HW 6 is due Thursday, April 16 and HW 7 is due Tuesday, April 21, both at the beginning of lecture.

HW6

Work in hyperbolic geometry.

1. Let and be two parallel lines admitting a common perpendicular: There are points and with perpendicular to both and . Suppose that are other points in with , that is, is between and . Let be the foot of the perpendicular from to , and let be the foot of the perpendicular from to .

Show that . That is, and drift apart away from their common perpendicular.

(Note that and are Lambert quadrilaterals, and therefore and . The problem is to show that .)

As an extra credit problem, show that for any number we can find (on either side of ) such that , that is, and not just drift apart but they do so unboundedly.

2. Now suppose instead that and are critical (or limiting) parallel lines, that is, they are parallel, and if and is the foot of the perpendicular from to , then on one of the two sides determined by the line , any line through that forms with a smaller angle than does, cuts at some point.

On the same side as just described, suppose that are points on with , that is, is between and . Let be the foot of the perpendicular from to , and let be the foot of the perpendicular from to .

Show that . That is, and approach each other in the appropriate direction.

As an extra credit problem, show that for any we can choose so that . That is, and are asymptotically close to one another. Do they drift away unboundedly in the other direction?

HW 7

Show that the critical function is continuous. Recall that measures the critical angle, that is, iff there are critical parallel lines and and a point such that if is the foot of the perpendicular from to , and , then and make an angle of measure in the appropriate direction.

(In lecture we verified that is strictly decreasing. This means that the only possible discontinuities of are jump discontinuities. We also verified that approaches as , and approaches as . It follows that to show that has no jump discontinuities, it suffices to verify that it takes all values between and , that is, one needs to prove that for any there is an such that .)

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