Download 1 Solution of Test

Math 3181
Dr. Franz Rothe
September 24, 2014
Name:
All3181\3181_fall14t1.tex
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Solution of Test
Problem 1.1. Encircle the statements below which are valid for any arbitrary incidence
plane; and scratch through the statements which are not always valid.
• There exists at most one parallel to a given line through a given point.
• X Any two different lines which are not parallel, have a unique point of intersection.
• X On any given line lie two or more points.
• X Through any point go at least two distinct lines.
• X Two different lines have at most one point in common.
• Any two lines have a unique point of intersection.
• X If there exist three distinct lines with an intersection point, there exist at least
four points.
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Figure 1: Five examples for incidence planes.
Problem 1.2. Match the figures on page 2 one-to-one with the following descriptions:
• a handshake model (e)
• a straight fan (c)
• the smallest affine plane (d)
• the smallest projective plane (b)
• the number of lines is one more than the number of points (a)
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Problem 1.3. As far as two-dimensional geometry is concerned, Hilbert’s Proposition 1 reduces to one simple statement: any two different lines either intersect in one
point, or are parallel. This statement can be rephrased in many formulations. Encircle
the statements below which are such equivalent reformulations; and scratch through the
statements which are not.
• There exists at most one parallel to a given line through a given point.
• Through any point go at least two distinct lines.
• X Any two different lines which are not parallel, have a unique point of intersection.
• On any given line lie two or more points.
• X Two different lines have at most one point in common.
• Any two different lines which are parallel, have a unique point of intersection.
• X If two lines have two or more points in common, they are equal.
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Figure 2: Two intersecting segments.
Problem 1.4. We assume A ∗ B ∗ C and A ∗ D ∗ E, and points A, C, E do not lie on
a line. Prove that the segments BE and CD intersect each other. Provide a drawing.
(i) Why does the segment BE intersect the line CD?
(ii) Why does the segment CD intersect the line BE?
(iii) Why are these two intersection points equal? Why do the segments BE and CD
intersect each other?
Answer. (i) We use Pasch’s axiom for triangle 4ABE and line CD. This line intersects
side AE, but not side AC. Hence line CD intersects the side BE.
(ii) We use Pasch’s axiom for triangle 4ACD and line BE. This line intersects side
AC, but not side AD. Hence line BE intersects the side CD.
(iii) The two lines CD and BE are distinct, otherwise all points A, B, C, D, E would
lie on a line. Hence lines CD and BE have unique intersection point X. By items
(i) and (ii), point X is the intersection of the segments BE and CD.
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Problem 1.5. Give exact definitions of the terms segment, ray, triangle in terms of the
order relation. Clarify the obvious questions. Give illustrations.
Answer.
Definition 1 (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 2 (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.
Definition 3 (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 or ray, 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.
For the congruence, similarity of two triangles, and for the Theorem of Desargues,
we assign a definite order to the vertices and sides of these triangles.
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Problem 1.6. Give exact definitions of the terms angle, interior and exterior of an
angle. Clarify the obvious questions. Give illustrations.
Answer.
−→
−→
Definition 4 (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 3: Interior and exterior of an angle
Definition 5 (Interior and exterior domain 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.
Problem 1.7. Give exact definitions of terms isosceles triangle, equilateral triangle,
and right triangle. For which geometries does exist an isosceles right triangle?
Answer.
Definition 6 (Isosceles triangle, equilateral triangle, right triangle). A triangle
with two congruent sides is called isosceles. A triangle for which all three sides are
congruent, is called equilateral. A triangle with at least one right angle is called a right
triangle.
An isosceles right triangle exists in any Hilbert plane, especially in Euclidean and
hyperbolic geometry. In spherical geometry, too, exists an isosceles right 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.
Problem 1.8. Formulate the theorem with specific quantities from a triangle 4ABC.
Provide a drawing. Give a detailed proof.
Figure 4: An isosceles triangle
Answer. For the triangle 4ABC as in the drawing on page 7, 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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Figure 5: Transfer of a triangle.
Proposition 2 (Transfer of a triangle). Any given triangle 4ABC can be transferred
into any given half-plane H0 of any given ray r, such that one obtains a congruent
−−→
triangle 4A0 B 0 C 0 lying in the prescribed half-plane, and the given ray is r = A0 B 0 ,
emanates from vertex A0 and lies on the side A0 B 0 .
Problem 1.9. Provide a drawing for Proposition 2 and prove the Proposition, starting
from the axioms and ASA-congruence. State clearly which axioms you use, and where
one has to use the ASA-congruence.
Proof. Using axiom (III.1), we transfer the segment AB onto the ray r to obtain the
congruent segment AB ∼
= A0 B 0 . Next we axiom (III.4), and transfer the angle ∠BAC
into the given half-plane H0 to obtain the congruent angle ∠BAC ∼
= ∠(r, k) . Thus
the newly produce angle has the given ray r as one side. Onto the other side k of
the newly produced angle, we transfer the segment AC and obtain the congruent
−−→
−−→
segment AC ∼
= A0 C 0 . Since by construction r = A0 B 0 and k = A0 C 0 , we see that
∠BAC ∼
= ∠B 0 A0 C 0 .
From axiom (III.5)—which we now use even literally!—we conclude the congruence
∠ABC ∼
= ∠A0 B 0 C 0 . We are now in the position to use the ASA-congruence for
the triangles 4ABC and 4A0 B 0 C 0 . Hence we conclude that the triangle congruence
4ABC ∼
= 4A0 B 0 C 0 , as to be shown.
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Problem 1.10 (Drop a Perpendicular). Given is a line OA and a point B not on
this line. We have to drop the perpendicular from point B onto line OA.
−−→
Construction 1. Draw ray OB. Transfer angle ∠AOB, into the half plane opposite
−→
to B, with ray OA as one side. On the newly produced ray, transfer segment OB to
produce a new segment OC ∼
= OB. The line BC is the perpendicular, dropped from
point B onto line OA.
Figure 6: Drop the perpendicular, the two cases
Proof of validity.
Question. Why do the line OA and the segment BC intersect.
Answer. Line OA and the segment BC intersect because points B and C lie on different
sides of line OA.
I call the intersection point M .
Question. It can happen that O = M . How does one proceed in this special case.
Answer. In the special case that O = M , we know that A 6= M and
(**)
∠AM B ∼
= ∠AM C
But these is a pair of congruent supplementary angles, because M lies between B and
C. By definition, an angle congruent to its supplementary angle is a right angle. Hence
∠AM B is a right angle.
In the generic situation O 6= M , we distinguish two cases:
(i) points M and A lie on the same ray with vertex O;
(ii) point O lies between M and A.
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In both cases we show the two triangles 4OM B and 4OM C are congruent. Indeed,
the angles ∠M OB ∼
= ∠M OC at vertex O are congruent, both in case (i) and (ii).
Question. In case (i), one gets this congruence directly from the construction since the
−→ −−→
rays OA = OM are equal. How does one proceed in case (ii).
Figure 7: Congruences needed in the two cases.
−→
−−→
Answer. In case (ii), the rays OA and OM are opposite. Thus ∠M OB and ∠AOB, as
well as ∠M OC and ∠AOC are supplementary angles. Again ∠AOB ∼
= ∠AOC because
of the angle transfer done in the construction. Now Theorem 14 of Hilbert tells that
supplements of congruent angles are congruent. Hence we get ∠M OB ∼
= ∠M OC once
again.
Question. Finish the argument showing the triangle congruence 4M OB ∼
= 4M OC.
∼
Answer. The adjacent sides OB = OC are congruent by construction, too. Finally
the common sides OM is congruent to itself. Hence the congruence follows by SAS
congruence. The drawing stresses the pieces matched to prove the congruence.
Because the two triangles are congruent, the corresponding angles at vertex M are
congruent, too. Hence
(*)
∠OM B ∼
= ∠OM C
which is a pair of congruent supplementary angles.
Question. How is a right angle defined. How do we see that ∠OM B is a right angle.
Answer. A right angle is, by definition, an angle congruent to its supplementary angle
. Hence ∠OM B is a right angle because of formula (*).
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