A Theorem of Hilbert. Mat 3271 Class

Transcription

1 Theorem of Hilbert Mat 3271 lass Theorem (Hilbert) ssume that there exists lines l and l parallel such that l is not asymptotic to l. Then there exists a unique common perpendicular to the given lines. Sketch of the Proof: The proof is obtained by justifying the following steps. 1. onsider points and on and drop perpendiculars and to. If either is a common perpendicular then we are done. If = then there exist a common perpendicular at the midpoint which is unique since there are no rectangles in H On the basis of part (1), We assume WOLOG that > and we construct E on such that E =. (segment duplication and the embedding relation) 3. t E construct the line EF such that F is on SS( ) as and F E = G where G. E F G Figure 1. onstruction of E = and F E = G. 1

2 2 4. We must show that EF. If H = l(ef ), then we can find K, such that K = EH. 5. If this is the case, we will show that HH = KK where HH and KK are perp to, which implies that the midpoint of HK on gives the point which produces the unique common perpendicular. We say that [ is a biangle if and and are on SS( ). We say that is the base of the biangle. In addition, we will say that the biangle is closed at if every interior ray emanating from intersects. Observe that is asymptotic to on the given side when the biangle is closed at. X Figure 2. n interior ray for the biangle [. To prove that such a point H exists we need the following lemmas.

3 3 Lemma 1 (Extension) ssume that [ is closed at. If P or P then the biangle [P is closed at. Proof: P P X Y Figure 3. iangle closed at implies the biangles are closed at P and P. 1. Extend to point P such that P 2. Show that P X 3. onstruct an angle at E by corresponding angles to P X 4. P X (orresponding ngles) 5. E P X 6. P X (Transitivity of Parallelism) ontradiction 7. hoose a point P such that P 8. onstruct an angle at E by corresponding angles to P Y 9. E (orresponding ngles) 10. P Y E 11. E (Transitivity of Parallelism) ontradiction 12. [ is closed at

4 4 Lemma 2 (iangle losure Symmetry) ssume that [ is closed at and show that the biangle is closed at. Proof: 1. ssume by way of contradiction that [ is not closed at. 2. Then some interior ray E does not intersect. 3. hoose E < E. This is possible by the corollary to ristotle s axiom in hapter E = because E. 5. Interior ray E intersects in a point F and E F because. 6. Since E is an exterior angle for EF, which gives us our contradiction. 7. Thus. E > EF = E E Figure 4. To show that the biangle is closed at.

5 5 Lemma 3 (Inner Transitivity) ssume that the biangles [EF and [EF are closed at their respective verticies and that and are on SS(l(EF )). Then the biangle [ is closed. Proof: If and are both limiting parallel to EF, then they are limiting parallel to each other. E G I H F Figure 5. losed biangles [EF and [EF. 1. and have no point in common, by etweenness xiom 4: Plane Separation along line EF. 2. Hence, there are two cases, depending on whether EF is between and or and are both on the same side of EF, by etweenness xiom 3: Trichotomy. 3. In case EF is between and, let G be the intersection of with EF, by etweenness xiom 4: Plane Separation. (To switch half-planes, it would have to cross EF.) 4. ny ray H interior to G must intersect EF in a point I. ecause is limiting parallel to EF, any interior ray must intersect EF, or is not a limiting parallel. 5. IH, lying interior to IF, must intersect, because EF is limiting parallel to. y symmetry of limiting parallelism, any interior ray must intersect. 6. Hence, any ray H interior to must intersect, so is limiting parallel to.

6 6 Lemma 4 (Outer Transitivity) ssume that the biangles [EF and [EF are closed at their respective verticies and that and are on OS( EF ). Then the biangle [ is closed. Proof: E F H G I K Figure 6. losed biangles [EF and [EF. 1. It suffices to show there is a line transversal to the three rays,, EF. 2. ase 1: and F are on the same side of E. 3. Then ray E is interior to E. Since, are SS( EF ) and, F are SS( E) 4. Then E intersects since [F E. Thus E is our transversal. 5. ase 2: (See figure 6) and F are on opposite sides of E. 6. Let G be the point at which F meets E. Since [F EJ. 7. dd H such that E F H by segment extension. 8. Thus [HF J Since [F EJ. 9. Now we have HF G > E y exterior angle theorem. 10. Now construct ray F I interior to HF such that HF I = E. y angle duplication. 11. Let J be the point that F I meets J. Since [HF J. J

7 7 12. Now F J E by alternate interior angle theorem. 13. Since E intersects side F and does not intersect side F J of F J it must intersect J by Pasch s Theorem. 14. Thus E is our transversal. E F H G I K J Figure 6. losed biangles [EF and [EF.

8 8 Lemma 5 (E for symp- ) ssume that P QΩ is a singly asymptotic triangle at Ω. Then the exterior angle at P is greater than the interior angle at Q. N.. by symmetry, the exterior angle at q is greater than the interior angle at P. Proof: 1. Extend PQ to a point R. 2. Given R * Q * P. We must show that RQΩ > QP Ω. 3. Let Q be the unique ray on the same side of P Q as QΩ such that RQ = QP Ω. (orresponding ngles) 4. Extend Q to a point U. 5. If U * Q * P, then UQP = QP Ω. (Vertical ngle Theorem) 6. y Exercise 14, Q and P Ω are divergently parallel. 7. Q must be between QR and QΩ. 8. If Q is between QΩ and P Ω, then Q meets P Ω which is a contradiction. 9. Therefore, RQ < RQΩ 10. Since QP Ω = RQ, onclusion, RQΩ > QP Ω. P Ω U Q R Figure 7. The exterior angle theorem.

9 9 Lemma 6 (ongruence of symp- ) ssume that the singly asymptotic triangles Ω and Ω satisfy Ω = Ω. Then Ω = δ Ω if and only if =. Proof: Ω Ω Figure 8. =. 1. ( )ssume = according to the hypothesis. 2. lso ssume Ω > 3. There exisits a uniuque ray such that = Ω 4. ray intersects Ω at pt 5. Let be the unique point on Ω such that = 6. = 7. = Ω = which gives a contradiction. 8. ( )ssume that Ω = Ω 9. ssume that < 10. Let be the point on such that =

10 let Ω be the limiting ray from to Ω 12. Ω is also a limiting parallel to Ω 13. Ω = Ω 14. Ω = Ω 15. Ω > Ω. Ω Ω Figure 8. =.

11 11 Lemma 7 ssume the conditions of the theorem. Then there exists a point H = l() l(ef ). Proof: E H J F N M G P Figure 9. emonstration of the existence of H. 1. Let M be limiting parallel to EF, N limiting parallel to G, and P limiting parallel to G 2. Since E = and EF = G, we have E M = P (side and angle gives us congruent triangles) 3. L differs from N (from previous results) L 4. M L = P L (by angle subtraction) 5. P is a limiting parallel to N 6. Hence N L is smaller than P L (since N L < M L) 7. M lies between N and, so it must intersect G at a point we ll call J (since closed at ) 8. ) J is on the same side of EF as, thus on the opposite side of

12 12 9. Thus J intersects EF in a point H which must be on EF because H is on the same side of as J E H J F N M G P Figure 9. emonstration of the existence of H. L