ca. 230 BC Eratosthenes

 Greek mathematician, astronomer, geographer  Chief librarian of the Library of Alexandria (ca BC)

 Circumference of the Earth (ca BC) Syene to Alexandria 7.2 ˚ = Earth’s circumference 360 ˚ Eratosthenes’ estimate: 24,466 miles Accepted value: 24,860 miles

 Also known for  Mapping of the world according to longitude and latitude  Divided the earth into climatic zones  Prime sieve  Poem “Hermes” (ca BC)

ca. 230 BC Eratosthenes ca. 225 BC Archimedes ca. 210 BC Apollonius Han dynasty ca. 202 BCca. 221 BC Qin dynasty Great Wall of China

 “The Great Geometer”  Conics (ca BC)

ca. 230 BC Eratosthenes ca. 225 BC Archimedes ca. 150 BCca. 210 BC Apollonius Han dynasty ca. 202 BCca. 221 BC Qin dynasty Great Wall of China Hipparchus

 First person documented to use trigonometry  Chord table  Catalogue of over 850 fixed stars (ca BC)

ca. 230 BC Eratosthenes ca. 225 BC Archimedes ca. 150 BCca. 210 BC Apollonius Posidonius ca. 1 AD Liu Hsin Han dynasty ca. 202 BCca. 221 BC Qin dynasty Great Wall of China ca. 146 BC Roman Aqueducts ca. 30 BC Roman’s take Egypt Caesar assassinated ca. 44 BC Trade along Silk Road ca. 110 BCca. 79 AD Colosseum Heron ca. 75 ADca. 100 BC Hipparchus Romans destroy Carthage

 Also known as Hero  Mathematician, physicist and engineer  Taught at Museum of Alexandria (ca. 75 AD ?)

 Mechanics › Mechanical machines, methods of lifting  Dioptra › Surveying, instruments for surveying  Pneumatica › Describes various types of machines and devices  Metrica › Most important geometric work, included methods of measurement

 Automatic opening of temple doors › Temple Doors opened by fire on an altar.  Earliest known slot machine › Sacrificial Vessel which flows only when money is introduced.

 “Wind ball” in Greek  Earliest recorded steam turbine › Regarded as a toy › Principle similar to jets

 Areas of triangles, polygons, surfaces of pyramids, spheres, cylinders  Volumes of spheres, prisms, pyramids  Divisions of areas and volumes in parts

 Heron’s method for the square root of a non square integer › If, is approximated by › Successive approximation gives better results  ie. If is the first approximation for is a better approximation, but is even better and so on.

 Why?  Uses SSS congruence  No intuitive appeal  Formula: ? a b c where

1. The bisectors of the angles of a triangle meet at a point that is the center of the triangle’s inscribed circle.

2. In a right-angled triangle, if a perpendicular is drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another.

1. The bisectors of the angles of a triangle meet at a point that is the center of the triangle’s inscribed circle. 2. In a right-angled triangle, if a perpendicular is drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another. 3. In a right triangle, the midpoint of the hypotenuse is equidistant from the three vertices. B D M CA

1. The bisectors of the angles of a triangle meet at a point that is the center of the triangle’s inscribed circle. 2. In a right-angled triangle, if a perpendicular is drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another. 3. In a right triangle, the midpoint of the hypotenuse is equidistant from the three vertices. 4. If AHBO is a quadrilateral with diagonals AB and OH and if < HAB and <HOB are right angles, then a circle can be drawn passing through the verticies A, O, B, and H. A B H O

1. The bisectors of the angles of a triangle meet at a point that is the center of the triangle’s inscribed circle. 2. In a right-angled triangle, if a perpendicular is drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle and to one another. 3. In a right triangle, the midpoint of the hypotenuse is equidistant from the three vertices. 4. If AHBO is a quadrilateral with diagonals AB and OH and if < HAB and <HOB are right angles, then a circle can be drawn passing through the vertices A, O, B, and H. 5. The opposite angles of a cyclic quadrilateral sum to two right angles.

 For a triangle having sides of length a, b, and c and area K, we have where is the triangle’s semi- perimeter.  PROOF: ABC is an arbitrary triangle configured so that side AB is at least as long as the other two

s is the semiperimeter

 ∆ OCE = ∆ OCE (sas)  ∆ OBE = ∆ OBD (sas)  ∆ OAF = ∆ OBD (sas)  Then extend BA such that BG = s  From these triangle congruence we have  s – c = CE = CF = AG  s – b = BD = BE  s – a = AD = AF

 From part a, we have that the area of ∆ ABC is r.s. Need:  rs = √s(s - a)(s - b )(s - c)  r²s² = s(s – a)(s – b)(s – c)  r²s = (s – a)(s – b)(s – c)  r²/ (s – b) = (s – a)(s – c)/ s (1)  From ∆ KOB, we have that OD² = DK.DB, so r² = DK(s – b)  so r²/ (s – b) = DK. (2)

 Equivalently, need: (s – a)(s – c) = DK.s if so AD.AG = DK.BG (3)  if so AD/ DK = BG/ AG (4)  if so AD/DK – 1 = BG/AG – 1  then AK/ DK = AB/AG (5)  Now it boils down to prove that  (i) ∆ HAK ~ ∆ ODK  (ii) ∆ OCE ~ ∆ AHB  (iii) angle ABH = ½ (angle ACB)

 Indeed:  (iii) angle ABH = ½ (angle ACB) by proposition of two opposite angles in a cyclic quadrilateral  (ii) ∆ OCE ~ ∆ AHB (a.a.a)  (i) ∆ HAK ~ ∆ ODK (a.a.a)  So: CE/AB = OE/ AH  CE = AG, OE = OD  Hence AG/ AB = OD/ AH.

 Now if we shuffle the steps we just went through … we realized that Heron’s proof utilizes many things about geometry, especially cyclic quadrilateral, triangle and circles, triangle congruence and similarity.  But there are more straightforward derivations.

 Consider the general triangle.  By Pythagorean theorem, b² = h² + u², c² = h² + v² so u² - v² = b² - c²  Dividing both sides by a = u + v …  Adding u + v = a to both sides and solving for u gives u = ( a^2 + b^2 - c^2 u)/ 2a  Now just take h = √(b² - u²) …

 What happens if we factor things inside the square root? Brahmagupta (620 AD) generalized the case beautifully by adding a 4 th side:  What happens if we factor out the term ab?

 This equation is the building block for the third proof:  Which is …?

Greece China Ancient Babylonia Arabia Egypt Rome India

Mathematics World History Greek Trigonometrist and geometer - first to recognize that curves were analogues of straight lines Eruption of Mt. Vesuvius destroyed cities of Pompeii and Herculaneum Nine Chapters on the Mathematical Art (Jiu Zhang Suanshu) - arithmetic and elementary algebra Theon of Smyrna- number theory and mathematic in music Claudius Ptolemaeus- famous theorem: Zhoubi Suanjing – created a visual proof for the Pythagorean Theorem Diophantus – father of algebra Sunzi Suanjing – 473- important book of problems: Ex. A woman aged 29 is 9 months pregnant. What sex is her baby? Pappus – developed theorem on volume of a solid of revolution Gives freedom of Religion in the Roman empire as the Emperor Constantine I converts to Christianity Hypatia - the first notable woman mathematician First of the Gothic Wars signaling the collapse of the Roman Empire Political division into the Western and Eastern Roman Empires as Christianity becomes the official religion of Rome Aryabhata I - solved basic algebra equations Ex. by = ac + c and by = ax – c where a,b,c are all integers Brahmagupta – One of the first to use negative numbers, described how to sum a series, created the rules for zero Tang Dynasty – period of high scholarship 15 Jul Muslim calendar is invented Wang Xiaotong – solved the cubic equation Xiahou Yan used zero as a placeholder Ja’far Muhammad ibn Musa al- Khwarizmi - algebra and algorithms Al-Battani - bsin(A) = asin(90 o -A) Abu Kamil Shuja – link between Arab and European math Ex. x 5 = x 2 x 2 x and x 6 =x 3 x 3 A B C D