\section{Intersections} It is possible to determine the coordinates of the points of intersection between two straight lines, a straight line and a circle, and two circles. The associated commands have no optional arguments and the user must determine the existence of the intersection points himself. \subsection{Intersection of two straight lines} \begin{NewMacroBox}{tkzInterLL}{\parg{$A,B$}\parg{$C,D$}}% Defines the intersection point \tkzname{tkzPointResult} of the two lines $(AB)$ and $(CD)$. The known points are given in pairs (two per line) in brackets, and the resulting point can be retrieved with the macro \tkzcname{tkzDefPoint}. \end{NewMacroBox} \subsubsection{Example of intersection between two straight lines} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[rotate=-45,scale=.75] \tkzDefPoint(2,1){A} \tkzDefPoint(6,5){B} \tkzDefPoint(3,6){C} \tkzDefPoint(5,2){D} \tkzDrawLines(A,B C,D) \tkzInterLL(A,B)(C,D) \tkzGetPoint{I} \tkzDrawPoints[color=blue](A,B,C,D) \tkzDrawPoint[color=red](I) \end{tikzpicture} \end{tkzexample} \subsection{Intersection of a straight line and a circle} As before, the line is defined by a couple of points. The circle is also defined by a couple: \begin{itemize} \item $(O,C)$ which is a pair of points, the first is the centre and the second is any point on the circle. \item $(O,r)$ The $r$ measure is the radius measure. The unit can be the \emph{cm} or \emph{pt}. \end{itemize} \begin{NewMacroBox}{tkzInterLC}{\oarg{options}\parg{$A,B$}\parg{$O,C$} or \parg{$O,r$} or \parg{$O,C,D$}}% So the arguments are two couples. \medskip \begin{tabular}{lll}% \toprule options & default & definition \\ \midrule \TOline{N} {N} { (O,C) determines the circle} \TOline{R} {N} { (O, 1 cm) or (O, 120 pt)} \TOline{with nodes}{N} { (O,C,D) CD is a radius} \bottomrule \end{tabular} \medskip The macro defines the intersection points $I$ and $J$ of the line $(AB)$ and the center circle $O$ with radius $r$ if they exist; otherwise, an error will be reported in the |.log| file. \end{NewMacroBox} \subsubsection{Simple example of a line-circle intersection} In the following example, the drawing of the circle uses two points and the intersection of the straight line and the circle uses two pairs of points: \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.75] \tkzInit[xmax=5,ymax=4] \tkzDefPoint(1,1){O} \tkzDefPoint(0,4){A} \tkzDefPoint(5,0){B} \tkzDefPoint(3,3){C} \tkzInterLC(A,B)(O,C) \tkzGetPoints{D}{E} \tkzDrawCircle(O,C) \tkzDrawPoints[color=blue](O,A,B,C) \tkzDrawPoints[color=red](D,E) \tkzDrawLine(A,B) \tkzLabelPoints[above right](O,A,B,C,D,E) \end{tikzpicture} \end{tkzexample} \subsubsection{More complex example of a line-circle intersection} Figure from \url{http://gogeometry.com/problem/p190_tangent_circle} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.75] \tkzDefPoint(0,0){A} \tkzDefPoint(8,0){B} \tkzDefMidPoint(A,B) \tkzGetPoint{O} \tkzDrawCircle(O,B) \tkzDefMidPoint(O,B) \tkzGetPoint{O'} \tkzDrawCircle(O',B) \tkzDefTangent[from=A](O',B) \tkzGetSecondPoint{E} \tkzInterLC(A,E)(O,B) \tkzGetSecondPoint{D} \tkzDefPointBy[projection=onto A--B](D) \tkzGetPoint{F} \tkzMarkRightAngle(D,F,B) \tkzDrawSegments(A,D A,B D,F) \tkzDrawSegments[color=red,line width=1pt, opacity=.4](A,O F,B) \tkzDrawPoints(A,B,O,O',E,D) \tkzLabelPoints(A,B,O,O',E,D) \end{tikzpicture} \end{tkzexample} \subsubsection{Circle defined by a center and a measure, and special cases} Let's look at some special cases like straight lines tangent to the circle. \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.5] \tkzDefPoint(0,8){A} \tkzDefPoint(8,0){B} \tkzDefPoint(8,8){C} \tkzDefPoint(4,4){I} \tkzDefPoint(2,7){E} \tkzDefPoint(6,4){F} \tkzDrawCircle[R](I,4 cm) \tkzInterLC[R](A,C)(I,4 cm) \tkzGetPoints{I1}{I2} \tkzInterLC[R](B,C)(I,4 cm) \tkzGetPoints{J1}{J2} \tkzInterLC[R](A,B)(I,4 cm) \tkzGetPoints{K1}{K2} \tkzDrawPoints[color=red](I1,J1,K1,K2) \tkzDrawLines(A,B B,C A,C) \tkzInterLC[R](E,F)(I,4 cm) \tkzGetPoints{I2}{J2} \tkzDrawPoints[color=blue](E,F) \tkzDrawPoints[color=red](I2,J2) \tkzDrawLine(I2,J2) \end{tikzpicture} \end{tkzexample} \subsubsection{More complex example} \tkzHandBomb\ Be careful with the syntax. First of all, calculations for the points can be done during the passage of the arguments, but the syntax of \tkzname{xfp} must be respected. You can see that I use the term \tkzname{pi} because \NamePack{xfp} can work with radians. You can also work with degrees but in this case, you need to use specific commands like |sind| or |cosd|. Furthermore, when calculations require the use of parentheses, they must be inserted in a group... \TEX \{ \dots \}. \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=1.25] \tkzDefPoint(0,1){J} \tkzDefPoint(0,0){O} \tkzDrawArc[R,line width=1pt,color=red](J,2.5 cm)(180,0) \foreach \i in {0,-5,-10,...,-85,-90}{ \tkzDefPoint({2.5*cosd(\i)},{1+2.5*sind(\i)}){P} \tkzDrawSegment[color=orange](J,P) \tkzInterLC[R](P,J)(O,1 cm) \tkzGetPoints{M}{N} \tkzDrawPoints[red](N) } \foreach \i in {-90,-95,...,-175,-180}{ \tkzDefPoint({2.5*cosd(\i)},{1+2.5*sind(\i)}){P} \tkzDrawSegment[color=orange](J,P) \tkzInterLC[R](P,J)(O,1 cm) \tkzGetPoints{M}{N} \tkzDrawPoints[red](M) } \end{tikzpicture} \end{tkzexample} \subsubsection{Calculation of radius example 1} With \tkzname{pgfmath} and \tkzcname{pgfmathsetmacro} The radius measurement may be the result of a calculation that is not done within the intersection macro, but before. A length can be calculated in several ways. It is possible of course, to use the module \tkzname{pgfmath} and the macro \tkzcname{pgfmathsetmacro}. In some cases, the results obtained are not precise enough, so the following calculation $0.0002 \div 0.0001$ gives $1.98$ with pgfmath while xfp will give $2$. \subsubsection{Calculation of radius example 2} With \tkzname{xfp} and \tkzcname{fpeval}: \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture} \tkzDefPoint(2,2){A} \tkzDefPoint(5,4){B} \tkzDefPoint(4,4){O} \edef\tkzLen{\fpeval{0.0002/0.0001}} \tkzDrawCircle[R](O,\tkzLen cm) \tkzInterLC[R](A,B)(O, \tkzLen cm) \tkzGetPoints{I}{J} \tkzDrawPoints[color=blue](A,B) \tkzDrawPoints[color=red](I,J) \tkzDrawLine(I,J) \end{tikzpicture} \end{tkzexample} \subsubsection{Calculation of radius example 3} With \TEX\ and \tkzcname{tkzLength}. This dimension was created with \tkzcname{newdimen}. 2 cm has been transformed into points. It is of course possible to use \TEX\ to calculate. \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture} \tkzDefPoints{2/2/A,5/4/B,4/4/0} \tkzLength=2cm \tkzDrawCircle[R](O,\tkzLength) \tkzInterLC[R](A,B)(O,\tkzLength) \tkzGetPoints{I}{J} \tkzDrawPoints[color=blue](A,B) \tkzDrawPoints[color=red](I,J) \tkzDrawLine(I,J) \end{tikzpicture} \end{tkzexample} \subsubsection{Squares in half a disc} A Sangaku look! It is a question of proving that one can inscribe in a half-disc, two squares, and to determine the length of their respective sides according to the radius. \begin{tkzexample}[latex=6cm,small] \begin{tikzpicture}[scale=.75] \tkzDefPoints{0/0/A,8/0/B,4/0/I} \tkzDefSquare(A,B) \tkzGetPoints{C}{D} \tkzInterLC(I,C)(I,B)\tkzGetPoints{E'}{E} \tkzInterLC(I,D)(I,B)\tkzGetPoints{F'}{F} \tkzDefPointsBy[projection = onto A--B](E,F){H,G} \tkzDefPointsBy[symmetry = center H](I){J} \tkzDefSquare(H,J)\tkzGetPoints{K}{L} \tkzDrawSector[fill=brown!30](I,B)(A) \tkzFillPolygon[color=red!40](H,E,F,G) \tkzFillPolygon[color=blue!40](H,J,K,L) \tkzDrawPolySeg[color=red](H,E,F,G) \tkzDrawPolySeg[color=red](J,K,L) \tkzDrawPoints(E,G,H,F,J,K,L) \end{tikzpicture} \end{tkzexample} \subsubsection{Option "with nodes"} \begin{tkzexample}[latex=8cm,small] \begin{tikzpicture}[scale=.75] \tkzDefPoints{0/0/A,4/0/B,1/1/D,2/0/E} \tkzDefTriangle[equilateral](A,B) \tkzGetPoint{C} \tkzDrawCircle(C,A) \tkzInterLC[with nodes](D,E)(C,A,B) \tkzGetPoints{F}{G} \tkzDrawPolygon(A,B,C) \tkzDrawPoints(A,...,G) \tkzDrawLine(F,G) \end{tikzpicture} \end{tkzexample} \subsection{Intersection of two circles} The most frequent case is that of two circles defined by their center and a point, but as before the option \tkzname{R} allows to use the radius measurements. \begin{NewMacroBox}{tkzInterCC}{\oarg{options}\parg{$O,A$}\parg{$O',A'$} or \parg{$O,r$}\parg{$O',r'$} or \parg{$O,A,B$} \parg{$O',C,D$}}% \begin{tabular}{lll}% options & default & definition \\ \midrule \TOline{N} {N} {$OA$ and $O'A'$ are radii, $O$ and $O'$ are the centres} \TOline{R} {N} {$r$ and $r'$ are dimensions and measure the radii} \TOline{with nodes} {N} { in (A,A,C)(C,B,F) AC and BF give the radii. } \bottomrule \end{tabular} \medskip This macro defines the intersection point(s) $I$ and $J$ of the two center circles $O$ and $O'$. If the two circles do not have a common point then the macro ends with an error that is not handled. \\ It is also possible to use directly \tkzcname{tkzInterCCN} and \tkzcname{tkzInterCCR}. \end{NewMacroBox} \subsubsection{Construction of an equilateral triangle} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[trim left=-1cm,scale=.5] \tkzDefPoint(1,1){A} \tkzDefPoint(5,1){B} \tkzInterCC(A,B)(B,A)\tkzGetPoints{C}{D} \tkzDrawPoint[color=black](C) \tkzDrawCircle[dashed](A,B) \tkzDrawCircle[dashed](B,A) \tkzCompass[color=red](A,C) \tkzCompass[color=red](B,C) \tkzDrawPolygon(A,B,C) \tkzMarkSegments[mark=s|](A,C B,C) \tkzLabelPoints[](A,B) \tkzLabelPoint[above](C){$C$} \end{tikzpicture} \end{tkzexample} \subsubsection{Example a mediator} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[scale=.5] \tkzDefPoint(0,0){A} \tkzDefPoint(2,2){B} \tkzDrawCircle[color=blue](B,A) \tkzDrawCircle[color=blue](A,B) \tkzInterCC(B,A)(A,B)\tkzGetPoints{M}{N} \tkzDrawLine(A,B) \tkzDrawPoints(M,N) \tkzDrawLine[color=red](M,N) \end{tikzpicture} \end{tkzexample} \subsubsection{An isosceles triangle.} \begin{tkzexample}[latex=7cm,small] \begin{tikzpicture}[rotate=120,scale=.75] \tkzDefPoint(1,2){A} \tkzDefPoint(4,0){B} \tkzInterCC[R](A,4cm)(B,4cm) \tkzGetPoints{C}{D} \tkzDrawCircle[R,dashed](A,4 cm) \tkzDrawCircle[R,dashed](B,4 cm) \tkzCompass[color=red](A,C) \tkzCompass[color=red](B,C) \tkzDrawPolygon(A,B,C) \tkzDrawPoints[color=blue](A,B,C) \tkzMarkSegments[mark=s|](A,C B,C) \tkzLabelPoints[](A,B) \tkzLabelPoint[above](C){$C$} \end{tikzpicture} \end{tkzexample} \subsubsection{Segment trisection} The idea here is to divide a segment with a ruler and a compass into three segments of equal length. \begin{tkzexample}[latex=9cm,small] \begin{tikzpicture}[scale=.8] \tkzDefPoint(0,0){A} \tkzDefPoint(3,2){B} \tkzInterCC(A,B)(B,A) \tkzGetPoints{C}{D} \tkzInterCC(D,B)(B,A) \tkzGetPoints{A}{E} \tkzInterCC(D,B)(A,B) \tkzGetPoints{F}{B} \tkzInterLC(E,F)(F,A) \tkzGetPoints{D}{G} \tkzInterLL(A,G)(B,E) \tkzGetPoint{O} \tkzInterLL(O,D)(A,B) \tkzGetPoint{J} \tkzInterLL(O,F)(A,B) \tkzGetPoint{I} \tkzDrawCircle(D,A) \tkzDrawCircle(A,B) \tkzDrawCircle(B,A) \tkzDrawCircle(F,A) \tkzDrawSegments[color=red](O,G O,B O,D O,F) \tkzDrawPoints(A,B,D,E,F,G,I,J) \tkzLabelPoints(A,B,D,E,F,G,I,J) \tkzDrawSegments[blue](A,B B,D A,D% A,F F,G E,G B,E) \tkzMarkSegments[mark=s|](A,I I,J J,B) \end{tikzpicture} \end{tkzexample} \subsubsection{With the option \tkzimp{with nodes}} \begin{tkzexample}[latex=6cm,small] \begin{tikzpicture}[scale=.5] \tkzDefPoints{0/0/a,0/5/B,5/0/C} \tkzDefPoint(54:5){F} \tkzDrawCircle[color=gray](A,C) \tkzInterCC[with nodes](A,A,C)(C,B,F) \tkzGetPoints{a}{e} \tkzInterCC(A,C)(a,e) \tkzGetFirstPoint{b} \tkzInterCC(A,C)(b,a) \tkzGetFirstPoint{c} \tkzInterCC(A,C)(c,b) \tkzGetFirstPoint{d} \tkzDrawPoints(a,b,c,d,e) \tkzDrawPolygon[color=red](a,b,c,d,e) \foreach \vertex/\num in {a/36,b/108,c/180, d/252,e/324}{% \tkzDrawPoint(\vertex) \tkzLabelPoint[label=\num:$\vertex$](\vertex){} \tkzDrawSegment[color=gray,style=dashed](A,\vertex) } \end{tikzpicture} \end{tkzexample} \endinput