Affine Planes with Polygons David Pierce Mimar Sinan Gzel Sanatlar - PowerPoint PPT Presentation

Affine Planes with Polygons David Pierce Mimar Sinan Güzel Sanatlar Üniversitesi mat.msgsu.edu.tr/~dpierce , polytropy.com Batumi ( batumi ), September,  sakartvelos matema�i�osta �av�iris X qovel�liuri saerta�oriso �onperen ia X Annual International Conference of the Georgian Mathematical Union

b b b b L Let V ∗ = O + a · − OV + b · − − → → OL V ∗ arbitrarily on the locus of P , where P = O + x · − OV + y · − − → → x 2 ± y 2 = 1 . OL, O Apollonius’s Theorem. The affinity fixing O and V interchanging V and V ∗ fixes the locus. Modern proof. The affinity corresponds to � x � � a � � x � ± b �→ . − a y b y

Apollonius’s Proof. The locus is given by XPY = V XY ∗ E ∗ , because this is true when P is V ∗ , and XPY ∝ XP 2 ∝ XV · XW ∝ V XY ∗ E ∗ . L E ∗ V ∗ Y ∗ Y W O X M V X ∗ E Adding XY X ∗ Y ∗ yields Y ∗ PX ∗ = V Y X ∗ E ∗ , P then Y ∗ PX ∗ = EY X ∗ V ∗ .

The plane is K 2 for some field K , The foregoing happens in an affine plane, satisfying if also, assuming AB � DE & AC � DF, ) two points determine a line; ) Desargues’s Theorem: ) through a point not on a line, a single parallel passes; BC � EF, ) there is a proper triangle. if AD , BE , and CF either a) are mutually parallel or C B A A b) have a common point; F ) Pappus’s Theorem: C B BF � CE, F D • D lies on BC and if E D E • A lies on EF .

Of Desargues, case (a), “Prism,” vectors, lets us define, for non-collinear OA : − − → OD : : − − → OB : − − → − → OE directed segments, ⇐ ⇒ AB � DE. − AD = − − → − → BE ⇐ ⇒ ABED is a parallelogram; O C C B B F A E A F E D D On the plane, ratios of vectors act as a field or skew-field. case (b), “Pyramid,” for non-parallel pairs of parallel With Pappus, it is a field.

For Pappus and Desargues to be Translation. ABED and Theorems, I propose axioms: BCFE being parallelograms, ABC = DEF. Addition. The polygons A D compose an abelian group where E − ABC · · · = · · · CBA, B AAB = 0 , A ∗ = ∗ A, G A ∗ B + B † A = A ∗ B † , C F ∗ and † being strings of vertices. Bisection. ABCG being a Linearity. parallelogram, ABC � = 0 ∧ BCD = 0 CGA = ABC. ∧ C � = D ⇒ ACD � = 0 . Halving. All nonzero polygons Parallels . . . have the same order, not 2 .

Hence Euclid i ., : by Addition, A B E F ACDB = ACGB + CDG = ACE − BGE + CDG = H G BDF − BGE + CDG = ECDF ; by Bisection, ACDB = 2 ACD, ECDF = 2 FCD ; C D by Halving, • Assuming AF � CD , by ACD = FCD. Parallels we may let • If AF ∦ CD , let AH � CD . AC � BD, CE � DF ; ACD = HCD, by Translation, FCD = HCD + FCH, ACE = BDF ; so ACD � = FCD by Linearity.

B F We can now prove: • Prism, by Translation, i ., and Parallels. A D • Pappus. From AB � DE and AC � DF , we obtain BF � CE , since, by i ., BFC = BFAD = BFE. C E E B • Pyramid, special case. Assume O – BE and CF meet at O , β – AB � DE and AC � DF , N A – AB � OC and AC � OB . C Then γ AD contains O ⇐ ⇒ β = γ D ⇐ ⇒ BC � DF. F M

E B Pyramid, less special case. If O – AD , BE , and CF meet at O , A – AB � DE and AC � DF , – AB � OC , then BC � DF . D General case. Assuming AC ′ � DF ′ , C ⇒ ABC ′ ∼ DEF ′ , ABC ∼ DEF = provided CC ′ � OA ; we can remove this; putting C ′ on OC is enough. F ′ F F C C ′ O A B D E

A C Given ABC ∼ DEF , we let BG � AC, DF, HG, DF ′ � AC ′ . C ′ H By Pappus, B ) from ACHGBC ′ , HC � BC ′ ; G ) from BCDFEG , DC � EG ; D ) from HCDF ′ EG , HC � EF ′ . F Thus BC ′ � EF ′ , F ′ ABC ′ ∼ DEF ′ ; also ADC ∼ BEG . E