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For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. *******************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 28104, 1022]*) (*NotebookOutlinePosition[ 34955, 1200]*) (* CellTagsIndexPosition[ 34870, 1194]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell[TextData[{ "Teaching geometry to my 12 years old grandson using ", StyleBox["Geometrica", FontSlant->"Italic"], " " }], "Title", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Bruno Autin", FontFamily->"Helvetica", FontSize->24, FontWeight->"Bold", FontColor->RGBColor[0.6, 0.4, 0.2]], StyleBox["\nbautin1@club-internet.fr", FontFamily->"Helvetica", FontSize->16, FontColor->RGBColor[0.6, 0.4, 0.2]] }], "Text", CellTags->"TeachingGeometrica"], Cell[TextData[{ "This is holiday time and my grandson, Jos, is curious about ", StyleBox["Geometrica", FontSlant->"Italic"], ". I explain a few things and we agree to survey special lines in the \ triangle called ", StyleBox["perpendicular bisectors", FontSlant->"Italic"], " or, more usually, ", StyleBox["bisectors", FontSlant->"Italic"], ". First, we learn how to call ", StyleBox["Geometrica", FontSlant->"Italic"], ", then to draw a triangle and to determine step by step the bisectors of \ the triangle. We finish with the circumcircle of a triangle. The naive \ questions asked by the child shed a light on the basic concepts of both ", StyleBox["Mathematica", FontSlant->"Italic"], " and ", StyleBox["Geometrica", FontSlant->"Italic"], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[CellGroupData[{ Cell["Calling Geometrica", "Section", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(<< Geometrica`Geometrica05`\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tI trust you but this is a strange way of using a program.\n", StyleBox["I", FontWeight->"Bold"], ".\tIn ", StyleBox["Mathematica", FontSlant->"Italic"], ", there is a syntax as in any language: english, french, spanish, chinese, \ ... The sign \"<<\" means \"Call\". What you call is the program \"", StyleBox["Geometrica", FontSlant->"Italic"], "05\" located in the folder ", StyleBox["Geometrica", FontSlant->"Italic"], ". " }], "Text", CellTags->"TeachingGeometrica"] }, Open ]], Cell[CellGroupData[{ Cell["Manipulating a triangle", "Section", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ".\tWe want to draw a triangle which has no special property. The function \ to do that is ", StyleBox["Triangle[] ", FontFamily->"Courier New"], StyleBox["and we give the name ", FontFamily->"Times New Roman"], StyleBox["t", FontFamily->"Times New Roman", FontSlant->"Italic"], StyleBox[" to that triangle.", FontFamily->"Times New Roman"] }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(t = Triangle[]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tI understand \"Triangle\" but why the brackets after it?\n", StyleBox["I", FontWeight->"Bold"], ".\tIn ", StyleBox["Mathematica", FontSlant->"Italic"], ", any statement is made of an ", StyleBox["operator", FontSlant->"Italic"], " which acts on ", StyleBox["variables", FontSlant->"Italic"], " enclosed in a pair of brackets", ". Here, the operator is ", StyleBox["Triangle", FontFamily->"Courier New"], ". The variables of ", StyleBox["Triangle", FontFamily->"Courier New"], " are the three vertices. In our case, they are hidden, one says that they \ are ", StyleBox["default", FontSlant->"Italic"], " variables. You can see what they are actually in the result returned by \ ", StyleBox["Geometrica", FontSlant->"Italic"], ". They are the three points of general form ", StyleBox["CPoint[x,y]", FontFamily->"Courier New"], " where x and y are the coordinates of the point.\n", StyleBox["Jos", FontWeight->"Bold"], ". \tWhat are the coordinates of a point?\n", StyleBox["I", FontWeight->"Bold"], ".\tTo know where a point is in the plane, one defines a ", StyleBox["reference system", FontSlant->"Italic"], " made of two perpendicular axes ", StyleBox["x", FontSlant->"Italic"], " and ", StyleBox["y", FontSlant->"Italic"], " which intersect at the ", StyleBox["origin ", FontSlant->"Italic"], "O", ". The coordinates of a point are the ", StyleBox["abscissa", FontSlant->"Italic"], " and the ", StyleBox["ordinate", FontSlant->"Italic"], ". The abscissa is the distance of the point to the y-axis and the ordinate \ the distance to the x-axis.\n", StyleBox["Jos", FontWeight->"Bold"], ". \tI do not see the interest of all that complications. At school, we \ define a triangle by three points. I put the points somewhere on a page of \ notebook and I join them by lines. Nothing more.\n", StyleBox["I", FontWeight->"Bold"], ".\tOK, this is the way Greeks dealt with geometry. When one works with a \ computer, we have to define a mechanism that the computer can understand. \ That mechanism is the ", StyleBox["analytic", FontSlant->"Italic"], " geometry invented by Pierre de Fermat and Ren\[EAcute] Descartes in the \ XVII-th century. But I go too far. You will learn that in more detail later \ on. Now, I shall type a semi-colon at the end of each statement, the result \ will not appear and we shall draw all we need. We continue?\n", StyleBox["Jos", FontWeight->"Bold"], ". \tNo there is still something bizarre: you defined a triangle and I see \ ", StyleBox["Segment", FontFamily->"Courier New"], " in the result.\n", StyleBox["I", FontWeight->"Bold"], ".\tIf you give me two points and ask me to join them by a piece of \ straight line, I draw a segment. True?\n", StyleBox["Jos", FontWeight->"Bold"], ". \tYes.\n", StyleBox["I", FontWeight->"Bold"], ".\tIf you give me three, four, five points, I shall draw a triangle, a \ quadrangle, a pentagon. And I can continue with as many points as you like. \ The operation is always the same. It consist of joining points with segments \ of straight lines. The figures so obtained belong to the family of the \ polygonal lines. I had to choose a name for the family, it could have been ", StyleBox["PolygonalLine", FontFamily->"Courier New"], " but it was long and not well adapted for two points, I prefered to take \ the most elementary one, the segment.\n", StyleBox["Jos", FontWeight->"Bold"], ". \tFine. Go ahead now and draw me figures.\n", StyleBox["I", FontWeight->"Bold"], ".\tThe simplest thing to do is to draw the triangle ", StyleBox["t", FontSlant->"Italic"], " with the function ", StyleBox["Draw", FontFamily->"Courier New"], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Draw[t]; \)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tNot bad but a bit dry. A triangle, I am used to call it ABC.\n", StyleBox["I", FontWeight->"Bold"], ".\tFor that, you have to add the function ", StyleBox["Legend", FontFamily->"Courier New"], ".to the arguments of ", StyleBox["Draw", FontFamily->"Courier New"], ". The variables are then the text you want to write, say \"ABC\" and the \ object to which you apply the legend, here the triangle ", StyleBox["t", FontSlant->"Italic"], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Draw[t, Legend["\", t]]; \)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tAnd some color?\n", StyleBox["I", FontWeight->"Bold"], ".\tEasy. You add the color in ", StyleBox["Draw", FontFamily->"Courier New"], ". Do it yourself." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Draw[blue, t, Legend["\", t]]; \)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ".\tAll ", StyleBox["Mathematica", FontSlant->"Italic"], " names start with a capital letter.\n", StyleBox["Jos", FontWeight->"Bold"], ". \tBut ", StyleBox["t", FontSlant->"Italic"], " is not a capital letter.\n", StyleBox["I", FontWeight->"Bold"], ".\t", StyleBox["t", FontSlant->"Italic"], " was defined by us, it does not belong to the catalog of ", StyleBox["Mathematica", FontSlant->"Italic"], " names. Redo the drawing." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Draw[Blue, t, Legend["\", t]]; \)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tThe", " letters stay black.\n", StyleBox["I", FontWeight->"Bold"], ".\tYes, because the color affects only graphical objects and the legend is \ just text. You can also paint the triangle." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Draw[Paint[t, Blue], Legend["\", t]]; \)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tCool!" }], "Text", CellTags->"TeachingGeometrica"], Cell["", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Bisectors", "Section", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ". \tYou remember what the bisectors are?\n", StyleBox["Jos", FontWeight->"Bold"], ". \tThere are bisectors for segments and for ... angles.\n", StyleBox["I", FontWeight->"Bold"], ". \tDealing with angles is quite possible with ", StyleBox["Geometrica", FontSlant->"Italic"], " but the definition of angles is somewhat involved and that took us too \ far. Let us limit ourselves to the bisector of a segment which is the \ perpendicular to the midpoint of a segment. The segments that we use are the \ sides of the triangle.\n", StyleBox["Jos", FontWeight->"Bold"], ". \tI guess how to find them." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Sides[t]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ". \tBravo. How many sides did you find?\n", StyleBox["Jos", FontWeight->"Bold"], ". \t3.\n", StyleBox["I", FontWeight->"Bold"], ". \tFor a start, I would like you to work with one side only, say the \ first one.\n", StyleBox["Jos", FontWeight->"Bold"], ". \tHow can I take it among the three?\n", StyleBox["I", FontWeight->"Bold"], ". \tThe result you got is a list. The three sides are inside a pair of \ curly brackets. You remember these basic notations: a pair of square brackets \ for variables of a function and a pair of curly brackets for a list of \ objects. A neat thing to do is to define the sides by a list of three objects \ also. The first, second and third object of the left hand list will be the \ name of the first, second and third side of the right hand list. Let me do it \ for you." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \({s1, s2, s3} = Sides[t]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tI would like to see what they are.\n", StyleBox["I", FontWeight->"Bold"], ". \tDraw them in different colors.\n", StyleBox["Jos", FontWeight->"Bold"], ". \tI try." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Draw[Red, s1, Green, s2, Blue, s3]; \)], "Input", CellTags->"TeachingGeometrica"], Cell[CellGroupData[{ Cell["Attaching a point to a segment", "Subsection", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ". \tYou become an expert! Now, we shall learn how to place a point on a \ segment. There is a general function called Pointer to attach a point to an \ object. Here the object is the segment ", Cell[BoxData[ \(TraditionalForm\`s\_1\)]], ". The position of the point on a segment ", StyleBox["AB", FontSlant->"Italic"], " is defined by a variable called a ", StyleBox["parameter", FontSlant->"Italic"], ". If the point is in ", StyleBox["A", FontSlant->"Italic"], ", the parameter is 0. If the point is in ", StyleBox["B", FontSlant->"Italic"], ", the parameter is 1. Do you guess the value of the parameter for the \ midpoint?", "\n", StyleBox["Jos", FontWeight->"Bold"], ". \t0.5.\n", StyleBox["I", FontWeight->"Bold"], ". \tRight or ", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], ". If you use 0.5, all the results will be given with decimal points. If \ you take ", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], ", the results will be exact. As our problem is simple, it is better to \ take ", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(m = Pointer[s1, 1/2]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tLet me check." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Draw[m, Red, s1]; \)], "Input", CellTags->"TeachingGeometrica"] }, Open ]], Cell[CellGroupData[{ Cell["First construct", "Subsection", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ". \tNow, we have to find the perpendicular to the segment ", Cell[BoxData[ \(TraditionalForm\`s\_1\)]], "at ", StyleBox["m", FontSlant->"Italic"], ".\n", StyleBox["Jos", FontWeight->"Bold"], ". \tI know, one takes a compass, places the center at one end of the \ segment and draw an arc of circle around the midpoint. One does the same \ thing at the other end and we join the intersections of the two arcs.\n", StyleBox["I", FontWeight->"Bold"], ". \tI give you the tool. It is the function ", StyleBox["ECircle", FontFamily->"Courier New"], ", its variables are the center of the circle and the radius. You will ask \ me why ", StyleBox["E", FontFamily->"Courier New", FontWeight->"Bold"], StyleBox["Circle", FontFamily->"Courier New"], " and not simply ", StyleBox["Circle", FontFamily->"Courier New"], ". This is because ", StyleBox["Circle", FontFamily->"Courier New"], " is already a ", StyleBox["Mathematica", FontSlant->"Italic"], " function which denotes a graphical object whereas ", StyleBox["ECircle", FontFamily->"Courier New"], " is a geometric object that you can draw, transform, intersect, etc. The \ ", StyleBox["E", FontWeight->"Bold"], " reminds that it is a Euclidean definition, from the name of the greek \ geometer Euclid.\n", StyleBox["Jos", FontWeight->"Bold"], ". \tI think I can do it. Let me try." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \({c1, c2} = {ECircle[]}\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tBut I do not have the ends of ", Cell[BoxData[ \(TraditionalForm\`s\_1\)]], "...\n", StyleBox["I", FontWeight->"Bold"], ". \tYou will learn a great feature of ", StyleBox["Mathematica", FontSlant->"Italic"], ": the change of head of an object. It is very natural to change the \ variables of a function. The name ", StyleBox["variable", FontSlant->"Italic"], " indicates that they may take any value. It is more surprising to maintain \ the variables fixed and to change the name of the function. To get the ends \ of ", Cell[BoxData[ \(TraditionalForm\`s\_1\)]], ", it would be ideal to replace the head ", StyleBox["Segment ", FontFamily->"Courier New"], "by the head ", StyleBox["List", FontFamily->"Courier New"], ". This is actually feasible using the notation @@. Let me show you." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \({a, b} = List @@ s1\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tIt works but I do not see the word ", StyleBox["List", FontFamily->"Courier New"], ".\n", StyleBox["I", FontWeight->"Bold"], ". \tIt is a matter of notations. The result in the form of curly brackets \ is interpreted by ", StyleBox["Mathematica", FontSlant->"Italic"], " as ", StyleBox["List[", FontFamily->"Courier New"], "a,b", StyleBox["]", FontFamily->"Courier New"], ".\n", StyleBox["Jos", FontWeight->"Bold"], ". \tI believe you. Let me continue my construction now." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \({c1, c2} = {ECircle[\(a\)\(,\)]}\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tAh, but what do I take for the radius?\n", StyleBox["I", FontWeight->"Bold"], ". \tEvaluate the distance from ", StyleBox["a", FontSlant->"Italic"], " to ", StyleBox["b", FontSlant->"Italic"], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(d = Distance[a, b]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tI should be able to do it now." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[{ \({c1, c2} = {ECircle[a, 3 d/4], ECircle[b, 3 d/4]}; \), "\n", \(Draw[s1, Blue, c1, c2]; \)}], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tHow to find the intersections of the two circles?\n", StyleBox["I", FontWeight->"Bold"], ". \tWith the function ", StyleBox["Intersections", FontFamily->"Courier New"], ".\n", StyleBox["Jos", FontWeight->"Bold"], ". \tLike that?" }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \({p, q} = Intersections[c1, c2]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tand to join ", StyleBox["pq", FontSlant->"Italic"], " by a straight line?\n", StyleBox["I", FontWeight->"Bold"], ". \tUse ", StyleBox["ELine", FontFamily->"Courier New"], ".\n", StyleBox["Jos", FontWeight->"Bold"], ". \tAlways Euclid!" }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[{ \(bi = ELine[p, q]; \), "\n", \(Draw[s1, Blue, c1, c2, Red, bi]; \)}], "Input", CellTags->"TeachingGeometrica"] }, Open ]], Cell[CellGroupData[{ Cell["Second construct", "Subsection", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ". \tYou are very good.", "\n", StyleBox["Jos", FontWeight->"Bold"], ". \tIt's fun and I did not use the point ", StyleBox["m", FontSlant->"Italic"], ".\n", StyleBox["I", FontWeight->"Bold"], ". \tTrue. It would have been useful if you had drawn the bisector as the \ line perpendicular to ", Cell[BoxData[ \(TraditionalForm\`s\_1\)]], "in ", StyleBox["m", FontSlant->"Italic"], ".", " Still with ", StyleBox["ELine", FontFamily->"Courier New"], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(bi1 = ELine[s1, m]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tI found a bug in your program!\n", StyleBox["I", FontWeight->"Bold"], ". \tYou have used ", StyleBox["ELine ", FontFamily->"Courier New"], "before with two points. You reused it with a segment and a point. This \ form is not recognized by ", StyleBox["Geometrica", FontSlant->"Italic"], ". Let us look at what ", StyleBox["Geometrica", FontSlant->"Italic"], " offers to you." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(\(?ELine\)\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ". \tThe next statement should work." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[{ \(d1 = ELine[a, b]; \), "\[IndentingNewLine]", \(bi1 = ELine[m, d1]\), "\[IndentingNewLine]", \(Draw[s1, m, Red, bi1]; \)}], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ". \tI am not convinced that ", StyleBox["bi", FontSlant->"Italic"], " and ", Cell[BoxData[ \(TraditionalForm\`bi\_1\)]], "are the same.\n", StyleBox["I", FontWeight->"Bold"], ". \tYou could draw them and observe that they are superimposed but this is \ not rigorous. Better use the function ", StyleBox["IdenticalQ", FontFamily->"Courier New"], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(IdenticalQ[bi, bi1]\)], "Input", CellTags->"TeachingGeometrica"] }, Open ]], Cell[CellGroupData[{ Cell["Quick construct", "Subsection", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ".\tA program is written not only to emulate what you learn at school but \ also to work quickly. Therefore, you can get the bisector of a segment \ without performing a detailed construction.\n", StyleBox["Jos", FontWeight->"Bold"], ". \tWith ", StyleBox["Bisector", FontFamily->"Courier New"], " and a segment?\n", StyleBox["I", FontWeight->"Bold"], ".\tYes." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Bisector[s1]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ".\tand for a triangle?\n", StyleBox["I", FontWeight->"Bold"], ". \tand for any polygonal line since I told you that the family of the \ polygonal lines was called ", StyleBox["Segment", FontFamily->"Courier New"], " in ", StyleBox["Geometrica", FontSlant->"Italic"], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Bisector[t]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ".", "\tCool. Let me draw everything. I give a name to the three bisectors \ first." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[{ \(bt = Bisector[t]; \), "\[IndentingNewLine]", \(Draw[Legend["\", t], Paint[t, Red], Blue, bt]; \)}], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ".\tThe bisectors are too big.\n", StyleBox["I", FontWeight->"Bold"], ". \tYou see, a line goes to infinity. If you do not say anything, the \ parameter of the line extends from -\[Pi] to \[Pi]. \n", StyleBox["Jos", FontWeight->"Bold"], ".\tWhat is the parameter of a line?\n", StyleBox["I", FontWeight->"Bold"], ". \tIt looks like the parameter of a segment. It is a number which gives \ the position of a point on the line. To reduce the length of the line on the \ figure, we can limit the range of variations of the parameter to (-1,1) using \ ", StyleBox["DrawRange", FontFamily->"Courier New"], ". I show you." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Draw[Legend["\", t], Paint[t, Red], DrawRange[\(-1\), 1], Blue, bt]; \)], "Input", CellTags->"TeachingGeometrica"] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Circumcircle", "Section", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ".", "\tIs it true that they pass all by the same point?\n", StyleBox["I", FontWeight->"Bold"], ". \tYes. ", StyleBox["Geometrica", FontSlant->"Italic"], " can tell you that the conjecture is true. You intersect two bisectors ... \ \n", StyleBox["Jos", FontWeight->"Bold"], ".", "\tOK, I know." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[{ \({b1, b2, b3} = bt; \), "\[IndentingNewLine]", \(x = Intersections[b1, b2]\)}], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["I", FontWeight->"Bold"], ".", "\tThe result is not pretty, try to simplify it using ", StyleBox["Simplify", FontFamily->"Courier New"], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(Simplify[x]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ".\tBingo!\n", StyleBox["I", FontWeight->"Bold"], ". \tAnd now you check that x belongs to the third bisector using ", StyleBox["OfQ", FontFamily->"Courier New"], "." }], "Text", CellTags->"TeachingGeometrica"], Cell[BoxData[ \(OfQ[x, b3]\)], "Input", CellTags->"TeachingGeometrica"], Cell[TextData[{ StyleBox["Jos", FontWeight->"Bold"], ".\tOur teacher asks us for more than True or False. He wants a proof.\n", StyleBox["I", FontWeight->"Bold"], ". \t", StyleBox["Geometrica", FontSlant->"Italic"], " has an internal procedure of validation based on solution of equations. \ It does not give you the sequence of statements which makes a proof but isn't \ it better to leave you the privilege of reasoning?\n", StyleBox["Jos", FontWeight->"Bold"], ".\tYou mean that, even with ", StyleBox["Geometrica", FontSlant->"Italic"], ", I shall still have to do my homework?\n", StyleBox["I", FontWeight->"Bold"], ". \tYes. Let me show you a last thing. This point, common to the \ bisectors, is the center of the circle circumscribed to the triangle.\n", StyleBox["Jos", FontWeight->"Bold"], ".\tI do that and we stop because I am a bit tired with Euclid, parameters, \ coordinates and all that stuff. 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