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Geometry And Topology

By E. H. Askwith

Initially released in 1917. This quantity from the Cornell college Library's print collections was once scanned on an APT BookScan and switched over to JPG 2000 structure through Kirtas applied sciences. All titles scanned hide to hide and pages may possibly contain marks notations and different marginalia found in the unique quantity.

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39 Chapter 3: Homogeneous Barycentric Coordinates Exercises 1. The centers of the three squares erected externally on the sides of triangle ABC form a triangle perspective with ABC. The perspector is called the (positive) Vecten point. Why is this a Kiepert perspector? Identify its Kiepert parameter, and write down its coordinates? 25 2. Let ABC be a given triangle. Construct a small semicircle with B as center and a diameter perpendicular to BC, intersecting the side BC. Animate a point T on this semicircle, and hide the semicircle.

29 This point is not in the current edition of ETC. 2 41 Chapter 3: Homogeneous Barycentric Coordinates Exercises 1. Let X ′ , Y ′ , Z ′ be respectively the pedals of X on BC, Y on CA, and Z on AB. Show that X ′ Y ′ Z ′ is a cevian triangle. 30 2. For i = 1, 2, let Xi Yi Zi be the triangle formed with given angles θi , ϕi and ψi . Show that the intersections X = X1 X2 ∩ BC, form a cevian triangle. 30 31 Y = Y1 Y2 ∩ CA, 31 Floor van Lamoen. Floor van Lamoen. X = (0 : Sψ1 − Sψ2 : Sϕ1 − Sϕ2 ). 1 Two-point form The equation of the line joining two points with coordinates (x1 : y1 : z1 ) and (x2 : y2 : z2 ) is x1 y1 z1 x2 y2 z2 = 0, x y z or (y1 z2 − y2 z1 )x + (z1 x2 − z2 x1 )y + (x1 y2 − x2 y1 )z = 0.

Examples (1) The orthocenter has coordinates 1 1 1 : : SA SB SC 15 P. Yff, An analogue of the Brocard points, Amer. Math. Monthly, 70 (1963) 495 – 501. 16 = (SBC : SCA : SAB ). L. Crelle, 1815. 34 YIU: Introduction to Triangle Geometry Note that in the last expression, the coordinate sum is SBC + SCA + SAB = S2. (2) The circumcenter, on the other hand, is the point O = (a2 SA : b2 SB : c2 SC ) = (SA (SB + SC ) : SB (SC + SA ) : SC (SA + SB )). Note that in this form, the coordinate sum is 2(SAB + SBC + SCA ) = 2S 2 .

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