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Midterm Exam review questions
Midterm Exam review questions

... Find the measures of each of the interior angles of ∆ABC. (draws on Unit 3) a. m ∠ CAB = __________ b. m ∠ ABC = __________ c. m ∠ BCA = __________ 10. On her way to school Rhonda likes to cut across a vacant lot, from the corner of Third and Elm Streets to the corner of Fourth and Spruce. How much ...
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Geometrization conjecture

In mathematics, Thurston's geometrization conjecture states that certain three-dimensional topological spaces each have a unique geometric structure that can be associated with them. It is an analogue of the uniformization theorem for two-dimensional surfaces, which states that every simply-connected Riemann surface can be given one of three geometries (Euclidean, spherical, or hyperbolic).In three dimensions, it is not always possible to assign a single geometry to a whole topological space. Instead, the geometrization conjecture states that every closed 3-manifold can be decomposed in a canonical way into pieces that each have one of eight types of geometric structure. The conjecture was proposed by William Thurston (1982), and implies several other conjectures, such as the Poincaré conjecture and Thurston's elliptization conjecture. Thurston's hyperbolization theorem implies that Haken manifolds satisfy the geometrization conjecture. Thurston announced a proof in the 1980s and since then several complete proofs have appeared in print.Grigori Perelman sketched a proof of the full geometrization conjecture in 2003 using Ricci flow with surgery.There are now several different manuscripts (see below) with details of the proof. The Poincaré conjecture and the spherical space form conjecture are corollaries of the geometrization conjecture, although there are shorter proofs of the former that do not lead to the geometrization conjecture.
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