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12.6. Lemma. Consider an h-plane with a unit circle as the absolute. Let O be the center of the absolute and P be another h-point. Let P′ denotes the inverse ofP in the absolute.
Then the circle Γwith the centerP′ and radius √1−OP2
OP is perpendic- ular to the absolute. Moreover,O is the inverse ofP in Γ.
P P′ O
T Γ Proof. Follows from Exercise 10.20. Assume Γis a circline that is perpen- dicular to the absolute. Consider the in- versionX 7→X′ inΓ, or ifΓis a line, set
X 7→X′ to be the reflection acrossΓ. The following observation says that the mapX7→X′ respects all the notions introduced in the previous section. To- gether with the lemma above, it implies that in any problem that is for- mulated entirely in h-terms we can assume that a given h-point lies in the center of the absolute.
12.7. Main observation. The map X7→X′ described above is a bijec- tion from the h-plane to itself. Moreover, for any h-pointsP,Q,R such thatP 6=QandQ6=R, the following conditions hold:
(a) The h-line(P Q)h, h-half-line[P Q)h, and h-segment[P Q]hare trans- formed into (P′Q′)h,[P′Q′)h, and[P′Q′]
h correspondingly. (b) δ(P′, Q′) =δ(P, Q)andP′Q′h=P Qh.
(c) ∡hP′Q′R′≡ −∡ hP QR.
It is instructive to compare this observation with Proposition 5.6. Proof. According to Theorem 10.15, the map sends the absolute to itself. Note that the points on Γ do not move, it follows that points inside of the absolute remain inside after the mapping. Whence theX 7→X′ is a bijection from the h-plane to itself.
Part(a)follows from 10.7 and 10.25. Part(b)follows from Theorem 10.6. Part(c)follows from Theorem 10.25.
12.8. Lemma. Assume that the absolute is a unit circle centered atO. Given an h-point P, setx=OP andy=OPh. Then
y= ln1 +x
1−x and x=
ey−1
ey+ 1.
Observe that according to lemma, OPh → ∞ as OP → 1. That is ifP approaches absolute in Euclidean sense, it escapes to infinity in the h-sense.
P O
A B
Proof. Note that the h-line (OP)h forms a diameter of the absolute. If Aand B are the ideal points as in the definition of h-distance, then OA=OB= 1, P A= 1 +x, P B= 1−x. In particular, y= lnAP·BO P B·OA = ln 1 +x 1−x.
Taking the exponential function of the left and the right hand side and applying obvious algebra manipulations, we get that
x= e y−1
ey+ 1.
12.9. Lemma. Assume the pointsP,Q, andRappear on one h-line in the same order. Then
P Qh+QRh=P Rh.
Proof. Note that
P Qh+QRh=P Rh is equivalent to
➊ δ(P, Q)·δ(Q, R) =δ(P, R).
LetAandB be the ideal points of(P Q)h. Without loss of generality, we can assume that the points A, P, Q, R, and B appear in the same
order on the circline containing(P Q)h. Then δ(P, Q)·δ(Q, R) =AQ·BP QB·P A· AR·BQ RB·QA = =AR·BP RB·P A = =δ(P, R). Hence➊follows.
LetP be an h-point andρ >0. The set of all h-points Qsuch that
P Qh=ρis called an h-circle with the centerP and theh-radius ρ. 12.10. Lemma. Any h-circle is a Euclidean circle that lies completely in the h-plane.
More precisely for any h-point P and ρ > 0 there is a ρˆ >0 and a pointPˆ such that
P Qh=ρ ⇐⇒ P Qˆ = ˆρ for any h-pointQ.
Moreover, if O is the center of the absolute, then 1. Oˆ =O for anyρand
2. Pˆ ∈(OP)for any P 6=O.
Proof. According to Lemma 12.8,OQh=ρif and only if
OQ= ˆρ= e ρ−1
eρ+ 1.
Therefore, the locus of h-points Q such that OQh = ρ is a Euclidean circle, denote it by∆ρ. O P ˆ P Q ∆′ ρ
If P 6=O, then by Lemma 12.6 and the main observation (12.7) there is inversion that respects all h-notions and sendsO7→P.
Let ∆′
ρ be the inverse of ∆ρ. Since the in- version preserves the h-distance,P Qh=ρif and only ifQ∈∆′
ρ.
According to Theorem 10.7,∆′
ρis a Euclidean circle. LetPˆ and ρˆdenote the Euclidean center and radius of∆′
ρ. Finally, note that∆′
ρreflects to itself across(OP); that is, the center ˆ
P lies on(OP).
12.11. Exercise. AssumeP,Pˆ, andO are as in the Lemma 12.10 and
Axiom I
Evidently, the h-plane contains at least two points. Therefore, to show that Axiom I holds in the h-plane, we need to show that the h-distance defined on page 91 is a metric on h-plane; that is, the conditions(a)–(d) in Definition 1.1 hold for h-distance.
The following claim says that the h-distance meets the conditions(a) and(b).
12.12. Claim. Given the h-points P and Q, we have P Qh > 0 and
P Qh= 0if and only if P=Q.
Proof. According to Lemma 12.6 and the main observation (12.7), we may assume thatQis the center of the absolute. In this case
δ(Q, P) = 1 +QP 1−QP >1
and therefore
QPh= ln[δ(Q, P)]>0. Moreover, the equalities holds if and only ifP =Q.
The following claim says that the h-distance meets the condition 1.1c. 12.13. Claim. For any h-pointsP andQ, we haveP Qh=QPh. Proof. Let A andB be ideal points of (P Q)h and A, P, Q, B appear on the circline containing(P Q)h in the same order.
A B P Q Then P Qh= ln AQ·BP QB·P A= = lnBP·AQ P A·QB = =QPh.
The following claim shows, in particular, that the triangle inequality (which is condition 1.1d) holds forh-distance.
12.14. Claim. Given a triple of h-points P,Q, andR, we have
P Qh+QRh>P Rh.
Moreover, the equality holds if and only if P,Q, and Rlie on one h-line in the same order.
Proof. Without loss of generality, we may assume thatP is the center of the absolute andP Qh>QRh>0.
Let∆ denotes the h-circle with the center Qand h-radiusρ=QRh. LetS andT be the points of intersection of(P Q)and∆.
By Lemma 12.9, P Qh > QRh. Therefore, we can assume that the pointsP,S,Q, andT appear on the h-line in the same order.
According to Lemma 12.10,∆ is a Euclidean circle; letQˆ denotes its Euclidean center. Note thatQˆ is the Euclidean midpoint of[ST].
P Q ˆ Q R S T ∆
By the Euclidean triangle inequality
➋ P T =PQˆ+ ˆQR>P R
and the equality holds if and only ifT =R. By Lemma 12.8, P Th= ln 1 +P T 1−P T, P Rh= ln 1 +P R 1−P R.
Since the function f(x) = ln1+1−xx is increasing forx∈[0,1), inequality➋implies
P Th>P Rh
and the equality holds if and only ifT =R. Finally, applying Lemma 12.9 again, we get that
P Th=P Qh+QRh. Hence the claim follows.
Axiom II
Note that once the following claim is proved, Axiom II follows from Corol- lary 10.18.
12.15. Claim. A subset of the h-plane is an h-line if and only if it forms a line for the h-distance in the sense of Definition 1.9.
Proof. Letℓ be an h-line. Applying the main observation (12.7) we can assume that ℓ contains the center of the absolute. In this case, ℓ is an intersection of a diameter of the absolute and the h-plane. LetAand B
Consider the mapι:ℓ→Rdefined as
ι(X) = lnAX
XB.
Note thatι: ℓ→Ris a bijection.
Further, ifX, Y ∈ℓ and the pointsA, X, Y, andB appear on[AB] in the same order, then
ι(Y)−ι(X) = lnAY Y B −ln AX XB = ln AY·BX Y B·XB =XYh.
We proved that any h-line is a line for h-distance. The converse follows from Claim 12.14.
Axiom III
Note that the first part of Axiom III follows directly from the definition of the h-angle measure defined on page 91. It remains to show that ∡h satisfies the conditions IIIa, IIIb, and IIIc on page 19.
The following two claims say that∡h satisfies IIIa and IIIb.
12.16. Claim. Given an h-half-line [OP)h and α∈ (−π, π], there is a unique h-half-line [OQ)h such that ∡hP OQ=α.
12.17. Claim. For any h-points P,Q, and R distinct from an h-point
O, we have
∡hP OQ+∡hQOR≡∡hP OR.
Proof of 12.16 and 12.17. Applying the main observation, we may assume thatO is the center of the absolute. In this case, for any h-pointP 6=O, the h-half-line [OP)h is the intersection of the Euclidean half-line [OP) with h-plane. Hence the claims 12.16 and 12.17 follow from the axioms IIIa and IIIb of the Euclidean plane.
The following claim says that∡h satisfies IIIc. 12.18. Claim. The function
∡h: (P, Q, R)7→∡hP QR
is continuous at any triple of points (P, Q, R) such thatQ6=P,Q6=R, and∡hP QR6=π.
Proof. LetO denotes the center of the absolute. We can assume that Q
Let Z denotes the inverse of Q in the absolute; let Γ denotes the circle perpendicular to the absolute and centered at Z. According to Lemma 12.6, the pointO is the inverse of Qin Γ.
LetP′ andR′ denote the inversions in Γof the points P andR cor- respondingly. Note that the point P′ is completely determined by the pointsQ and P. Moreover, the map (Q, P)7→ P′ is continuous at any pair of points (Q, P) such that Q 6= O. The same is true for the map (Q, R)7→R′
According to the main observation
∡hP QR≡ −∡hP′OR′.
Since ∡hP′OR′ = ∡P′OR′ and the maps (Q, P) 7→ P′, (Q, R) 7→ R′ are continuous, the claim follows from the corresponding axiom of the Euclidean plane.
Axiom IV
The following claim says that Axiom IV holds in the h-plane.
12.19. Claim. In the h-plane, we have △hP QR ∼=△hP′Q′R′ if and only if
Q′Ph′ =QPh, Q′R′h=QRh and ∡hP′Q′R′=±∡P QR.
Proof. Applying the main observation, we can assume that Q and Q′ coincide with the center of the absolute; in particular Q =Q′. In this case
∡P′QR′=∡
hP′QR′=±∡hP QR=±∡P QR. Since
QPh=QPh′ and QRh=QRh′,
Lemma 12.8 implies that the same holds for the Euclidean distances; that is,
QP =QP′ and QR=QR′.
By SAS, there is a motion of the Euclidean plane that sendsQto itself,
P to P′ andR toR′
Note that the center of the absolute is fixed by the corresponding motion. It follows that this motion gives also a motion of the h-plane; in particular, the h-triangles△hP QRand△hP′QR′ are h-congruent.
Axiom h-V
Finally, we need to check that the Axiom h-V on page 86 holds; that is, we need to prove the following claim.
P ℓ 12.20. Claim. For any h-lineℓ and any h-point
P /∈ℓ there are at least two h-lines that pass thru
P and have no points of intersection with ℓ.
Instead of proof. Applying the main observation we can assume that P is the center of the absolute.
The remaining part of the proof can be guessed from the picture
12.21. Exercise. Show that in the h-plane there are 3 mutually paral- lel h-lines such that any pair of these three lines lies on one side of the remaining h-line.