Math 3181 Name:Dr. Franz RotheMay 13, 2016

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Solution of Final

10 Problem 1 (Building up geometry). For an axiomatically built-up geom-etry, 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 Order Congruence Parallelism Circles Continuityincidence yes no no no no 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

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Incidence and order

10 Problem 2. Give exact definitions of the terms supplementary angles andvertical 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 othersides are the opposite rays on a line.

Definition 2 (Vertical Angles). Two angles are called vertical angles, iff they have acommon 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 isthe interior of the vertical angle corresponding to the given angle ∠(h, k).

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Proposition 1 (About n points in a plane). Among any n ≥ 3 points lying in anordered incidence plane but not on a line, there exist three points P,Q,R such that alln points lie in the interior or on the boundary of the angle ∠PQR.

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, hencethe angle ∠PQR 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 intoan ordered list P1 ∗ P2 ∗ · · · ∗ Pn. By assumption point Pn+1 does not lie on theline l. We put P := P1, the first item in the list; R := Pn, the !! last item in thelist; and choose Q := Pn+1 as !! vertex of the angle. Thus all n + 1 points lie!! inside or on the boundary of the angle ∠PQR.

(ii) The points P1 . . . Pn do not lie on a line l. By the induction assumption, there arepoints A,B,C among them such that points P1 . . . Pn lie in the interior or on theboundary of the angle ∠ABC.

• In case that the extra point Pn+1 lies in the interior or on the boundary ofthe angle ∠ABC, we are ready.

• In case that the extra point Pn+1 lies in the interior of a supplementaryangle of the angle ∠ABC, all points P1 . . . Pn+1 lie in the interior or on theboundary of either angle ∠ABPn+1 or angle ∠Pn+1BC.

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Figure 1: Any n points lie in the interior or on the boundary of an angle.

• We choose any points A ∗B ∗A′ and C ∗B ∗C ′. Assume that the extra point

Q := Pn+1 lies in the interior of the vertical angle ∠A′BC ′ or on the ray−−→BA′.

This case in shown in the figure on page 4. We draw the segments PiPn+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 ofAB as point Pn+1, all these segments !! intersect the line !! AB .

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+1S1, and

R be a point among P1 . . . Pn on the !! last ray−−−−→Pn+1Sn.

Now all points P1 . . . Pn+1 lie in the interior or on the boundary of the angle∠PQR.

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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 bisectsthe angle into two congruent angles is called the angular bisector.

Construction 1 (Construction of the angular bisector). One transfers two congru-ent segments AB and AC onto the two sides of the angle, both starting from the vertexA of the angle. The perpendicular, dropped from the vertex A onto the segment BC, isthe angular bisector.

Figure 2: The angular bisector

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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 notationfrom your proof.

Answer. In 4ABC, we assume for sides AB and BC that c = AB > BC = a. Theissue 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. Thusone gets a segment BD ∼= BC, with point D between B and A. Because the 4BCD isisosceles, 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 αacross 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

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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’snotation 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 linesform congruent z-angles with a transversal, they are parallel.

10 Problem 8. For the statements below write a short remark to convince yourselfthey are true, or tell they are not valid.

• X Referring to the pair of z-angles shown in the figure on page 7, proposition 4tells that

α ∼= β ⇒ a ‖ b

Answer. Yes. The next two items make clear that this true.

• X This statement is equivalent to its contrapositive

a ∦ b⇒ α 6∼= β

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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 exteriorangular theorem for triangle 4ABC implies α � β.

• X The exterior angle theorem is valid in neutral geometry: both in Euclidean andhyperbolic 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 oneparallel 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 ”dou-ble perpendicular” gives a parallel.

• X ”For every line l and for every point P lying not on l, there exists exactly oneparallel 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 foran affine plane.

• The converse of Euclid I.27 holds in hyperbolic geometry, too.

Answer. No. The existence of multiple parallels in hyperbolic geometry confirmsthat a ‖ 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 ‖ b⇒ α ∼= β.

• X The unique transport of an angle is postulated in neutral geometry,—independentlyof 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 alwaysexists.

Answer. Yes. Indeed any perpendicular dropped from any point of line a onto theparallel line b ‖ a is perpendicular to both lines.

• X In hyperbolic geometry, a common perpendicular of two parallel lines exists ifand 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 ornot exist.

Answer. Yes. There is no common perpendicular of an asymptotic parallel rayand the base line.

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Euclidean geometry

10 Problem 9. For a right triangle with the angles 30◦, 60◦, 90◦, the hypothenusehas twice the length of the shorter leg. Give any convincing reason you want for thisfact.

Answer. Here are several possible answers;—a drawing instead of the exact explanationwould be acceptable, too. Note that one cannot get this simple result directly fromPythagoras’ Theorem.

(i) Reflect the triangle across its longer leg. Together with the reflected image, oneobtains a triangle with three angles of 60◦, which is known to be equilateral. Itssides are congruent to the hypothenuse of the given triangle. The reflection axisbisects one of the sides of the equilateral triangle. The shorter leg of the originaltriangle is one half of this bisected side. Hence the shorter leg is half of thehypothenuse.

(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 threevertices lie on the semicircle. Half of the hypothenuse and the shorter leg of thegiven triangle are sides of an isosceles triangle 4OAC. This triangle has twocongruent base angles at vertices A and C, which we know to measure 60◦. Henceall 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 hypothenuseAB .

(iii) From the definition of the sin function, we see that sin 30◦ is the ratio of the shorterleg 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.

1We see that this answer depends on previous knowledge of trigonometry.

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10 Problem 10. Construct a right triangle with projections p = 1 and q = 3 ofthe legs onto the hypothenuse. Use a construction based on Thales’ theorem and describeyour 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 |FB| = p = 1,adjacent to each other on one line. Erect the perpendicular on line AB at point F . Drawa semicircle with diameter AB. The semicircle and the perpendicular intersect at pointC. The triangle 4ABC is a right triangle with hypothenuse AB, and the projectionsq = AF and FB = p have the lengths as required.

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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, oppositeangle 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.

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Ratios of segments

10 Problem 12. Draw two circles of equal radii, the center of one lying on the cir-cumference of the other one. Draw the line through the two centers. Use some of the in-tersection 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 sidesof the smaller one be a′ and c′. Determine the ratio a

a′= c

c′.

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 Dand E, too. In the figure on page 12, these four points have the order A∗D ∗E ∗B. LetC be an intersection point of the two circles. For example, we get two isosceles similartriangles 4ABC ∼ 4ACD. Both have the base angles 30◦ and top angle 120◦.

We determine the ratio aa′

= cc′

of the sizes of the two triangles. Since c′ = a andc = 3a′, we get

a

a′=c

c′=

3a′

aa2

a′2= 3 and

a

a′=√

3

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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 Aand B and have the lengths ha and hb, respectively. Use similar triangles to show

hab

=hba

Complete the paragraph below. For a proof of the proportion it is not possible to use thearea. 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 altitudeis AD, were D is the foot-point of the perpendicular dropped from vertex A onto sideBC.

As a second possibility, we can take side AC as base. The corresponding altitudeis BE, were E is the foot-point of the perpendicular dropped from vertex B onto sideAC.

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

hab

=|AD||AC|

=|BE||BC|

=hba

= sin γ

Question. Which formula for the area of the triangle in terms of a, b, γ can one obtain?

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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 ofhalf 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 havelength |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 frompoints A and B.

Answer.

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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 legsonto the hypothenuse. What are the lengths of the two legs of the triangle. Use the legtheorem 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.

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Figure 11: Two equally good runners start at O and R. Do they better meet at point S orpoint T .

10 Problem 16. Two equally fast runners start at the opposite points O and Rof 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. Canthey meet quicker at point S or at point T .

(i) Calculate the distance |OS| for which |OS| = |SR|.

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10 Problem 17. Two equally fast runners start at the opposite points O and Rof 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. Canthey meet quicker at point S or at point T .

(ii) Let the distance |OT | = x. Calculate the coordinates of the meeting point T . anddistances |TC| and |CR|.

(iii) Set up the equation |OT | = |TC| + |CR| and check that it reduces to a linearequation for x.

(iv) Calculate the distance |OT |,—at least numerically,—and decide which distance isshorter, |OT | or |OS|.

Answer. (i) Let S = (x, 0) be the coordinates of the meeting point S. Since the dis-tances |OS| = |SR| are equal, one calculates

x =√

(3− x)2 + 22

x2 = 9− 6x+ x2 + 4

x =13

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(ii) Let the distance |OT | = x. The coordinates of the meeting point are T =(√x2 − 12, 1). We get |TC| = 2 −

√x2 − 1 and |CR| =

√2. Since the distances

|OT | = |TC|+ |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

x =2 +√

2

2+

1

2(2 +√

2)=

2 +√

2

2+

2−√

2

4=

6 +√

2

4= 1.85355

Since |OS| = 2.1667, we see that point T is where to meet quicker.

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Figure 12: Two equally good runners start at O and R. Construction of the meeting pointsS and T .

10 Problem 18. Continuing the last problem, I have shown a purely geometricalsolution. Describe and justify the construction of the two possible meeting points S andT done in the figure on page 18.

Answer. The point S is the intersection of the perpendicular bisector of OR with thelower 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 ofthe middle horizontal boundary of the place with the perpendicular bisector of OE.

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Figure 13: Another proof of the parallelogram equation.

10 Problem 19 (Parallelogram equation). The sum of the squares of thediagonals of a parallelogram equals the sum of the squares of its four sides. Completethe 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 theangle at B to be obtuse. We drop the perpendiculars from vertex C onto the extensionof side AB, and from vertex B onto side CD. The foot points are F and G, respectively.Let the rectangle �BFCG have side lengths |BG| = |CF | = h and |BF | = |CG| = k.Apply Pythagoras’ Theorem three times:

|AC|2 = |AF |2 + |FC|2 = (a+ k)2 + h2 for triangle 4AFC;

|BD|2 = |BG|2 + |GD|2 = h2 + (a− k)2 for triangle 4BGD;

h2 + k2 = |BF |2 + |FC|2 = |BC|2 = b2 for triangle 4BFC.

Adding the first two equations yields

|AC|2 + |BD|2 = (a+ k)2 + (a− k)2 + 2h2 = 2a2 + 2k2 + 2h2 = 2a2 + 2b2

We see that the sum of the squares of the diagonals of parallelogram �ABCD equalsthe sum of the squares of its four sides.

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10 Problem 20. A parallelogram has sides of length 3 and 4, and one diagonalhas 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 thesecond 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 =√

14.

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Trigonometry

10 Problem 21. For a triangle are given

γ = 45◦ ,c

a=

4

5, α < 90◦ and b = 2 000

Calculate the angles α, β, and sides a and c.

Answer. The sin Theorem yields

sinα =a sin γ

c=

5

4√

2= 0.8839

We get α = 62.11◦ since this angle is acute. The angle β is obtained from the anglesum. One gets

β = 180◦ − α− γ = 72.89◦

Now we determine a and c by the sin theorem:

a = sinαb

sin β=

5

4√

2· 2000

sin 72.89◦= 1849.7

c = sin γb

sin β=

1√2· 2000

sin 72.89◦= 1479.8

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