E-320: Teaching Math with a Historical Perspective Oliver Knill, 2012Lecture 3: GeometryThe butterfly theoremDraw an arbitrary chord AB in a circle. Now draw two new arbitrary chords P Q, RSthrough the center M of AB. The line segments P R and QS now cut the chord AB inequal distance.We will look at some aspects of this theorem in class.Morley’s miracleThe following theorem was discovered in 1899 by Frank Morley at Haverford college nearPhiladelphia.If one trisects the angles of a triang le, the corresponding trisector intersections form anequilateral triangle.It is a beautiful result because it is not obvious, or even surprising. John Conway foundan elegant proof: write a′for the angle a + π/3 and a′′for a + 2π/3. Build 7 triangles withangles (0′, 0′, 0′), (a, b′, c′), (a′, b, c′), (a′, b′, c), (a, b, c′′), (a, b′′, c), (a′′, b, c) and cyclic. The triangles(a, b′, c′), (a′, b, c′), (a′, b′, c) are determined by a ssuming their shortest side length is the one fromthe equilateral t r iangle (0′, 0′, 0′). The other three are required to have the same height than thetriangle (0′, 0′, 0′). These 7 tria ngles can be put together to a large triangle with angles (a, b, c).ΑΑΑΒΒΒΓΓΓPascal’s mystic hexagramThe following result has been found in 1640 by Pascal, when he was 17. He probably got theproblem from his father who was a friend of Desargues. See Stillwell ”mathematics by its history”page 95 .Pairs of opposite sides of a hexagon inscribed in a conic section meet in three collinearpoints.Also this result is not obvious. Pascal probably proved it first for circles. Applying a lineartransformation on the picture preserves the linear incidence structure and gets it for all. We canalso been seen as a consequence o f the Pappus-Pascal theorem.Pythagoras theoremFor all right angle triangles of side length a, b, c, the quantity a2+ b2− c2is zero.As shown in class, there are rearrangment proofs.An other beautiful result is:Given a circle of radius 1 and a point P inside the circle. For any line through P whichintersects the circle at points A, B we have |P O|2− |P A||P B| = 1.This is a consequence of Pythagoras. By scaling tr anslation and rotation we can assume the circleis at the origin and that the line through the point P = (a, b) is horizontal. The intersectionpoints are then (±√1 − a2, a). Now (b −√1 − a2)(b +√1 − a2) = b2− 1 + a2.Pappus theoremPappus of Alexandria (290- 350) showed:Take three points P1, P2, P3on a first line and three points Q1, Q2, Q3on a second line.Draw all possible connections PiQjwith i 6= j. The intersection points of the lines PiQjand PjQiare on a line.P1P2P3Q1Q2Q3Thales TheoremThales of Miletus (625 BC -546 BC) got the following beautiful r esultA triangle inscribed in a fixed circle is defor med by moving one of its points on the circle,then t he angle at this point does not change.The result is relevant also because tThales is considered the first modern Mathematician.Thales theorem is a prototype of a stability result. We look a t a slightly more general case thanusual treated which is called the ”Fass kreis” theorem in Europe. In this worksheet we want tounderstand it and prove it.Lets look first at the case when o ne side of the triangle goes throug h the center. a) The tr iangleBCO is an isosceles tria ngle.b) The central angle AOB is twice the angle ACB.OCBA11ODCBAHere are the steps to see the theorem: c) What is the relation between the angles AOD and ACD?d) What is the relation between the angles DOB and DCB?e) Find a relation between the central angle AOB and the angle ACB?f) Why does the angle ACB not change if C moves on the circle?Hyppocrates TheoremThe quadrature of the Lune is a result of Hippocrates of Chios (470 BC - 400 BC) and alsocalled Hyppocrates theorem. It is the first rigorous quadrature of a curvilinear area. It states:The sum L + R of the area L of the left moon and the area R of the right moon is equalto the area T of the triangle.RLTIf A, B, C are the areas of the half circles build over t he sides of the triangle, then A + B = C. IfU is the area of the intersection of A with the upper half circle C. and let V be the area of theintersection of B with C. Let T be the area of the triangle. ThenU + V + T = C . InterpretL = A − U and R = B − V are the moon ar eas we can add them up a nd use the just shownrelation to see L + R = T .Feuerbach’s TheoremThe 3 midpoints of each side, the 3 f oots of each altitude and the three midpoints of theline segments from the vertices to the orthocenter lie on a common circle.This result is attributed to Karl Wilhelm Feuerbach (1800-1834), who found a partial resultof this in 1822. We will prove it with the computer in class. In the case of an equilateral triang lethe midpoints and the height bases are the same and we have only 6 points. The Feuerbach circleis the circle inscribed into the triangle.The centroidThe centroid of a triangle is the intersection of the lines which connect the vertices of a trianglewith the midpo ints of the opposite side. It is not at all trivial t hat these three lines intersect inone point. It is a stability pr operty of the triangle. Deforming a triangle does not change thisproperty. If A, B, C are the coordina tes of the vertices, then (A + B)/2, (A + C)/2 and ( B + C)/2are the midpoints of the sides. To verify the property just check that with P = (A + B + C)/3,the points A, P, (B + C)/2 are on a line, the points (B, P, (A + C)/2 ar e on a line and the points(C, P, (A + B)/2 are on a line. There is an easier but more advanced way to see this: checkit first for the equilateral triangle. Now, any triangle can be mapped into any other by a lineartransformation. Because linear transformations preserve lines and ratios, the intersection propertywill stay true for all triangles.To the left, we seethe situation as we would expect it without ”knowing” that the three intersection points agree.To the right, we wee the actual situation.The orthocenterThe orthocenter is the intersection of the three alt itudes of a triangle. Also here - a priory -we have three different points the intersection, for each pair of a lt itudes. Why do they meet inone point? It is no t obvious and was not proven by the Gr eeks for example. One can take theintersection of two altitudes, get a point P and form the line from P to the third point in thetriangle. The fact that this line is