The Beltrami-Klein model of hyperbolic geometry is a version of plane geometry where the points are the interior of the unit circle in R2, and the straight lines are the usual straight lines of Cartesian geometry (y = mx + b). There are complicated formulas for the distance between points and the angle between lines (different from the usual notion of distance and angle in the plane), and any straight line through that, you can move arbitrarily far in either direction without ever reaching the edge of the disk (the points are the interior of the disk, and do not include the unit circle). Since we haven't discussed angles in enough detail to check this take as given that Euclid's 4th postulate is true in this context. 1. For each of Euclid's 5 postulates, explain whether they do or do not hold in the Beltrami-Klein model. 2. Given a line L and a point p not on L. every line through p makes an acute angle (or two right angles) with the unique line M perpendicular to through p (it's a result that such an Mexists and is unique in the BK model). A line is called critically parallel if it is parallel to L, and any line through p that makes a smaller angle with M is not parallel. Show that in usual Euclidean geometry, every parallel line is critically parallel. Show that in the Beltrami-Klein model, a non-critically parallel line exists for every pair (L,p) as above. 3. With L, P, M as above, there is a unique line N through p which is perpendicular to M. Is there any point p where N is not parallel to L? Is there any point where N is critically parallel to L? Is there any point where N is non-critically parallel to L? Keep in mind that L.M and N are not perpendicular under the usual angle in the plane. (Hint: all of the triangle congruence theorems (SSS, ASA, SAS, etc.) still hold in the Beltrami-Klein model.) Addendum: Since it seems like many students are struggling with the fact that I didn't explicitly write out how to find orthogonals in the BK plane, let me explain it here. Using it in the problem above is optional, but might be helpful. Every line L in the BK plane has an associated point PL called its polar there are two tangents at the two points where the line meets the unit circle, and these have a unique point of intersection (unless L is a diameter, though you can interpret pr. In that case as a point at infinity). The line M in the BK plane is perpendicular to L if it passes through the polar PL (so for a diameter, it must be parallel to the tangents and thus perpendicular to L in the usual sense). Time left Show The Beltrami-Klein model of hyperbolic geometry is a version of plane geometry where the points are the interior of the unit circle in R2 and the straight lines are the usual straight lines of Cartesian geometry (y = mx + b). There are complicated formulas for the distance between points and the angle between lines (different from the usual notion of distance and angle in the plane), and any straight line through that, you can move arbitrarily far in either direction without ever reaching the edge of the disk (the points are the interior of the disk, and do not include the unit circle). Since we haven't discussed angles in enough detail to check this, take as given that Euclid's 4th postulate is true in this context 1. For each of Euclid's 5 postulates, explain whether they do or do not hold in the Beltrami-Klein model 2. Given a line L and a point p not on L, every line through p makes an acute angle (or two right angles) with the unique line M perpendicular to L through p (it's a result that such an M exists and is unique in the BK model). A line is called critically parallel if it is parallel to L, and any line through p that makes a smaller angle with M is not parallel. Show that in usual Euclidean geometry, every parallel line is critically parallel. Show that in the Beltrami-Klein model, a non-critically parallel line exists for every pair (Lp) as above 3. With Lp, M as above, there is a unique line through p which is perpendicular to M. Is there any point p where N is not parallel to L? Is there any point where N is critically parallel to L? Is there any point where N is non-critically parallel to L? Keep in mind that L M and N are not perpendicular under the usual angle in the plane. (Hint: all of the triangle congruence theorems (SSS. ASA, SAS, etc.) still hold in the Beltrami-Klein model.) Addendum: Since it seems like many students are struggling with the fact that I didn't explicitly write out how to find orthogonals in the BK plane, let me explain it here. Using it in the problem above is optional, but might be helpful. Every line L in the BK plane has an associated point pe called its polar there are two tangents at the two points where the line meets the unit circle, and these have a unique point of intersection (unless L is a diameter, though you can interpret PL In that case as a point at infinity). The line M in the BK plane is perpendicular to Lif it passes through the polar PL (so for a diameter, it must be parallel to the tangents and thus perpendicular to L in the usual sense). Time lot