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by: Cassidy Grimes


Cassidy Grimes

GPA 3.51

Ma Filaseta

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About this Document

Ma Filaseta
Class Notes
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This 2 page Class Notes was uploaded by Cassidy Grimes on Monday October 26, 2015. The Class Notes belongs to MATH 532 at University of South Carolina - Columbia taught by Ma Filaseta in Fall. Since its upload, it has received 12 views. For similar materials see /class/229525/math-532-university-of-south-carolina-columbia in Mathematics (M) at University of South Carolina - Columbia.




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Date Created: 10/26/15
MATH 5327361 LECTURE NoTEs 8 mquot 1 2A M 5 be mwm m c m m m My gm 5 0 mt Wm mmmc1 7 MM rmquot 3 M 5m empammmmmm mm m was man zyz o and 445445 MA 5 m Comm mn Dmnimn 1m mmch AABC 1nd AA B C m WWWmm paw X x 1h hm TA 5 md lupmwx Sm gwc 1 De nikinn No gas AABC 1nd AA B39C39 m WWW m a me n p Q 1nd R m uhMum P 13 1h mmsemmm 75 mdA39T Q 13 m m asechun mfdm 39 m R 1h m asechun of 27339 1nd A 3 g 2 A 5 Figun z 955w Thmrzm ma Warm m WWMMD pm m m m WWW m a In Cmnmzm Eclw 3 mm mmxmnx P Q 1nd a It pummm my plum 1nd m palms A 5 a Au 3 Ind v m mm 1nd M m m mm plum Plant EyThwzcm um mumbashk Inde smmu x1kac A39pkgswwqugwuw We show that k1 7 k2 Assume k1 k2 Observe that k1 7 0 since otherwise we would have X A B contradicting that A and B are distinct points Also k1 7 1 since otherwise we would have X A B contradicting that A and B are distinct points We get that 1 i k1A i 1 i k2B kZB 7 MA 7 7 Hence 1 7 k1BA klA B and the vectors BA and AB either have the same direction or the exact opposite direction This contradicts that the point P exists Hence k1 7 kg From we obtain that l 7 k k 7 l k 7k 1 2 2 B 1 kgikl kgikl kzikl By Theorem 1 with t kg 7 1 kg 7 k1 we see that the expression on the left is a point on line lt AB By Theorem 1 with t 7k1k2 7 k1 we see that the expression on the right is a point on lt line A B Therefore szklA ilikl 271 Pikg kl kgile39 Hence k2 k1P 1 WA k2 1B 1 From we use now that l 7 k2B 717 k3O kgO 7 kgB Similarly to the above we obtain that k2 7 k3 that likz ksil k3 kz O B krkz krkz krkz k37k2 and that k3 k2Q 1 k2B k3 1O 2 Using once again we obtain 1 7 k3O 717 k1A MA 7 kgO Similarly to the above we deduce that k1 7 k3 that 1 7 k3 k1 71 k1 7kg A A 0 klikg klikg klikg k1 k3 7 and that k1 k3R 1 500 k1 DA 3 Take AP BQ OR k2k17 yk3k27 and 2k1k3 in Theorem 3 Note that z 7 0 We deduce that P Q and R are collinear nishing the proof I


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