SOLUTIONS FOR MTH 338 Final, June 2009 A 3 5 note card allowed. Closed book. No calculator. You may also use your list of the first 15 SMGS axioms. 1. In neutral geometry and without a compass, prove that given a line m and a point P not on m, there exists a point Q on m such that l(p, Q) is perpendicular to m. Sol n. Let R be any point on m. If the line through P and R happens to be perpendicular to m, then we may set Q = R. Otherwise, use the angle construction axiom to construct a ray from R on the other side of m from P making the same angle measure with m as does the ray from R through P. Use the ruler axiom to determine the point S so that RP = RS. Let Q be the point of intersection of P Q and m. By SAS, we have P RQ = SRQ. By the supplement axiom, P QR is a right angle. 1
2. In neutral geometry and without a compass, prove that given a line m and a point P not on m, there exists a line through P parallel to m. Sol n. Let l 1 be any line through P. If l 1 happens to be parallel to m, then are done. Otherwise, let Q be the point of intersection of l 1 and m. Use the angle construction axiom to construct another line l 2 through P a so that the alternate interior angles are equal as in the figure. The alternate interior angle theorem, then tells us that l 2 and m are parallel. 3. Can one successfully define cartesian coordinates in hyperbolic geometry? If so, give your definition and explain why it is a success. If not, why not? Sol n. Student opinions varied. I liked the No answer that said, Since there are no rectangles in hyperbolic geometry, one will definitely not be able to use a rectangular grid to define cartesian coordinates. Many Yes answers also received full marks. 4. Define the defect of a convex polygon. Sol n. The defect of a convex polygon with n sides is (n 2) 180 minus the sum of the interior angles of the polygon. 2
5. In hyperbolic geometry, prove the validity of the side-angle-defect congruence principle, SAD, for triangles, i.e., If ABC and DEF satisfy AB = DE, ABC = DEF, and defect ( ABC) = defect ( DEF ), then ABC = DEF. Sol n. If BC = EF, then the triangles are congruent by SAS. Otherwise, we may assume without loss of generality that BC < EF. Using the ruler axiom, we find a point F with E F F and BC = EF. Then ABC = DEF. Since defects add and the original two triangles had the same defect, we conclude that the defect of DF F is zero, which is a contradiction. 3
6. Assume we are in hyperbolic geometry. Suppose ABC is given and X is a point in the segment BC with B X C. Prove that defect ( ABC) = defect ( ABX) + defect ( AXC) Sol n. By the supplement axiom, we have m BXA + m AXC = 180. By angle addition, we have m XAB + m CAX = m CAB. Also we have ABX = ABC and XCA = BCA. Thus we have defect ( ABX) + defect ( AXC) = [ 180 m ABX m BXA m XAB ] + [ 180 m AXC m XCA m CAX ] = 180 + 180 [ m BXA + m AXC ] m ABX m XCA [ m XAB + m CAX ] = 180 m ABC m BCA m CAB = defect ( ABC). 4
7. Show that the quadrilateral shown in the figure is gleichwertig to a right isosceles triangle. Sol n. Construct the segment BD. By SSS, we have BCD = BDA. So BAD is also a right angle. The quadrilateral is a square! The geometry is Euclidean! Cut the square in half and reassemble as an isosceles right triangle. 5
8. Give a construction in hyperbolic geometry and without a compass, for a quadrilateral like that shown in the figure. Sol n. Pick any point P and construct any line through P. Using the angle construction axiom, construct a second line through P perpendicular to the first line. Using the ruler four times, construct the points A, B, C, and D all at the same distance from P in the four distinct directions. There are now four right triangles, ABP, BCP, CDP, DAP, that are congruent by SAS, so AB = BC = CD = DA and ABC = BCD = CDA = DAB. 6