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Dimensions | Chapter 9

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half an
hour and a half of doing mathematics means above all proving what 1 phase we have seen that the stereographic projection sense circles on the sphere not going through the pole to circles in the plane or and now we're going to prove it even though this has been known for many centuries it is i dont have been and who will be sent this proof due to find frequently on it since 1 speaks today of the renin sphere proving is much more than showing you is not enough to see in a movie that some curve looks like a called to be sure that it is indeed a so called on a mathematical proof must use reasoning to be convincing way and has to explain why it is indeed a circle the great you played during the 3rd century before christ formulated the rules of the mathematical game in his book the elements in the the proof has to rely on facts themselves have to be proved and not 1 has to start with something so that some statements have to be accepted without proof these are the axioms therefore mathematics appears as a gigantic construction foundations consist of the axioms and such that each break rests on the previous 1 In order to prove the theorem about the stereographic projection of circles we should in principle start with the axioms of course we have no time for that now we will assume that we already know the theorems of geometry which is studied say secondary school and we will prove this theorem but I have a book that I and start with something simple the intersection of the sphere and a plane but as we see that if a plane cops sphere and if it is not tangent to the sphere in the intersection is a circle but we can see it but why is it true how do we prove this so that in a good mood well let's consider an arbitrary plane colored in blue we control the perpendicular from the center see of the sphere to the plane that's called p the footer this perpendicular consider 2 points a and b on the intersection of the sphere and the plane at the and let's look at the 2 triangles CP and
you can see PB or they share a common side CPE that would both have a right
angle since the angle that p is of course
right and that since the plane is perpendicular to CP at my now enacted the hypotenuse is AC and BC have the same length because a and B here on the sphere and I hands in the same distance from the center C but remember Pythagoras's theorem since are 2 right-angle triangles had 2 sides of the same length the 3 sides must have the same length of hence we have proved that PA and PB have the same length that is that a and B are on the same circle with center pivot that in the blue plane therefore we have proved that all points which are both on the sphere and the plane belonged to some circle that does that imply that all points on the circle on the sphere and on the plane view a
priority to know we still have to prove it uh and and uh uh this a a point which is on the sphere and the plane you consider the circle in the blue plane with center p and the goes through any but we will prove that this circle is contained in the sphere a next BEB some point on the circle the uh look at the 2 triangles CPA and CPB you they share side cp and both a
right angle triangles since the angle at P has a right angle but the lengths of P a and P b r equal since a and B are on the same circle with center p again using by the aggressors theorem we conclude the hypothenuse is have the same lengths c n equals C B at this means that the point be also lies on the sphere since it is at the same distance from C as but 6 we have proved that when a plane cops
sphere the cross-section is a circle but what now let's look at the diameter APB of our circle and let's place it in the plane of the screen the blue plains appear as a straight line on the screen and the sphere appears as a circle and and physical the tangent to the circle of a and be the they intersect in a point x the 1 on the right of course the line cs is again the symmetry axis for a figure that why well because the triangle CAS and CBS equal do you know why they because they're both right angled triangles having a common hypothenuse and the side CA and CB have the same length why well because these are 2 radio and course we see if we had to go right to the end of all the arguments this movie would be the longest in the history of the cinema reviews look we've just proved that any circle drawn on the sphere can always be thought of as the contact locus between coming of revolution and a tendency to the if you like a sphere is like ice cream in a kind of uh well we mustn't forget what our aim is to you have to show the stereographic projection carry circles on 2 circles you this 1st prove what mathematicians call a lemma moon member in the here is the tangent plane to the sphere at some point they seen from side fat and can't get out of that now here's the tangent plane at some other point B and also seems so these 2 planes intersect on aligned deep the but at present we only see 1 point since this line is perpendicular to the screen or the figure that you're looking at is symmetric with respect to the bisecting line of the 2 lines that we see this three-dimensional picture is symmetric with respect to the bisecting plane of the 2 pension plan of the world and the the back of the room uh choose some plane containing segment AB it intersects the line D N A . M holistic is parallel to d of course on the symmetry of the figure with respect to the bisecting plane shows that a M and B M have the same length but her the triangle ABM is subsidies here it is that that was our lemma that well now we can prove our fear of using what we have just learned so that uh consider a circle on the sphere which does not go to the north pole we want to show that its projection is a circle and if you a th a
room room
in moved to expand if instead of projecting onto the tangent plane to the south pole we projected onto some other parallel plane the famous theorem of phase would imply that all the projections a similar what ends in order to prove all theorem we may choose the projection plane as we wish of course as long as it is parallel to the tangent plane to the south pole
of well let's place yellow circle in a can what remember the ball of ice cream in cone with vertex s well we're going to project onto the horizontal plane through S the yes the point B projects onto a point D
but look at the figure the triangles
a and B and D S B a
similar uh what well
again face there do agree
now remember on them the triangle
ABM is our subsidies hence the same is true for the triangle the x so that B S has the same length as the x so the only thing you can do it in when the moves along the yellow circle the segment BS keeps tangent to the sphere the its length is therefore constant so that seem to be S and D S have the same length as the moving segment the has also retains a concept that let's see saying that DS has a constant length means precisely that d describes the circle with center yes so the projection of our yellow circle on the horizontal plane through this is contained in a circle around to uh we have seen that my face there and this implies that the projection onto the tangent plane to the south pole is also contained in this is going on so I have that all QED quod erat the and the the you if the we have the the the the
the the and the it theory would you know there are
a lot of time in my mind I'm a bit of a I my the yeah
Ebene
Punkt
Kreisfläche
Kurve
Mathematik
Schlussregel
Element <Mathematik>
Dreieck
Computeranimation
Stereographische Projektion
Polstelle
Kugel
Theorem
Beweistheorie
Vorlesung/Konferenz
Ordnung <Mathematik>
Drei
Axiom
Phasenumwandlung
Geometrie
Unterhaltungsmathematik
Rechter Winkel
Winkel
Computeranimation
Ebene
Länge
Kreisfläche
Kugel
Punkt
Theorem
Pivot-Operation
Abstand
Dreieck
Computeranimation
Ebene
Länge
Punkt
Tangentialraum
Computeranimation
Streuquerschnitt
Stereographische Projektion
Kugel
Symmetrie
Lemma <Logik>
Rotationsfläche
Abstand
Tangente <Mathematik>
Figurierte Zahl
Gerade
Parametersystem
Kreisfläche
Durchmesser
Winkel
Dreieck
Polstelle
Differenzkern
Rechter Winkel
Mathematikerin
Projektive Ebene
Ebene
Polstelle
Theorem
Tangentialraum
Projektive Ebene
Ordnung <Mathematik>
Phasenumwandlung
Computeranimation
Ebene
Knotenmenge
Kreisfläche
Punkt
Projektive Ebene
Figurierte Zahl
Dreieck
Computeranimation
Drucksondierung
Computeranimation
Ebene
Polstelle
Länge
Kugel
Kreisfläche
Tangentialraum
Projektive Ebene
Dreieck
Computeranimation
Grothendieck-Topologie
Physikalische Theorie
Computeranimation

Metadaten

Formale Metadaten

Titel Dimensions | Chapter 9
Serientitel Dimensions
Teil 9
Anzahl der Teile 9
Autor Leys, Joe (Images and Animations)
Ghys, Étienne (Scenario and Mathematics)
Alvarez, Aurélien (Image Rendering and Post-production)
Mitwirkende Bullet, Shaun (Speaker)
Delong, Matt (Speaker)
Guaschi, John (Speaker)
Grant, John Lewis (Music)
Lizenz CC-Namensnennung - keine kommerzielle Nutzung - keine Bearbeitung 3.0 Unported:
Sie dürfen das Werk bzw. den Inhalt in unveränderter Form zu jedem legalen und nicht-kommerziellen Zweck nutzen, vervielfältigen, verbreiten und öffentlich zugänglich machen, sofern Sie den Namen des Autors/Rechteinhabers in der von ihm festgelegten Weise nennen.
DOI 10.5446/14675
Herausgeber Joe Leys, Étienne Ghys, Aurélien Alvarez
Erscheinungsjahr 2008
Sprache Englisch
Produzent École Normale Supérieure de Lyon (ENS-Lyon)

Inhaltliche Metadaten

Fachgebiet Mathematik

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