Math 3181 Name:Dr. Franz RotheMay 12, 2014

All3181\3181_spr14fs.tex

Use the back pages for extra space18 problems are required for full credit

Solution of Final

10 Problem 1. 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 2 (Another triangle). Construct a triangle with angles of 30◦, 45◦

and 105◦, using only compass and straightedge, but no protractor. Describe your con-struction.

Figure 1: Construction of a triangle with angles of 45◦, 30◦ and 105◦.

Answer. On an arbitrary segment OB, an equilateral triangle 4OBD is erected. Asdescribed in Euclid I.1, D is an intersection point of a circle around B through O witha circle around O through B. Let A be the second endpoint of diameter AOB.

At point O, we erect the perpendicular onto diameter AB. Let E be the intersectionof the perpendicular with the left circle, lying on the same side of AB as point D. The

rays−−→AD and

−−→BE intersect in point C.

We thus get a triangle 4ABC with the angles 30◦, 45◦ and 105◦ at its vertices A,Band C. The angle ∠BAD = 30◦ since triangle 4OBD is equilateral and ∠ADB = 90◦.The angle ∠ABC = 45◦ is obtained from the right isosceles triangle 4OBE. Finally,one calculates the third angle ∠BCA = 180◦ − 30◦ − 45◦ = 105◦ by the angle sum oftriangle 4ABC.

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10 Problem 3. Construct a right triangle with projections p = 5 and q = 4 of thelegs onto the hypothenuse. Use Thales’ theorem. Describe your construction. Measurethe lengths of the two legs of your triangle.

Figure 2: Construction of a right triangle with projections p = 5, q = 4.

Answer. Adjacent to each other on one line, we draw segments with the given lengths|AF | = q = 4 and |FB| = p = 5. We erect the perpendicular on line AB at point F anddraw a semicircle with diameter AB. The semicircle and the perpendicular intersectat point C. The triangle 4ABC is a right triangle with hypothenuse AB, and theprojections q = AF and FB = p have the lengths as required.I measure the two legs as a = 6.7 and b = 6.

10 Problem 4. What are the lengths of the two legs of the triangle from the lastproblem. Use the leg theorem to calculate exact expressions.

Answer. The leg theorem gives the squares a2 = (p + q)p = 45 and b2 = (p+ q)q = 36.Hence the lengths of the legs are

a =√

45 = 3√

5 and b =√

36 = 6

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Angles in a circle

10 Problem 5 (A construction using an altitude). Using Euclid III.21,construct a triangle 4ABC with the three given pieces: side c = |AB| = 6, oppositeangle γ = ∠BCA = 30◦, and altitude hc = 4.(hc is the altitude dropped from vertex C onto the opposite side AB).

Question (a). Do the construction as exact as possible.

Figure 3: A triangle construction

Answer.

Question (b). Estimate the length of the shortest side of the triangle.

© between 3.0 and 4.0 X © between 4.0 and 5.0

© between 5.0 and 6.0 © between 6.0 and 7.0

Question (c). Describe the steps for your construction.

Answer. Draw side AB = 6 and its perpendicular bisector p. Let M be the midpointof AB. The center O of the circum circle lies on the perpendicular bisector. Thecenter angle is double the circumference angle γ. Hence ∠AOB = 2γ = 60◦, and∠AOM = 30◦. For the example given, the point O is especially easy to find becausethe 4AOB is equilateral. Next, we draw the circle around O through A and B. Thisis the circum circle of 4ABC, on which vertex C lies.

Secondly, vertex C lies on a parallel q to AB of distance |MD| = hc = 4, because ofthe given altitude hc = 4. Hence vertex C is an intersection point of this parallel withthe circum circle. One may choose any one of these two intersection points.

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A proof of commutativity of segment arithmetic

Figure 4: Prove commutativity ab = ba.

The following proof of commutativity of segment arithmetic does not refer to Pappus’

theorem. Instead, we use angles in a circle directly. Indeed, only one circle is needed. We usemeasurements along the horizontal and vertical axes crossing at point O perpendicularly.The given positive segment |OA| = a > 0 is transferred to the vertical axis. Thesegments |OE| = 1 and |OB| = b > 0 are transferred to the horizontal axis.

10 Problem 6. Answer the following questions:

Question (a). Explain how the product ab has been constructed.

Answer. We draw the line through B parallel to EA, and get its intersection point withthe vertical axis at point ab with distance |Oab| = ab from the origin. As next step, thetriangle 4OAE is reflected across the −45◦ line.

Question (b). Explain why, because of these constructions, there appear three congruentangles.

Answer. The angles ∠OAE ∼= ∠OabB are congruent because the vertical axis traversesa pair of parallel lines. The angles ∠OAE ∼= ∠OA′E ′ are congruent by SAS congruenceof the corresponding triangles. We have marked the three congruent angles by α.

Question (c). We now draw a circle through points A′, E ′ and B. Explain why thiscircle goes through the point ab.

Answer. The angles ∠E ′A′B ∼= ∠E ′abB are congruent, and lie both on the same sideof E ′B. They are hence circumference angles in the circle drawn. Hence ab lies on thiscircle. The triangle 4OBE ′ is reflected across the +45◦ line, as needed in the next step.

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Question (d). Explain how the product ba has to be constructed, in the left upperquadrant and complete the drawing.

Answer. We draw the line through A′ parallel to E ′′B′, and get its intersection pointwith the vertical axis at point ba with distance |Oba| = ba from the origin.

Question (e). Explain why, because of these constructions, there appear three congruentangles.

Answer. The angles ∠OB′E ′′ ∼= ∠ObaA′ are congruent because the vertical axis tra-verses a pair of parallel lines. The angles ∠OBE ′ ∼= ∠OB′E ′′ are congruent because ofSAS congruence of the corresponding triangles. We have marked the three congruentangles by β.

Question (f). Explain why the circle through A′, E ′ and B—as already drawn—goesthrough the point ba.

Answer. The angles ∠E ′BA′ ∼= ∠E ′baA′ are congruent, and lie both on the same sideof E ′A′. They are hence circumference angles in the circle drawn. Hence ba lies on this

circle. Since the circle has only one intersection point with the vertical ray−→OA, and we

have obtained that both ab and ba is this intersection point, we conclude that ab = ba.2

Figure 5: Commutativity ab = ba follows from both lying on the circle.

2In some former drawing—which is not from Leibniz’ ”best of all worlds”—the construction doesnot come out right. Would Leibniz have believed the proof or the construction?

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Theorems of Pappus

10 Problem 7 (Theorem of Pappus). Formulate the Theorem of Pappus andexplain:

• Explain in general words what the Pappus configuration is.

• Provide a drawing, using a straightedge.

• State the Theorem of Pappus in parallel setting. What does it say about three pairsof opposite sides.

• Name the points of the relevant hexagon in your drawing.

• Restate the Theorem of Pappus using the names of the points as you have put theminto your drawing.

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Answer. • We call the system of two intersecting lines, with three points on both ofthem, and the M-W hexagon zig-zaging between those triplets the configurationof Pappus.

Figure 6: For Pappus’ configuration: If BC ′ ‖ B′C and AC ′ ‖ A′C, then AB′ ‖ A′B.

• If the hexagon of Pappus’ configuration contains two pairs of opposite parallelsides, the third pair of opposite sides is parallel, too.

• In the figure on page 8, we have two lines intersecting at point O, we have putthree points A,B,C on the horizontal line, and three points A′, B′, C ′ on the otherline

• there are three pairs of lines

AB′ and A′B

BC ′ and B′C

C ′A and CA′

the parallelism of which has to be investigated. With the names from the figureon page 8, we state:

Theorem 1 (Pappus Theorem). Let A,B,C and A′, B′, C ′ be both three pointson two intersecting lines, all different from the intersection point. If the lines BC ′

and B′C are parallel, and the lines AC ′ and A′C are parallel, then the lines AB′

and A′B and parallel, too.

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Figure 7: Explain the Theorem of Pappus, in the general setting.

Theorem 2 (Pappus’ Theorem in general setting). Let A,B,C and A′, B′, C ′

be three points on two intersecting lines, respectively. The three intersection pointsX = BC ′ ∩B′C, Y = AC ′ ∩A′C and Z = AB′ ∩A′B of three pairs of opposite sides ofhexagon AB′CA′BC ′ lie on a line.

10 Problem 8. Mark the three points X, Y, Z and put the line through them intothe figure on page 9.

Answer.

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Figure 8: The Theorem of Pappus in the general setting.

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Similarity

We use Euler’s notation: angle α lies opposite to side a, angle β lies opposite to side b,and angle γ lies opposite to side c.

10 Problem 9. For any two triangles it is assumed

α = α′ ,b

c=b′

c′

Are the two triangles always similar or not? Depending on what is appropriate, giveexamples, counterexamples, or a reason based on my script, the lecture, or Euclid’s bookVI.

Answer. As stated in the section about similar triangles, Euclid VI.6 tells us:If two triangles have one pair of congruent angles, and the sides containing these

pairs are proportional, then the triangles are similar.For the present example, the angles α = α′ are congruent. The sides adjacent to

angle α are b and c, and to angle α′ are b′ and c′. These sides are proportional for thetwo triangles due to the assumption b

c= b′

c′. Hence Euclid VI.6 tells the two triangles

are similar.

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10 Problem 10. 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

Figure 9: For the calculation of its area, any side of a triangle may be used as its base.

Answer. For the triangle 4ABC, we can take side BC as base. The correspondingaltitude is AD, were D is the foot-point of the perpendicular dropped from vertex Aonto side BC.

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.

The two triangles 4CAD and 4CBE are equiangular, and hence similar. We getthe proportion

hab

=|AD||AC|

=|BE||BC|

=hba

= sin γ

Remark 1. By multiplication with the denominators, we obtain

aha = |AD| · |BC| = |BE| · |AC| = bhb = ab sin γ

which is both the double area. We see that the area of a triangle is well defined. For theproduct of half base times height, it does not matter which side one chooses as base.

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10 Problem 11. A candle is standing at distance a = 3 from a lens with focuslength f = 2. At which distance b behind the lens has one to put a screen to get a sharpimage of the candle flame. Give a construction and calculate the distance b.

Answer. The construction is shown in the figure on page 13. One measures that aboutb = 6.

Figure 10: Construct the appropriate position of the screen.

From the lens equation one gets 1/b = 1/f − 1/a = 1/2− 1/3 = 1/6 and hence b = 6 isthe appropriate distance from lens to screen.

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Trigonometry

10 Problem 12. For a triangle are given

γ = 45◦ ,c

a=

4

5, α < 90◦ and b = 20

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· 20

sin 72.89◦= 18.497

c = sin γb

sin β=

1√2· 20

sin 72.89◦= 14.798

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Figure 11: The common notation for a right triangle

Figure 12: How does Euclid III.36 imply the leg theorem?

Pythagoras group

Theorem 3 (Theorem of chord and tangent (Euclid III.36)). From a pointoutside a circle, a tangent and a second are drawn. The square of the tangent segmentequals the product of the segments on the chord, measured from the point outside to thetwo intersection points with the circle.

10 Problem 13 (From Euclid III.36 to the leg theorem). Complete thedrawing on page 15, and explain how you get the leg theorem a2 = pc.

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Figure 13: The leg theorem.

Answer. We choose point A to be the other endpoint of diameter CA. We see that4ACB is a right triangle, because tangent and radius are !! perpendicular to eachother. The segment AB intersects the circle in a second point F . The triangle 4ACBhas altitude CF , because !! Thales’ theorem shows that the angle !! ∠AFCis right. Now Euclid III.36 tells that the square of the leg CB equals the product of thehypothenuse BA time the projection BF of that leg onto the hypothenuse. One getsthe statement in its usual form a2 = pc.

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Figure 14: How does Euclid III.36 imply the Pythagorean Theorem?

10 Problem 14 (From Euclid III.36 to the Pythagorean Theorem). Com-plete the drawing on page 17, and explain how you get the Pythagorean Theorem.

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Figure 15: The Pythagorean Theorem

Answer. We draw the ray−−→BO from the point B outside the given circle through the

center O. We put A = O. We get a right angle at the touching point C of the tangentsince tangent and radius are !! perpendicular to each other. Let points E and Dbe the endpoints of the diameter on the ray already drawn. Now Euclid III.36, as givenin theorem 3, tells that

|BC|2 = |BD| · |BE|

In terms of the sides of the triangle 4ABC, this shows that

a2 = (c+ b)(c− b) = c2 − b2 and a2 + b2 = c2

We got the Pythagorean Theorem.

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Figure 16: Two equally good runners start at O and R. Do they better meet at point S orpoint T .

10 Problem 15. Two equally fast runners start at the opposite points O and Rof the place shown in the figure on page 19. 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|.

(ii) Calculate the point T such that |OT | = |TC|+ |CR|.

(iii) Calculate the distance |OT | and decide whether which distance is shorter |OT | or|OS|.

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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 T = (x, 1) be the coordinates of the meeting point T . Since the distances|OT | = |TC|+ |CR| are equal, one calculates

√x2 + 12 = 2− x+

√2

x2 + 1 = (2 +√

2)2 − 2(2 +√

2)x+ x2

2(2 +√

2)x = (2 +√

2)2 − 1

x =2 + 3

√2

4= 1.56066

(iii) We calculate the distance 3

|OT | = |TC|+ |CR| = (2− x) +√

2 = 1.85355

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

10 Problem 16. 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 21.

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.

3Note that it is not sufficient to just compare the values of x from parts (i) and (ii).

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Figure 17: Two equally good runners start at O and R. Construction of the meeting pointsS and T .

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10 Problem 17. 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 22. One measures that thesecond diagonal has length about 3.7. By the parallelogram equation the length x of

Figure 18: 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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10 Problem 18. In the coordinate plane are given: the circle with radius 3 andcenter (0, 0) and the line through points P = (−1, 0) and Q = (0, 1).

1. What are the equations for the circle and the line.

2. What are the coordinates (x, y) for the intersection points of the circle and theline.

3. Give the exact root expressions for both coordinates of one intersection point. Forthe construction of which regular polygon could this be a useful first step?

Answer. The equations for the circle and the line are

x2 + y2 = 9 and y = x+ 1

To get the coordinates (x, y) of the intersection points of the circle and the line, we plugy = x+ 1 into the equation of the circle and get

x2 + (x+ 1)2 = 9

2x2 + 2x− 8 = 0

x2 + x− 4 = 0

x1,2 =−1±

√17

2

The two intersection points have coordinates(−1−

√17

2,1−√

17

2

)and

(−1 +

√17

2,1 +√

17

2

)

These values could be used as a first step in a construction of a regular 17-gon.

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