Download 1 Solution of Test I

Math 3181
Dr. Franz Rothe
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All3181\3181_spr13t1.tex
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Solution of Test I
Definition 1 (Hand-shake model). A hand shake model is an incidence geometry for
which every line has exactly two points.
Definition 2 (Straight fan). A straight fan is an incidence geometry with all but one
point lying on one line.
10 Problem 1.1. How many lines does the hand-shake incidence geometry with
n points have. How many lines does the straight fan with n points have.
Answer. The hand-shake incidence geometry with n points has
1 + 2 + · · · + (n − 1) =
(n − 1)n
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lines. We see this as follows. We can connect the first point to the other (n − 1) points.
Disregarding this point, we connect the second point to (n − 2) different points, and so
on. The last line to be drawn is between the (n − 1)-th and the n-th point.
The straight fan with n points has n lines. There is one long line with n − 1 points,
and only one point P not on this line. There are n − 1 lines with two points each of
which connects P to a different point on the long line.
10 Problem 1.2. Which parallel property holds for the hand-shake model with
4 points. Which parallel property holds for the hand-shake model with n ≥ 5 points.
Which parallel property holds for a straight fan.
Answer. The hand-shake model with 4 points has the Euclidean parallel property. All
the hand-shake models with n ≥ 5 points have the hyperbolic parallel property. For all
straight fans, the elliptic parallel property holds.
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10 Problem 1.3 (A six-point incidence geometry). Find an incidence geometry with six points and seven lines. Mark all three-point lines with different blue
shades. Describe the properties of the points and lines. Count the lines.
Answer. A model easy to draw has three intersecting three-point lines forming a triangle
and the in-circle as the fourth three-point line. There are three lines with two points
each of which connects a vertex with the midpoint of the opposite side. Together we
get 4 + 3 = 7 lines.
Figure 1: A six-point incidence geometry with seven lines
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10 Problem 1.4 (Scheduling problem). Make a 4 day schedule for a group
of nine participants. Each day the students are divided in a different way into 3 groups
of 3 students. Never are two students in the same group more than one time during the
four days.
Use the coordinate plane Z3 × Z3 , with addition and multiplication modulo 3. Make
a picture of the schedule by drawing the 3 × 3 pattern of dots separately for every day.
The main point is to show the partition into the groups clearly, in a separate drawing
for every day.
Figure 2: Scheduling four days with differently chosen groups 3 + 3 + 3.
Answer.
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Definition 3 (Subplane). A subplane of an incidence plane is an incidence plane
which has as "points" a subset P 0 ⊆ P from the points P of the given plane, and as
"lines" the nonempty intersections of the lines of the original plane with the subset
P 0 of the remaining points. The incidence relation is the induced relation. The axioms
of incidence (I.1), (I.2),(I.3) have still to hold for the subplane.
10 Problem 1.5. Given is a subplane of the affine coordinate plane Z7 × Z7
which contains the 21 points on three parallel lines l1 , l2 , l3 .
The lines of the subplane are these three parallel lines with 7 points on each, and
a lot of lines with three points. How many lines ki with 3 points do exist? How many
parallels through a given point P has a three point line l.
Figure 3: In a subplane there exist multiple parallels.
Answer. The original Z7 × Z7 plane has 72 + 7 = 56 lines. Four lines are deleted. There
remain 52 lines, of which 49 have three points each one.
In the subplane there exist many parallel to a given line through a given point.
The parallels through point P to the three-point line l in the subplane are obtained by
cutting down the original parallel m and the lines P A, P B, P C, P D. Here A, B, C, D
be the four points deleted from the extended line l+ .
In the subplane, there are five parallels through a given point to a given three point
line.
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10 Problem 1.6. Give exact definitions of the terms segment, ray, triangle in
terms of the order relation. Clarify the obvious questions. Give illustrations.
Answer.
Definition 4 (Segment). Let A and B be two distinct points. The segment AB is the
set consisting of the points A and B and all points lying between A and B. The points
A and B are called the endpoints of the segment, the points between A and B are called
the interior points, and the remaining points on the line AB are called the exterior
points of the segment.
Definition 5 (Triangle). We define a triangle to be union of the three segments AB,
BC and CA. The three points A, B and C are assumed not to lie on a line. These three
points are the vertices, and the segments BC, AC, and AB are the sides of the triangle.
For a segment, it is assumed that the two endpoints A and B are different. For a
triangle 4ABC, it is assumed that the three vertices do not lie on a line.
−→
Definition 6 (Ray). Given two distinct points A and B, the ray AB is the set consisting
of the points A and B, the points inside the segment AB, and all points P on the line
AB such that the given point B lies between A and P . The point A is called the vertex
of the ray.
−→
The axiom of order II.2 tells that the ray AB contains points not lying in the segment
AB.
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10 Problem 1.7. Give exact definitions of the terms angle, interior and exterior
of an angle. Clarify the obvious questions. Give illustrations.
Answer.
−→
−→
Definition 7 (Angle). An angle ∠BAC is the union of two rays AB and AC with
common vertex A not lying on one line. The point A is called the vertex of the angle.
−→
−→
The rays AB and AC are called the sides of the angle.
Figure 4: Interior and exterior of an angle
Definition 8 (Interior and exterior of an angle). The interior of an angle lying in
a plane A is the intersection of two corresponding half planes—bordered by the sides
of the angle, and containing points on the other side of the angle, respectively. The
exterior of an angle is the union of two opposite half planes—-bordered by the sides of
the angle, and not containing the points neither in the interior nor on the legs of the
angle. Half planes, interior and exterior of an angle all do not include the lines or rays
on their boundary.
Thus the interior of ∠BAD is the intersection of the half plane of AB in which D lies,
and the half plane of AD in which B lies. The exterior of ∠BAD is the union of the
half plane of AB opposite to D, and the half plane of AD opposite to B.
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10 Problem 1.8. Given is a triangle and a line through an interior point of
the triangle on which no vertex of the triangle lies. Show that the line either intersects
exactly two sides of the triangle.
Figure 5: A line through an interior point of a triangle intersects either two sides, or goes
through a vertex and intersects the opposite side.
Answer. Let the line l go the point P in the interior of triangle 4ABC. We have
assumed that the line does not go through any vertex of the triangle. We draw the ray
−→
AP . Because point P lies in the interior of angle ∠BAC, the crossbar theorem shows
−→
that the ray AP intersects the opposite side BC at some point, say Q.
We now apply Pasch’s axiom to the triangle 4ABQ and the line l. This line intersects side AQ at point P . Hence it intersects a second side, either or BQ ⊂ BC or AB,
say at point D. In both cases, the line l intersects one side of the given triangle 4ABC.
By Pasch’s axiom it intersects a second side, say at point E. By Bernays’ lemma, the
line l intersects exactly two sides of the given triangle.
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Proposition 1 (Isosceles Triangle Proposition). [Euclid I.5, and Hilbert’s theorem
11] An isosceles triangle has congruent base angles.
10 Problem 1.9. Formulate the theorem with specific quantities from a triangle
4ABC. Provide a drawing. Give a detailed proof.
Figure 6: An isosceles triangle
Answer. For the triangle 4ABC as in the drawing on page 8, we have to show:
If a ∼
= b, then α ∼
= β.
This is an easy application of SAS-congruence. Assume that the sides AC ∼
= BC are
congruent in 4ABC. We need to show that the base angles α = ∠BAC and β = ∠ABC
are congruent. Define a second triangle 4A0 B 0 C 0 by setting
A0 := B , B 0 := A , C 0 := C
(It does not matter that the second triangle is just ”on top” of the first one.) To apply
SAS congruence, we match corresponding pieces:
(1) ∠ACB = ∠BCA = ∠A0 C 0 B 0 because the order of the sides of an angle is arbitrary.
By axiom III.4, last part, an angle is congruent to itself. Hence ∠ACB ∼
= ∠A0 C 0 B 0 .
(2) AC ∼
= A0 C 0 .
Question. Explain why this holds.
Answer. AC ∼
= BC because we have assumed the triangle to be isosceles, and
BC = A0 C 0 by construction. Hence AC ∼
= A0 C 0 .
(3) Similarly, we show that (3): BC ∼
= B 0 C 0 : Indeed, BC ∼
= AC because we have
assumed the triangle to be isosceles, and congruence is symmetric, and AC = B 0 C 0
by construction. Hence BC ∼
= B0C 0.
Finally, we use axiom III.5. Items (1)(2)(3) imply ∠BAC ∼
= ∠B 0 A0 C 0 = ∠ABC. But
this is just the claimed congruence of base angles.
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