Post on 18-Dec-2015
Hyperbolic Lines and Segments
• Poincaré disk model Line = circular arc, meets fundamental circle
orthogonally
• Note: Lines closer to
center of fundamentalcircle are closer to Euclidian lines
Why?
Poincaré Disk Model
• Model of geometric world Different set of rules apply
• Rules Points are interior to fundamental circle Lines are circular arcs orthogonal to
fundamental circle Points where line meets fundamental circle
are ideal points -- this set called • Can be thought of as “infinity” in this context
Poincaré Disk Model
Euclid’s first four postulates hold
1.Given two distinct points, A and B, a unique line passing through them
2.Any line segment can be extended indefinitely A segment has end points (closed)
3.Given two distinct points, A and B, a circle with radius AB can be drawn
4.Any two right angles are congruent
Hyperbolic Triangles
• Recall Activity 2 – so … how do you find measure?
• We find sum of angles might not be 180
Hyperbolic Triangles
• Lines that do not intersect are parallel lines
• What if a triangle could have 3 vertices on the fundamental circle?
Hyperbolic Triangles
• Generally the sum of the angles of a hyperbolic triangle is less than 180
• The difference between the calculated sum and 180 is called the defect of the triangle
• Calculatethe defect
Hyperbolic Polygons
• What does the hyperbolic plane do to the sum of the measures of angles of polygons?
Hyperbolic Circles
• A circle is the locus of points equidistant from a fixed point, the center
• Recall Activity 9.5
What seems “wrong”
with these results?
Hyperbolic Circles
• What happens when the center or a point on the circle approaches “infinity”?
• If center could beon fundamentalcircle “Infinite” radius Called a horocycle
Distance on Poincarè Disk Model
• Rule for measuring distance metric
• Euclidian distance
Metric Axioms
1.d(A, B) = 0 A = B
2.d(A, B) = d(B, A)
3.Given A, B, C points, d(A, B) + d(B, C) d(A, C)
2 2
1 1 2 2,d A B a b a b
Distance on Poincarè Disk Model
• Formula for distance
Where AM, AN, BN, BM are Euclidian distances
M
N
/( , ) ln ln
/
AM AN AM BNd A B
BM BN AN BM
Distance on Poincarè Disk Model
Now work through axioms
1.d(A, B) = 0 A = B
2.d(A, B) = d(B, A)
3.Given A, B, C points, d(A, B) + d(B, C) d(A, C)
/( , ) ln ln
/
AM AN AM BNd A B
BM BN AN BM
Circumcircles, Incircles of Hyperbolic Triangles
• Consider Activity 9.3a Concurrency of perpendicular bisectors
Circumcircles, Incircles of Hyperbolic Triangles
• Conjecture Three perpendicular bisectors of sides of
Poincarè disk are concurrent at O Circle with center O, radius OA also contains
points B and C
Circumcircles, Incircles of Hyperbolic Triangles
• Note issue of bisectors sometimes not intersecting
Circumcircles, Incircles of Hyperbolic Triangles
• Recall Activity 9.4 Concurrence of angle bisectors
Circumcircles, Incircles of Hyperbolic Triangles
• Conjecture Three angle bisectors of sides of Poincarè
disk are concurrent at O Circle with center O, radius tangent to one
side is tangent to all three sides
Congruence of Triangles in Hyperbolic Plane
• Visual inspection unreliable
• Must use axioms, theorems of hyperbolic plane First four axioms are available
• We will find that AAA is now a valid criterion for congruent triangles!!
Parallel Postulate in Poincaré Disk
• Playfair’s Postulate
Given any line l and any point P not on l,
exactly one line on P that is parallel to l
• Definition 9.4
Two lines, l and m are parallel if the do not intersect
l
P
Parallel Postulate in Poincaré Disk
• Playfare’s postulate Says exactly one line through point P, parallel to line
• What are two possible negations to the postulate?
1. No lines through P, parallel
2. Many lines through P, parallel
Restate the first – Elliptic Parallel Postulate
There is a line l and a point P not on l such that
every line through P intersects l
Elliptic Parallel Postulate
• Examples of elliptic space Spherical geometry
• Great circle “Straight” line on the sphere Part of a circle with center at
center of sphere
Elliptic Parallel Postulate
• Elliptic Parallel Theorem
Given any line l and a point P not on l every
line through P intersects l• Let line l be the equator
All other lines (great circles) through any pointmust intersect the equator
Hyperbolic Parallel Postulate
• Hyperbolic Parallel Postulate
There is a line l and a point P not on l such that …
more than one line through P is parallel to l
Hyperbolic Parallel Postulate
• Result of hyperbolic parallel postulateTheorem 9.4 There is at least one triangle whose angle
sum is less than the sum of two right angles
Hyperbolic Parallel Postulate
• Proof: We know at least two lines parallel to l Note to l, PQ Also to PQ, m
and thus || to l Note line n also
|| to l
Hyperbolic Parallel Postulate
XPY > 0 Not R on l such that we have PQR QPR < QPY Move R towards
fundamental circle, QRP 0
Thus QRP < XPY And PQR has one
rt. angle and the other two sum < 90 Thus sum of angles < 180
Parallel Lines, Hyperbolic Plane
• Theorem 9.5 Hyperbolic Parallel TheoremGiven any line l, any point P, not on l, at leas two lines through P, parallel to l Remember
parallel meansthey don’tintersect
Parallel Lines, Hyperbolic Plane
• Lines outside the limiting rays will beparallel to line AB
Calledultraparallel orsuperparallel orhyperparallel
Note line ED is limiting parallel with D at
Parallel Lines, Hyperbolic Plane
• Angles DCE & FCD are called the angles of parallelism The angle between
one of the limitingrays and CD
• Theorem 9.6The two anglesof parallelismare congruent
Parallel Lines, Hyperbolic Plane• Note results of Activity 9.8
CD is a commonperpendicular tolines AB, HF
• Can be proved inthis context If two lines do not
intersect then eitherthey are limiting parallelsor have a commonperpendicular
Quadrilaterals, Hyperbolic Plane
• Recall results of Activity 9.10
• 90 angles at B, A, and D only• Called a Lambert quadrilateral
Quadrilaterals, Hyperbolic Plane
• Saccheri quadrilateral A pair of congruent sides Both perpendicular to a third side
Quadrilaterals, Hyperbolic Plane
• Angles at A and B are base angles
• Angles at E and F aresummit angles Note they are congruent
• Side EF is the summit
• You should have foundnot possible to constructrectangle (4 right angles)