MATH 3181 - 001 FALL 1996

Homework Assignment 4
Due Monday, October 21, 1996

  1. Suppose that lines and have a common perpendicular MM. Let A and B be points on such that M is not the midpoint of segment AB. Show that A and B are not equidistant from .
  2. We have seen in hyperbolic geometry the defect of any triangle is positive. In Euclidean geometry all triangles have the same defect, namely zero. In hyperbolic geometry could all triangles have the same defect? Assume that they do and use the additivity of the defect to derive a contradiction. Is there an upper bound for the defect of a triangle?
  3. Let ABC be any triangle, and let L, M, and N be the midpoints of BC, AB, and AC, respectively. Prove that AMN is not similar to ABC.
  4. This problem has five parts. In the first part we will construct a Saccheri quadrilateral associated to any triangle; we will then apply this construction.
    1. Given triangle ABC, let I, J, and K be the midpoints of BC, CA, and AB, respectively. Drop perpendiculars AD, BE, and CF from the vertices to . Prove that , and hence, that EDAB is a Saccheri quadrilateral. Note: It has the same area as ABC, but you do not have to prove this.
    2. Prove that the perpendicular bisector of AB, is also perpendicular to , and, hence, that is hyperparallel to .
    3. Recall that we denote the length of a segment AB by |AB|. Prove that . Deduce that in hyperbolic geometry .
    4. Suppose now that is a right angle. Prove that the Pythagorean theorem does not hold in hyperbolic geometry. (Hint: If the theorem were valid for right triangles BCA and ICJ, then |IJ| = ½|AB| could be proved, contradicting part (c) above.
    5. Suppose instead that AC BC. Prove that K, F, and C are collinear, but F is not the midpoint of CK.


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Last updated 10/18/96 by
David Royster droyster@math.uncc.edu