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Math 3181
Name:
Dr. Franz Rothe
May 13, 2016
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Solution of Final
10 Problem 1 (Building up geometry). For an axiomatically built-up geometry, six groups of axioms needed:
I. Axioms of incidence
II. Axioms of order
III. Axioms of congruence
IV. Axiom of parallelism
V. Axioms about circles
VI. Axioms of continuity
Which groups of axioms are valid for the following kind of planes
(a) an incidence plane;
(b) an affine plane;
(c) neutral geometry;
(d) a Pythagorean plane;
(e) a Euclidean plane;
(f ) a Cartesian plane.
Answer.
incidence
Incidence
yes
Order Congruence
no
no
Parallelism
no
Circles
no
Continuity
no
affine
yes
no
no
yes
no
no
neutral
yes
yes
yes
no
no
no
Pythagorean yes
yes
yes
yes
no
no
Euclidean
yes
yes
yes
yes
yes
no
Cartesian
yes
yes
yes
yes
yes
yes
1
Incidence and order
10 Problem 2. Give exact definitions of the terms supplementary angles and
vertical angles, depending only on the axioms of incidence and order.
Answer.
Definition 1 (Supplementary Angles). Two angles are called supplementary angles,
iff they have a common vertex, both have one side on a common ray, and the two other
sides are the opposite rays on a line.
Definition 2 (Vertical Angles). Two angles are called vertical angles, iff they have a
common vertex, and their sides are two pairs of opposite rays on two lines.
10 Problem 3. Given an angle ∠(h, k) lying in a plane A.
Let K be the intersection of the half plane of h in which k does not lie with the half
plane of k in which h does not lie. Explain what the set K is like and provide a drawing.
Give a short exact description for this set.
Answer. From a drawing, one sees that set K is the interior of an angle. Indeed, it is
the interior of the vertical angle corresponding to the given angle ∠(h, k).
2
Proposition 1 (About n points in a plane). Among any n ≥ 3 points lying in an
ordered incidence plane but not on a line, there exist three points P, Q, R such that all
n points lie in the interior or on the boundary of the angle ∠P QR.
10 Problem 4. Complete the reason for this little proposition given below.
Induction start for n = 3: The three given points P, Q, R do not lie on a line, hence
the angle ∠P QR exists.
Induction step ”n 7→ n + 1”: Assume that proposition holds for any n points.
Given are now these points, and an extra point Pn+1 . We distinguish these cases:
(i) The points P1 . . . Pn lie on a line l. By the n-point proposition, they can be put into
an ordered list P1 ∗ P2 ∗ · · · ∗ Pn . By assumption point Pn+1 does not lie on the
line l. We put P := P1 , the first item in the list; R := Pn , the !! last item in the
list; and choose Q := Pn+1 as !! vertex
of the angle. Thus all n + 1 points lie
boundary of the angle ∠P QR.
!! inside or on the
(ii) The points P1 . . . Pn do not lie on a line l. By the induction assumption, there are
points A, B, C among them such that points P1 . . . Pn lie in the interior or on the
boundary of the angle ∠ABC.
• In case that the extra point Pn+1 lies in the interior or on the boundary of
the angle ∠ABC, we are ready.
• In case that the extra point Pn+1 lies in the interior of a supplementary
angle of the angle ∠ABC, all points P1 . . . Pn+1 lie in the interior or on the
boundary of either angle ∠ABPn+1 or angle ∠Pn+1 BC.
3
Figure 1: Any n points lie in the interior or on the boundary of an angle.
• We choose any points A ∗ B ∗ A0 and C ∗ B ∗ C 0 . Assume that the extra point
−−→
Q := Pn+1 lies in the interior of the vertical angle ∠A0 BC 0 or on the ray BA0 .
This case in shown in the figure on page 4. We draw the segments Pi Pn+1
for i = 1 . . . n, obviously including the segments APn+1 , BPn+1 and CPn+1 .
Since the points P1 . . . Pn either lie on the line AB, or the opposite side of
the line !! AB .
AB as point Pn+1 , all these segments !! intersect
By the n-point theorem we may order these intersection points into a list
−−−−→
S1 ∗ · · · ∗ Sn . Let P be a point among P1 . . . Pn on the first ray Pn+1 S1 , and
−−−−→
R be a point among P1 . . . Pn on the !! last ray Pn+1 Sn .
Now all points P1 . . . Pn+1 lie in the interior or on the boundary of the angle
∠P QR.
4
Neutral geometry
10 Problem 5 (The angular bisector).
.
• Give the definition of the (interior) angular bisector of an angle.
• Describe its construction.
• Provide a drawing for the construction.
Answer.
Definition 3 (The angular bisector). The ray in the interior of an angle which bisects
the angle into two congruent angles is called the angular bisector.
Construction 1 (Construction of the angular bisector). One transfers two congruent segments AB and AC onto the two sides of the angle, both starting from the vertex
A of the angle. The perpendicular, dropped from the vertex A onto the segment BC, is
the angular bisector.
Figure 2: The angular bisector
5
Proposition 2 (Comparison of sides implies comparison of angles). [Euclid I.18,
Theorem 23 of Hilbert] In any triangle, across the longer side lies the greater angle.
10 Problem 6. Give a proof of Euclid I.18. Provide a drawing, using the notation
from your proof.
Answer. In 4ABC, we assume for sides AB and BC that c = AB > BC = a. The
issue is to compare the angles α = ∠CAB and γ = ∠ACB across these two sides.
We transfer the shorter side BC at the common vertex B onto the longer side. Thus
one gets a segment BD ∼
= BC, with point D between B and A. Because the 4BCD is
isosceles, it has two congruent base angles
δ = ∠CDB ∼
= ∠DCB
Because B ∗ D ∗ A, we get by angle comparison at vertex C
δ = ∠DCB < γ = ∠ACB
Now we use the exterior angle theorem for 4ACD. Hence
α = ∠CAB < δ = ∠CDB
By transitivity, these three equations together imply that α < γ. Hence the angle α
across the smaller side CB is smaller than the angle γ lying across the greater side AB.
In short, we have shown that c > a ⇒ γ > α.
Figure 3: Across the longer side lies the greater angle
6
Proposition 3 (Comparison of angles implies comparison of sides). [Euclid I.19]
In any triangle, across the greater angle lies the longer side.
10 Problem 7. Write Euclid I.18 and Euclid I.19 in shorthand, using Euler’s
notation for triangles. Explain how Euclid I.18 and Euclid I.19 are logically related.
Does Euclid I.19 follow from Euclid I.18 by pure logic? Why not?
Answer.
Euclid I.18 in shorthand: c > a ⇒ γ > α.
Euclid I.19 in shorthand: γ > α ⇒ c > a.
Euclid I.19 is the converse of Euclid I.18.
No, the converse does not follow purely by logic.
Figure 4: A pair of z-angles.
Proposition 4 (Congruent z-angles imply parallels). [Euclid I.27] If two lines
form congruent z-angles with a transversal, they are parallel.
10 Problem 8. For the statements below write a short remark to convince yourself
they are true, or tell they are not valid.
• X Referring to the pair of z-angles shown in the figure on page 7, proposition 4
tells that
α∼
=β⇒akb
Answer. Yes. The next two items make clear that this true.
• X This statement is equivalent to its contrapositive
a∦b⇒α∼
6= β
7
Answer. Yes, each statement is logically equivalent to its contrapositive.
• X The contrapositive relates even more directly to the exterior angle theorem.
Thus we see Proposition 4 is a consequence of the exterior angle theorem.
Answer. The intersecting lines a ∦ b intersect at point C = a ∩ b. The exterior
angular theorem for triangle 4ABC implies α β.
• X The exterior angle theorem is valid in neutral geometry: both in Euclidean and
hyperbolic geometry.
Answer. Yes. Indeed, it has been proved in neutral geometry.
• X ”For every line l and for every point P lying not on l, there exists at least one
parallel m to l through point P .” This statement is true in neutral geometry.
Answer. Yes. Because of the exterior angle theorem, the construction of the ”double perpendicular” gives a parallel.
• X ”For every line l and for every point P lying not on l, there exists exactly one
parallel m to l through point P .” This statement is true in any affine plane.
Answer. Yes. By definition, existence and uniqueness of a parallel is required for
an affine plane.
• The converse of Euclid I.27 holds in hyperbolic geometry, too.
Answer. No. The existence of multiple parallels in hyperbolic geometry confirms
that a k b does not imply α ∼
= β.
• X The converse of Euclid I.27 holds in Euclidean geometry.
Answer. Yes. The uniqueness of parallel implies that a k b ⇒ α ∼
= β.
• X The unique transport of an angle is postulated in neutral geometry,—independently
of whether parallels are unique.
Answer. Yes. That is part of axiom of congruence (III.4).
• X In Euclidean geometry, a common perpendicular of two parallel lines always
exists.
Answer. Yes. Indeed any perpendicular dropped from any point of line a onto the
parallel line b k a is perpendicular to both lines.
• X In hyperbolic geometry, a common perpendicular of two parallel lines exists if
and only if they do not contains asymptotic parallel rays.
Answer. Yes. That is proved in the part on hyperbolic geometry.
• X In neutral geometry, a common perpendicular of two parallel lines may exist or
not exist.
Answer. Yes. There is no common perpendicular of an asymptotic parallel ray
and the base line.
8
Euclidean geometry
10 Problem 9. For a right triangle with the angles 30◦ , 60◦ , 90◦ , the hypothenuse
has twice the length of the shorter leg. Give any convincing reason you want for this
fact.
Answer. Here are several possible answers;—a drawing instead of the exact explanation
would be acceptable, too. Note that one cannot get this simple result directly from
Pythagoras’ Theorem.
(i) Reflect the triangle across its longer leg. Together with the reflected image, one
obtains a triangle with three angles of 60◦ , which is known to be equilateral. Its
sides are congruent to the hypothenuse of the given triangle. The reflection axis
bisects one of the sides of the equilateral triangle. The shorter leg of the original
triangle is one half of this bisected side. Hence the shorter leg is half of the
hypothenuse.
(ii) We draw the semicircle with the hypothenuse AB as diameter. From Thales’
Theorem, we know that the vertex C with the right angle, and hence all three
vertices lie on the semicircle. Half of the hypothenuse and the shorter leg of the
given triangle are sides of an isosceles triangle 4OAC. This triangle has two
congruent base angles at vertices A and C, which we know to measure 60◦ . Hence
all three angles of triangle 4OAC measure 60◦ , and this triangle is equilateral.
For the original triangle, we see that the shorter leg AC is half of the hypothenuse
AB .
(iii) From the definition of the sin function, we see that sin 30◦ is the ratio of the shorter
leg across to the 30◦ angle to the hypothenuse. We know that sin 30◦ = 1/2. 1
Hence the shorter leg is half of the hypothenuse.
1
We see that this answer depends on previous knowledge of trigonometry.
9
10 Problem 10. Construct a right triangle with projections p = 1 and q = 3 of
the legs onto the hypothenuse. Use a construction based on Thales’ theorem and describe
your construction.
Figure 5: Construction of a right triangle with projections p = 1, q = 3.
Answer. We draw segments of the lengths as given, |AF | = q = 3 and |F B| = p = 1,
adjacent to each other on one line. Erect the perpendicular on line AB at point F . Draw
a semicircle with diameter AB. The semicircle and the perpendicular intersect at point
C. The triangle 4ABC is a right triangle with hypothenuse AB, and the projections
q = AF and F B = p have the lengths as required.
10
Figure 6: A triangle construction
10 Problem 11 (A construction using an altitude). Using Euclid III.21,
construct a triangle 4ABC with the following pieces given: side c = AB = 6, opposite
angle γ = ∠BCA = 60◦ , and altitude hc = 4.
(hc is the altitude dropped from vertex C onto the opposite side AB).
Do the construction, using compass and straightedge, and measure the angles α and β.
I want just a good drawing, no description is to be considered.
Answer.
11
Ratios of segments
10 Problem 12. Draw two circles of equal radii, the center of one lying on the circumference of the other one. Draw the line through the two centers. Use some of the intersection points you have just obtained, and draw two triangles with angles 30◦ , 30◦ , 120◦
of different sizes, in two different colors.
Let the lengths of sides of the larger triangle be a and c, and the lengths of the sides
of the smaller one be a0 and c0 . Determine the ratio aa0 = cc0 .
Figure 7: Two isosceles similar triangles with angles of 30◦ and 120◦ .
Answer. The line through the two centers A and B intersects the two circles in points D
and E, too. In the figure on page 12, these four points have the order A ∗ D ∗ E ∗ B. Let
C be an intersection point of the two circles. For example, we get two isosceles similar
triangles 4ABC ∼ 4ACD. Both have the base angles 30◦ and top angle 120◦ .
We determine the ratio aa0 = cc0 of the sizes of the two triangles. Since c0 = a and
c = 3a0 , we get
c
3a0
a
=
=
a0
c0
a
2
√
a
a
3
=
3
and
=
a02
a0
12
Figure 8: For the calculation of its area, any side of a triangle may be used as its base.
10 Problem 13. In triangle 4ABC, the altitudes are dropped from vertices A
and B and have the lengths ha and hb , respectively. Use similar triangles to show
hb
ha
=
b
a
Complete the paragraph below. For a proof of the proportion it is not possible to use the
area. It is the other way around, only because of the proportion the area is well defined!
For the triangle 4ABC, we can take side BC as base. The corresponding altitude
is AD, were D is the foot-point of the perpendicular dropped from vertex A onto side
BC.
As a second possibility, we can take side AC as base. The corresponding altitude
is BE, were E is the foot-point of the perpendicular dropped from vertex B onto side
AC.
Question. Which two right equiangular triangles have we obtained?
Answer. The triangles 4CAD and 4CBE are equiangular, and hence similar.
Question. Which proportion do we obtain, with the ratio sin γ?
Answer. We get the proportion
ha
|AD|
|BE|
hb
=
=
=
= sin γ
b
|AC|
|BC|
a
Question. Which formula for the area of the triangle in terms of a, b, γ can one obtain?
13
Answer. By multiplication with the denominators, we obtain
aha = |AD| · |BC| = |BE| · |AC| = bhb = ab sin γ
which is both the double area.
Remark. We see that the area of a triangle is well defined and equal to the product of
half base times height. It does not matter which side one chooses as base.
10 Problem 14. Given is the segment AB, for simplicity we assume it to have
length |AB| = 3. Provide a drawing and short explanations for the following:
• find the point X in the segment AB for which |AX| = 2|BX|;
−→
• find the point Y on the ray AB but outside the segment AB for which |AY | =
2|BY |;
• construct the Apollonius circle of all points Z for which |AZ| = 2|BZ|.
Figure 9: The points on this Apollonius circle have all the ratio 2 : 1 of distances from
points A and B.
Answer.
14
Figure 10: The common notation for a right triangle
Pythagoras’ group of Theorems
10 Problem 15. A right triangle has the projections p = 9 and q = 16 of the legs
onto the hypothenuse. What are the lengths of the two legs of the triangle. Use the leg
theorem to calculate exact expressions.
Answer. The leg theorem gives the squares a2 = (p + q)p = 225 and b2 = (p + q)q = 400.
Hence the lengths of the legs are a = 15 and b = 20.
15
Figure 11: Two equally good runners start at O and R. Do they better meet at point S or
point T .
10 Problem 16. Two equally fast runners start at the opposite points O and R
of the place shown in the figure on page 16. They are allowed to run across the place,
but cannot enter any space outside the place. They want to meet on the boundary. Can
they meet quicker at point S or at point T .
(i) Calculate the distance |OS| for which |OS| = |SR|.
16
10 Problem 17. Two equally fast runners start at the opposite points O and R
of the place shown in the figure on page 16. They are allowed to run across the place,
but cannot enter any space outside the place. They want to meet on the boundary. Can
they meet quicker at point S or at point T .
(ii) Let the distance |OT | = x. Calculate the coordinates of the meeting point T . and
distances |T C| and |CR|.
(iii) Set up the equation |OT | = |T C| + |CR| and check that it reduces to a linear
equation for x.
(iv) Calculate the distance |OT |,—at least numerically,—and decide which distance is
shorter, |OT | or |OS|.
Answer. (i) Let S = (x, 0) be the coordinates of the meeting point S. Since the distances |OS| = |SR| are equal, one calculates
p
x = (3 − x)2 + 22
x2 = 9 − 6x + x2 + 4
13
x=
6
(ii) Let
point are T =
√
√ the distance |OT | = x. The √coordinates of the meeting
2
2
2
( x − 1 , 1). We get |T C| = 2 − x − 1 and |CR| = 2. Since the distances
|OT | = |T C| + |CR| are equal, one calculates
√
√
x = 2 − x2 − 1 + 2
√
(x − 2 − 2)2 = x2 − 1
√
√
x2 − 4x − 2 2x + (2 + 2)2 = x2 − 1
√
√
2(2 + 2)x = (2 + 2)2 + 1
(iii) We get the distance OT
√
√
√
√
2+ 2
1
2+ 2 2− 2
6+ 2
√ =
x=
+
+
=
= 1.85355
2
2
4
4
2(2 + 2)
Since |OS| = 2.1667, we see that point T is where to meet quicker.
17
Figure 12: Two equally good runners start at O and R. Construction of the meeting points
S and T .
10 Problem 18. Continuing the last problem, I have shown a purely geometrical
solution. Describe and justify the construction of the two possible meeting points S and
T done in the figure on page 18.
Answer. The point S is the intersection of the perpendicular bisector of OR with the
lower horizontal boundary of the place.
To construct the point T , we extend the middle horizontal boundary of the place,
starting at the corner C by a segment CE ∼
= CR . The point T is the intersection of
the middle horizontal boundary of the place with the perpendicular bisector of OE.
18
Figure 13: Another proof of the parallelogram equation.
10 Problem 19 (Parallelogram equation). The sum of the squares of the
diagonals of a parallelogram equals the sum of the squares of its four sides. Complete
the proof of the parallelogram equation
|AC|2 + |BD|2 = 2|AB|2 + 2|BC|2
using the case and the notation of the figure on page 19.
Proof. For the parallelogram ABCD, we may assume a = |AB| > |BC| = b, and the
angle at B to be obtuse. We drop the perpendiculars from vertex C onto the extension
of side AB, and from vertex B onto side CD. The foot points are F and G, respectively.
Let the rectangle BF CG have side lengths |BG| = |CF | = h and |BF | = |CG| = k.
Apply Pythagoras’ Theorem three times:
|AC|2 = |AF |2 + |F C|2 = (a + k)2 + h2
|BD|2 = |BG|2 + |GD|2 = h2 + (a − k)2
h2 + k 2 = |BF |2 + |F C|2 = |BC|2 = b2
for triangle 4AF C;
for triangle 4BGD;
for triangle 4BF C.
Adding the first two equations yields
|AC|2 + |BD|2 = (a + k)2 + (a − k)2 + 2h2 = 2a2 + 2k 2 + 2h2 = 2a2 + 2b2
We see that the sum of the squares of the diagonals of parallelogram ABCD equals
the sum of the squares of its four sides.
19
10 Problem 20. A parallelogram has sides of length 3 and 4, and one diagonal
has length 6.
• Construct the parallelogram and measure its second diagonal.
• Calculate the length of the second diagonal exactly.
Answer. The construction is shown in the figure on page 20. One measures that the
second diagonal has length about 3.7. By the parallelogram equation the length x of
Figure 14: A parallelogram with sides 3 and 4 and one diagonal of length 6.
the second diagonal satisfies 2 · 32 + 2 · 42 = 62 + x2 and hence x2 = 14 and x =
20
√
14.
Trigonometry
10 Problem 21. For a triangle are given
γ = 45◦ ,
c
4
= , α < 90◦ and b = 2 000
a
5
Calculate the angles α, β, and sides a and c.
Answer. The sin Theorem yields
sin α =
a sin γ
5
= √ = 0.8839
c
4 2
We get α = 62.11◦ since this angle is acute. The angle β is obtained from the angle
sum. One gets
β = 180◦ − α − γ = 72.89◦
Now we determine a and c by the sin theorem:
b
5
2000
= √ ·
= 1849.7
sin β
4 2 sin 72.89◦
b
1
2000
c = sin γ
=√ ·
= 1479.8
sin β
2 sin 72.89◦
a = sin α
21