New PDF release: Algebraic geometry, Oslo 1970; proceedings

By F. Oort

ISBN-10: 9001670806

ISBN-13: 9789001670801

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2◦ . We first show that triangles EAI, ABC are congruent. Indeed, we have AE = BA and EI = AG = AC. Now AEI is supplementary to EAG (they are consecutive angles in parallelogram EAGI), as is BAC (since the four angles about point A add up to 360◦ , and two of them are right angles). Hence AEI = BAC, and the triangles are congruent. These two triangles also have the same sense of rotation, and EA ⊥ AB, so AI ⊥ BC (by our Lemma). This means that point I lies on the altitude of triangle ABC 3◦ . Triangles BCD and AIB are congruent with the same sense of rotation, and their sides BD and AB are perpendicular, so CD and BI are equal and perpendicular.

SOLUTIONS FOR BOOK I Exercise 44. Let ABCD, DEF G be two squares placed side by side, so that sides DC, DE have the same direction, and sides AD, DG are extensions of each other. On AD and on the extension of DC, we take two segments AH, CK equal to DG. Show that quadrilateral HBKF is also a square. K B A C H E F D G Figure t44 Solution. It is not hard to see that the triangles ABH, CBK, EKF , GHF (Figure t44) are all congruent, and therefore HB = BK = KF = F H. Angle BKF is a right angle, since it is equal to the sum of the acute angles of right triangle BKC.

D) Keeping AB large, what happens as CD grows? What happens when both are diameters of the original circle? (e) When does the locus pass through points A and B? (Answer: Always. ) When does the locus pass through the center of the original circle? 2◦ . If points C and D lie on arc AP B, then AKB = 1 ◦ 2 (360 1 2 1 2 AQB + 12 CD= − α + β) (Figure t62c). If points C and D both lie on AQB, then AKB = AP B − 12 CD= 12 (α − β), and the same expression represents angle AKB if even one of the points C or D lies on AQB.

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Algebraic geometry, Oslo 1970; proceedings by F. Oort

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