Download Section 22.1

1. An equilateral triangle in hyperbolic geometry has angles of
measure 50 each. Find the defect.
δ(ABC) = 180 - ∡A - ∡ B - ∡ C = 180 – (3)(50) = 180 – 150 = 30.
§ 22.1
2. Show why congruent triangles in hyperbolic geometry have equal
areas. Is the converse proposition true?
If two triangles are congruent then they have equal angles. This
implies equal defects which in turn implies equal areas.
The converse is not true since equal areas imply equal defects but
equal defects do not imply equal angles just equal sum of angles.
3. Given the figure below. Calculate the defects of the four interior polygons.
Test the validity of the additive property of defects by calculating the defect of
B
quadrilateral ABCD.
30
A
δ 1 = 180 -178 = 2
50
45
δ2
45
δ 2 = 360 -354 = 6
δ 3 = 180 -174 = 6
δ1
88
δ 4 = 360 -354 = 6
δ 1 + δ 2 + δ 3 + δ 4 = 20
136
143
136
143
74
δ3
δ4
45
45
50
D
30
C
δABCD = 360 – (2)(90) – (2)(80) = 360 – 340 = 20
4. The acute angle of a certain Lambert Quadrilateral (You may have to look this
up.) has measure of 83. Find its defect? What is the maximum defect of a
Lambert Quadrilateral?
C
D
A Lambert Quadrilateral is a
quadrilateral with three 90 degree
angles.
δ(ABCD) = 360 – (3)(90) – 83 = 360 – 353 = 7
A
B
The maximum defect would occur when the third angle is 0 giving a defect of 90.
δABCD = 360 – (3)(90) – 0 = 360 – 270 = 90
Do you have an idea of what that quadrilateral would look like?
5. The summit angles of a certain Saccheri Quadrilateral (You may have to
look this up.) has measure of 83. Find the defect of the quadrilateral. Why
should the answer of this problem be exactly twice as much as the answer to
the previous problem
D
C
A Saccheri Quadrilateral is a
quadrilateral with two base 90 degree
angles. Sides AD and BC have the
same measure. It is two Lambert
Quadrilaterals with a common side.
A
δ(ABCD) = 360 – (2)(90) – (2)(83) = 360 – 180 - 166 = 14
B
6. The defect of a certain regular dodecagon in hyperbolic geometry is 12. if O
is the
center, find
a. the measure of each angle of the polygon.
b. The defect of the subtriangle ABO where O is the center and A and
B are adjacent vertices.
a. δ(P1 P2 . . . Pn ) = 180(12 - 2) – P1 – P2 . . . – P12
12 = 1800 – P1 – P2 . . . – P12
P1 + P2 . . . + P12 = 1800 – 12 = 1788
And since it is regular all the angles are the same so that
12 P = 1788
P = 149
b. Since there are 12 sub triangles and they are all congruent (it is a
regular dodecagon) each of the sub triangles would have defect of 1.
7. Two regular pentagons have equal defects. Show they must be
congruent
Central angles form five congruent triangles in each pentagon. This
implies than any two corresponding angles from either pentagon must
have equal angle measure. By AA criteria the triangles are all congruent
implying that the sides and angles of the two pentagons are congruent.
8. Show that if the defect of a regular pentagon is 90 then it can be used as a
fundamental region to tile the hyperbolic plane.
δ = 90 = ( 5 – 2) 180 – 5m A. which implies that m A = 90. thus each
vertex of the pentagon is a right angle. This means that four of the pentagons
will exactly fit at a vertex of the tiling.
This implies that since all the pentagons are congruent, the distance covered by
n tiles in a row = n diameters of the tile, which is unbounded, implying that the
are of the hyperbolic plane is infinite.
Note that the area of the hyperbolic plane is covered by congruent pentagons
while the are of the Euclidean plan is covered by congruent squares. It gets
even more interesting if you wish to look into it.