default values \texttt{0,0}, which evidently produce strange 
+ results in the output document, so as to remind the user to give a look 
+ to the \file{.log} file and to review his/her data.
+\item It is difficult to numerically determine the coordinates of the 
+ intersection points of two circles; it becomes easier if one of the 
+ intersections is known; To this end, a macro to draw a circle with a 
+ given center and passing through a given point is handy:
+\begin{ttsyntax}
+ \cs{CircleThrough}\meta{point}WithCenter\marg{center}
+\end{ttsyntax}
+ draws such circumference.
+\item With the above macro it becomes easy to draw two circumferences 
+ with different centers and passing through the same point; therefore 
+ it becomes determine the other intersection point by means of the 
+ following macro:
+\begin{ttsyntax}
+\cs{Segment}\parg{endpoint1}\parg{endpoint2}SymmetricPointOF\meta{p1} 
+ to\meta{p2}
+\end{ttsyntax}
+ The computation is simple, because the second intersection is the 
+ symmetrical point \meta{p2} of \meta{p1} with respect to the segment 
+ that joins the centers of the given circles intersecting one another 
+ in \meta{p1}.
+\item It would be interesting to solve the same problem with help of 
+ the following macro relating to right triangles identified with their 
+ hypothenuse and one of its legs; the other leg is found by means of this 
+ macro:
+\begin{ttsyntax}
+\cs{LegFromHypothenuse}\meta{length1} AndOtherLeg\meta{length2} 
+ to\marg{length3}
+\end{ttsyntax}
+ since  the intersection points of two circles define their common chord;
+  this chord and the two circle centers define two isosceles triangles 
+  with the same base; therefore the segment joining the circle radii, 
+  coincides whit the chord axis and divides each isosceles triangle in 
+  two right triangles, where the hypotenuse is one of two radii and the 
+  first leg is the distance from the chord middle point, intersection of 
+  the chord and the segment joining the circle centers; at this point the 
+  distance of the second point intersection from the chord middle point 
+  and the coordinates of the second intersection may be easily computed; 
+  of course this is a much clumsier way to determine the second 
+  intersection, but it is useful to solve this right triangle  easy problem.
 \item Command \cs{ThreePointCircle} draws a circle that goes through three
  given points; the syntax is the following:
 \begin{ttsyntax}
@@ -908,7 +960,6 @@ z^2      &= m^2\eu^{\iu 2\phi}\\
  the coordinates of the center of the circle, that might be useful even
  if the circle is not drawn; the optional asterisk, if present, does not
  draw the circle, but the center is available.
-
 \item Alternatively
 \begin{ttsyntax}\setfontsize{10.5}
  \cs{ThreePointCircleCenter}\parg{point1}\parg{point2}\parg{point3}to\meta{vector}
@@ -931,7 +982,7 @@ z^2      &= m^2\eu^{\iu 2\phi}\\
 \end{ttsyntax}
  Apparently these two commands do the same, but, no, they behave
  differently: in the former command the \meta{Radius} is a vector the
- modulus of which si computed and used as the radius; in the latter
+ modulus of which is computed and used as the radius; in the latter
  command the \meta{radius} is a scalar and (its magnitude) is directly
  used.
 
@@ -1497,6 +1548,64 @@ y(t) &= \dfrac{13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)}{16}
 \end{figure*}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\subsection{ Intersections involving circles}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+Determining the intersection of two circles is difficult; algebraically it requires the solution of second degree equations whose coefficients are sort of complicated expressions of the centers and radii; furthermore such equations might not have real roots. The problems are much simpler even geometrically when one circle is intersected by a line, and when two circles share a common point. In the first case there are two intersections, possibly coincident, if the centers distance is shorter than the sum of their radii; in the second case its easy to determine the chord common to both circles and the problem of finding the second intersection becomes that of finding the second end point of the chord.
+
+Figure~\ref{fig:circle-and-line} shows the simple geometrical construction that leads to the determination of the intersections; one of the lines is tangent to the circle and the intersection points $P_3$ and $P_4$ coincide.
+
+\begin{figure}
+\begin{Esempio}[\setfontsize{9}]
+\unitlength=0.007\linewidth
+\begin{picture}(100,100)
+\AutoGrid
+\IntersectionsOfLine(100,50)(0,20)%
+  WithCircle(40,40){30}to\Puno and\Pdue
+\Pbox(\Puno)[tl]{P_1}[2]
+\Pbox(\Pdue)[t]{P_2}[2]
+\Dotline(\C)(\Pt){2}[1.5]
+\Dotline(\C)(\Pq){2}[1.5]
+\Pbox(\Int)[t]{M}[2]
+\Dotline(\C)(\Int){2}[1.5]
+%
+\IntersectionsOfLine(0,70)(100,70)%
+  WithCircle(40,40){30}to\Ptre and\Pquat
+\Pbox(\Ptre)[bl]{P_3}[2]
+\Pbox(\Pquat)[br]{P_4}[2]
+\IntersectionsOfLine(0,40)(100,100)%
+  WithCircle(40,40){30}to\Pcin and\Psei
+\Pbox(\Pcin)[br]{P_5}[2]
+\Pbox(\Psei)[t]{P_6}[2]
+\end{picture}
+\end{Esempio}
+\caption{Intersection of a circle with several lines}\label{fig:circle-and-line}
+\end{figure}
+
+Figure~\ref{fig:two-intersecting-circles} shows the geometrical construction to determine the second intersection point $P_2$ of two circles that already have a first common point $P_1$. The common chord and the segment joining the centers are not shown, but the code, although with “strange” point names, shows all the steps necessary to find the second intersection point
+
+
+\begin{figure}\centering
+\begin{Esempio}[\setfontsize{9}]
+\unitlength0.007\linewidth
+\begin{picture}(100,100)
+\AutoGrid
+\edef\PCCuno{30,40}
+\edef\PCCdue{70,60}
+\edef\PCCzero{40,55}
+\Pbox(\PCCuno)[t]{C_1}[2]
+\Pbox(\PCCdue)[t]{C_2}[2]
+\Pbox(\PCCzero)[l]{P_1}[2.5]
+\CircleThrough\PCCzero WithCenter\PCCuno
+\CircleThrough\PCCzero WithCenter\PCCdue
+\Segment(\PCCuno)(\PCCdue)%
+  SymmetricPointOf\PCCzero to\PCCquat
+\Pbox(\PCCquat)[l]{P_2}[2.5]
+\end{picture}
+\end{Esempio}
+\caption{Second intersection point $P_2$ of two circles sharing a first intersection point $P_1$}\label{fig:two-intersecting-circles}
+\end{figure}
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \subsection{Triangles and their special lines}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
@@ -1704,6 +1813,51 @@ y(t) &= \dfrac{13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)}{16}
 \end{figure}
 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+\subsection{The tangent to an ellipse}\label{ssec:diretriz}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+Ellipses have mani interesting properties. One that I was not able to find anywhere in the documentation I examined (of course not the totality  of available on ellipses, except an on-line document written in Portuguese by Sergio Alvez, \url{https://docplayer.com.br/345411-Elipses-inscritas-num-triangulo.html}, is the \emph{director circumference}, literal translation of the Portuguese definition \emph{circunfência diretriz}.
+
+Consider an ellipse with its foci $F$ en $F'$, and a generic point $P$ on its contour.
+Trace a segment from $F\,P$  and another segment for $P\,F'$; these segments measure the distanced from point $P$ to each focus: their sum is the length of the main ellipse axis $2a$, being $a$ the semi axis. Now lengthen the segment $P\,F'$ to point $S$ by the length of $F\,P$; the length of $S\,F'$ is therefore equal to $2a$; now trace the circumference with center in $F'$ and radius $2a$; this is the \emph{director circumference}, that is labelled with~$\Gamma$.
+
+By construction, then, the circle $\gamma$ centred in $F$ and radius equal to $F\,P$ is tangent to $\Gamma$ in $S$ and passes through $F$. This allows to say that:
+\begin{itemize}[noitemsep]
+\item the ellipse is the locus of the centres of all circles passing through  focus $F$ and internally tangent to the circle $\Gamma$ centred in the other focus $F'$ and with radius $2a$;
+\item the axis of segment $S\,F$ is tangent to the ellipse;
+\item the tangency point is the point $P$;
+\item since this axis  passes through the midpoint $M$ of segment $S\,F$ and it is perpendicular to it, the segment $M\,P$ has the same direction of the tangent to the ellipse;
+\item notice that points $S$ and $F$ are symmetrical with respect to the tangent in point $P$.
+\end{itemize}
+Such properties can be viewed and controlled in figure~\ref{fig:diretriz}.
+
+\begin{figure}[!tb]\centering
+\begin{Esempio}*[\setfontsize{8.2}](0.40)
+\unitlength=0.005\linewidth
+\begin{picture}(170,160)(-60,-80)
+\GraphGrid(-60,-80)(170,160)
+\VECTOR(0,-80)(0,80)\Pbox(0,80)[r]{y}[0]
+\VECTOR(-60,0)(110,0)\Pbox(110,0)[t]{x}[0]
+\edef\A{40}\edef\B{30}\Xellisse{\A}{\B}[\thicklines]
+\edef\X{\fpeval{\A*cosd(120)}}\edef\Y{\fpeval{\B*sind(120)}}
+\edef\P{\X,\Y}\Pbox(\P)[b]{P}[3]
+\edef\C{\fpeval{sqrt(\A**2-\B**2)}}% Inter focal semi distance
+\CopyVect-\C,0 to\F \CopyVect\C,0 to\Fp\Pbox(\Fp)[t]{F'}[3]
+\Pbox(\F)[t]{F}[3]\Pbox(0,0)[tr]{O}[3]
+\edef\Raggio{\fpeval{2*\A}}\Circlewithcenter\Fp radius\Raggio
+\SegmentLength(\P)(\F)to\raggio\Circlewithcenter\P radius\raggio
+\SegmentArg(\Fp)(\P)to\Arg\AddVect\Fp and\Arg:\Raggio to\S
+\segment(\Fp)(\S)\Pbox(\S)[br]{S}[3]
+\segment(\F)(\S)\SegmentCenter(\F)(\S)to\M
+\Pbox(\M)[r]{M}[3]\SegmentArg(\F)(\S)to\Arg
+\edef\Arg{\fpeval{\Arg-90}}\AddVect\M and \Arg:50 to\D\segment(\M)(\D)
+\end{picture}
+\end{Esempio}
+\caption{The director circumference}\label{fig:diretriz}
+\end{figure}
+
+Of course the geometrical construction of figure~\ref{fig:diretriz} can be used also in reverse order; for example it may be given a line to play the role of the tangent, a point on  this line to play the role of tangency, and a point not belonging to the line to play the role of a focus, then it is possible to find the other focus laying on a horizontal line passing through the given focus. It suffices to find the symmetrical point of the first focus with respect with the given line, and to draw a line passing through this symmetrical point and the point of tangency that intersects the  horizontal line through the first focus, concluding that this is the second focus and that the ellipse major axis length is that of the segment joining this second focus with the above mentioned symmetrical point.
+
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \subsection{A triangle internally tangent ellipse given one of its foci}
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
@@ -1715,6 +1869,8 @@ y(t) &= \dfrac{13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)}{16}
  one point of tangency, because the focal distance and the sum of
  distances of that tangency point from the foci, is sufficient to
  determine all the parameters required to draw the ellipse.
+ 
+ As seen in the previous subsection~\ref{ssec:diretriz}, it is sufficient to find the threes symmetrical point of The given focus with respect to to the three sides, i.e. the three lines that pass through the triangle vertices; and the constructions gives simultaneously the major axis length and the three tangency points, therefore all the elements required to draw the ellipse. to draw
 
 \begin{figure}[!tb]\centering
 \begin{Esempio}[\setfontsize{7}](0.375)
@@ -1740,27 +1896,27 @@ y(t) &= \dfrac{13\cos(t)-5\cos(2t)-2\cos(3t)-\cos(4t)}{16}
 \end{figure}
 
 
- The geometrical construction is rather complicated; the steps to follow
+ The geometrical construction with help of what has been explained in 
+ subsection~\ref{ssub:diretriz} is easy; the steps to follow, therefore 
  are the following:\enlargethispage*{\baselineskip}
 \begin{itemize}[noitemsep]
 
 \item draw the triangle and the given focus $\mathsf{F}$;
 
-\item Find the symmetrical points $\mathsf G_i$ of this focus with 
- respect to the sides of the triangle;
+\item Find the symmetrical points $\mathsf S_i$ of this 
+focus with respect to the sides of the triangle;
 
-\item use these three points $\mathsf G_i$ as the vertices of a triangle
- with which to draw its circumcircle, actually only its center is of
+\item use these three points $\mathsf S_i$ as the vertices of a triangle
+ with which to draw its circumcircle that turns to be the \emph{director circumference}; actually only its center is of
  interest, because it represents the second focus $\mathsf F'$; the
  inter focal distance  $2c$; is just le length of vector $F' -F$;
 
-\item join with segments each symmetrical point $\mathsf G_i$ with the
+\item join with segments each symmetrical point $\mathsf S_i$ with the
  second focus $\mathsf F'$ and find their intersections $\mathsf T_i$
  with the triangle sides; they represent the tangency points of the
  ellipse to be drawn;
 
-\item use one of these tangency points to find its distances from the
- foci; their sum gives the ellipse larger axis length $2a$;
+\item the radius of the director circumference is the ellipse major axix;
 
 \item equation~\eqref{equ:axes-foci} allows to find the second axis
  length; the segment that joins the foci has the required inclination
diff --git a/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry.pdf b/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry.pdf
index 1665a1b1f00..06df57e151e 100644
--- a/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry.pdf
+++ b/Master/texmf-dist/doc/latex/euclideangeometry/euclideangeometry.pdf