Solution Manual for Foundations of Geometry, 2nd Edition

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’SSOLUTIONSMANUALFOUNDATIONS OFGEOMETRYSECONDEDITIONGerard A. VenemaCalvin College

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ContentsSolutions to Exercises in Chapter 11Solutions to Exercises in Chapter 26Solutions to Exercises in Chapter 314Solutions to Exercises in Chapter 427Solutions to Exercises in Chapter 550Solutions to Exercises in Chapter 672Solutions to Exercises in Chapter 786Solutions to Exercises in Chapter 896Solutions to Exercises in Chapter 10106Solutions to Exercises in Chapter 11127Solutions to Exercises in Chapter 12137Bibliography143ii

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Solutions to Exercises in Chapter 11.6.1Check that the formulaAD14.aCc/.bCd/works for rectangles but not forparallelograms.bdacabdcFIGURE S1.1:Exercise 1.6.1. A rectangle and a parallelogramFor rectangles and parallelograms,aDcandbDdandArea = base∗height.For a rectangle, the base and the height will be equal to the lengths of two adjacentsides. ThereforeADa∗dD12.aCa/∗12.bCb/D14.aCc/.bCd/In the case of a parallelogram, the height is smaller than the length of the side sothe formula does not give the correct answer.1.6.2The area of a circle is given by the formulaADπ.d2/2. According the Egyptians,Ais also equal to the area of a square with sides equal to89d; thusAD.89/2d2.Equating and solving forπgivesπD.89/2d214d2D648114D25681L3:160494:1.6.3The sum of the measures of the two acute angles in^ABCis 90◦, so the first shadedregion is a square. We must show that the area of the shaded region in the firstsquare.c2/is equal to the area of the shaded region in the second square.a2Cb2/.The two large squares have the same area because they both have side lengthaCb.Also each of these squares contains four copies of triangle^ABC(inwhite). Therefore, by subtraction, the shadesd regions must have equal area and soa2Cb2Dc2.1.6.4(a)SupposeaDu2−v2,bD2uvandcDu2Cv2.We must show thata2Cb2Dc2. First,a2Cb2D.u2−v2/2C.2uv/2Du4−2u2v2Cv4C4u2v2Du4C2u2v2Cv4and, second,c2D.u2Cv2/2Du4C2u2v2Cv4Du4C2u2v2Cv4. Soa2Cb2Dc2.(b)Letuandvbe odd. We will show thata,bandcare all even. Sinceuandvareboth odd, we know thatu2andv2are also odd. ThereforeaDu2−v2is even(the difference between two odd numbers is even). It is obvious thatbD2uvis even, andcDu2Cv2is also even since it is the sum of two odd numbers.(c)Suppose one ofuandvis even and the other is odd. We will show thata; b,andcdo not have any common prime factors. Nowaandcare both odd, so 2is not a factor ofaorc. SupposexZ2 is a prime factor ofb. Then eitherxdividesuorxdividesv, but not both becauseuandvare relatively prime. Ifxdividesu, then it also dividesu2but notv2. Thusxis not a factor ofaorc.1

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2Solutions to Exercises in Chapter 1Ifxdividesv, then it dividesv2but notu2. Againxis not a factor ofaorc.Therefore.a; b; c/is a primitive Pythagorean triple.1.6.5LethCxbe the height of the entire (untruncated) pyramid. We know thathCxxDab(by the Similar Triangles Theorem), soxDhba−b(algebra). The volume of thetruncated pyramid is the volume of the whole pyramid minus the volume of the toppyramid. ThereforeVD13.hCx/a2−13xb2D13.hChba−b /a2−13h.b3a−b /Dh3.a2Ca2ba−b /−h3.b3a−b /Dh3.a2Ca2b−b3a−b/Dh3.a2C.a−b/.abCb2/.a−b//Dh3.a2CabCb2/:aabbhxFIGURE S1.2:Exercise 1.6.5. A truncated pyramid.1.6.6Constructions using a compass and a straightedge.There are numerous ways inwhich to accomplish each of these constructions; just one is indicated in each case.(a)The perpendicular bisector of a line segmentAB.Using the compass, construct two circles, the first aboutAthroughB, thesecond aboutBthroughA.Then use the straightedge to construct a linethrough the two points created by the intersection of the two circles.(b)A line through a pointPperpendicular to a line`.Use the compass to construct a circle about P, making sure the circle is bigenough so that it intersects`at two points,AandB.Then construct theperpendicular bisector of segmentABas in part (a).

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Solutions to Exercises in Chapter 13ABFIGURE S1.3:Exercise 1.6(a) Construction of a perpendicular bisectorPABℓFIGURE S1.4:Exercise 1.6.6(b) Construction of a line through P, perpendicular to`(c)The angle bisector ofjBAC.Using the compass, construct a circle aboutAthat intersectsABandAC.Call those points of intersectionDandErespectively.Then construct theperpendicular bisector ofDE. This line is the angle bisector.1.6.7(a)No. Euclid’s postulates say nothing about the number of points on a line.(b)No.(c)No. The postulates only assert that there is a line; they do not say there is onlyone.1.6.8The proof of Proposition 29.1.6.9LetnABCDbe a rhombus (all four sides are equal), and letEbe the point ofintersection betweenACandDB.1We must show that^AEB>^CEB>^CED>^AED. NowjBAC>jACBandjCAD>jACDby Proposition 5. By additionwe can see thatjBAD>jBCDand similarly,jADC>jABC. Now we know that^ABC>^ADCby Proposition 4. Similarly,^DBA>^DBC. This implies thatjBAC>jDAC>jBCA>jDCAandjBDA>jDBA>jBDC>jDBC.1In this solution and the next, the existence of the pointEis taken for granted. Its existenceis obvious from the diagram. Proving thatEexists is one of the gaps that must be filled in theseproofs. This point will be addressed in Chapter 6.

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4Solutions to Exercises in Chapter 1ABCDEFIGURE S1.5:Exercise 1.6.6(c) Construction of an angle bisectorThus^AEB>^AED>^CEB>^CED, again by Proposition 4.2ABCDEFIGURE S1.6:Exercise 1.6.9 RhombusnABCD1.6.10LetnABCDbe a rectangle, and letEbe the point of intersection ofACandBD. Wemust prove thatAC>BDand thatACandBDbisect each other (i.e.,AE>ECandBE>ED).By Proposition 28,('DA‖('CBand('DC‖('AB.Therefore, byProposition 29,jCAB>jACDandjDAC>jACB. Hence^ABC>^CDAand^ADB>^CBDby Proposition 26 (ASA). Since those triangles are congruentwe know that opposite sides of the rectangle are congruent and^ABD>^BAC(by Proposition 4), and thereforeBD>AC.Now we must prove that the segments bisect each other.By Proposition 29,jCAB>jACDandjDBA>jBDC. Hence^ABE>^CDE(by Proposition 26)which implies thatAE>CEandDE>BE. Therefore the diagonals are equal andbisect each other.1.6.11The argument works for the first case. This is the case in which the triangle actuallyis isosceles. The second case never occurs (Dis never inside the triangle). The flawlies in the third case (Dis outside the triangle). If the triangle is not isosceles theneitherEwill be outside the triangle andFwill be on the edgeAC, orEwill be onthe edgeABandFwill be outside. They cannot both be outside as shown in thediagram. This can be checked by drawing a careful diagram by hand or by drawingthe diagram using GeoGebra (or similar software).2It should be noted that the fact about rhombi can be proved using just propositions that comeearly in Book I and do not depend on the Fifth Postulate, whereas the proof in the next exerciserequires propositions about parallelism that Euclid proves much later in Book I using his FifthPostulate.

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Solutions to Exercises in Chapter 15ABCDEFIGURE S1.7:Exercise 1.6.10 RectanglenABCD

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Solutions to Exercises in Chapter 22.4.1This is not a model for Incidence Geometry since it does not satisfy IncidenceAxiom 3. This example is isomorphic to the 3-point line.2.4.2One-point Geometry satisfies Axioms 1 and 2 but not Axiom 3.Every pair ofdistinct points defines a unique line (vacuously—there is no pair of distinct points).There do not exist three distinct points, so there cannot be three noncollinear points.One-point Geometry satisfies all three parallel postulates (vacuously—there is noline).2.4.3It helps to draw a schematic diagram of the relationships.(a)(b)(c)ABCABCABCDDDFIGURE S2.1:A schematic representation of the committee structures(a)Not a model.There is no line containing B and D. There are two linescontaining B and C.(b)Not a model. There is no line containing C and D.(c)Not a model. There is no line containing A and D.2.4.4(a)The Three-point plane is a model for Three-point geometry.(b)Every model for Three-point geometry has 3 lines. If there are 3 points, thenthere are also 3 pairs of points(c)Suppose there are two models for Three-point geometry,model A andmodel B. Choose any 1-1 correspondence of the points in model A to thepoints in model B. Any line in A is determined by two points.These twopoints correspond to two points in B. Those two points determine a line in B.The isomorphism should map the given line in A to this line in B. Then thefunction will preserve betweennes. Therefore models A and B are isomorphicto one another.2.4.5Axiom 1 does not hold, but Axioms 2 and 3 do. The Euclidean Parallel Postulateholds. The other parallel postulates are false in this interpretation.2.4.6See Figure S2.2.2.4.7Fano’s Geometry satisfies the Elliptic Parallel Postulate because every line sharesat least one point with every other line; there are no parallel lines. It does not satisfyeither of the other parallel postulates.2.4.8The three-point line satisfies all three parallel postulates (vacuously).2.4.9If there are so few points and lines that there is no line with an external point,then all three parallel postulates are satisfied (vacuously). If there is a line with anexternal point, then there will either be a parallel line through the external point orthere will not be. Hence at most one of the parallel postulates can be satisfied in6

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Solutions to Exercises in Chapter 27FIGURE S2.2:Five-point Geometrythat case. Since every incidence geometry contains three noncollinear points, theremust be a line with an external point. Hence an incidence geometry can satisfy atmost one of the parallel postulates.2.10Start with a line with three points on it. There must exist another point not on thatline (Incidence Axiom 3). That point, together with the points on the original line,determines three more lines (Incidence Axiom 1).But each of those lines musthave a third point on it. So there must be at least three more points, for a total ofat least seven points. Since Fano’s Geometry has exactly seven points, seven is theminimum.2.4.11See Figures S2.3 and S2.4.FIGURE S2.3:An unbalanced geometryFIGURE S2.4:A simpler example2.4.12(a)The three-point line (Example 2.2.3).(b)The square (Exercise 2.4.5) or the sphere (Example 2.2.9).

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8Solutions to Exercises in Chapter 2(c)One-point geometry (Exercise 2.4.2).2.4.13(a)The dual of the Three-point plane is another Three-point plane. It is a modelfor incidence geometry.(b)The dual of the Three-point line is a point which is incident with 3 lines. Thisis not a model for incidence geometry.(c)The dual of Four-point Geometry has 6 points and 4 lines.Each point isincident with exactly 2 lines, and each line is incident with 3 points. It is not amodel for incidence geometry because it does not satisfy Incidence Axiom 1.(d)The dual of Fano’s Geometry is isomorphic to Fano’s Geometry, so it is amodel for incidence geometry.2.5.1(a)H : it rainsC : I get wet(b)H : the sun shinesC : we go hiking and biking(c)H :x >0C :Eaysuch thaty2D0(d)H : 2xC1D5C :xD2 orxD32.5.2(a)Converse : If I get wet, then it rained.Contrapositive : If I do not get wet, then it did not rain.(b)Converse : If we go hiking and biking, then the sun shines.Contrapositive : If we do not go hiking and biking, then the sun does not shine.(c)Converse : IfEaysuch thaty2D0, thenx >0.Contrapositive : If5y,y2Z0, thenx≤0.(d)Converse : IfxD2 orxD3, then 2xC1D5.Contrapositive : IfxZ2 andxZ3, then 2xC1Z52.5.3(a)It rains and I do not get wet.(b)The sun shines but we do not go hiking or biking.(c)x >0 and5y,y2Z0.(d)2xC1D5 butxis not equal to 2 or 3.2.5.4(a)If the grade is an A, then the score is at least 90%.(b)If the score is at least 50%, then the grade is a passing grade.(c)If you fail, then you scored less than 50%.(d)If you try hard, then you succeed.2.5.5(a)Converse: If the score is at least 90%, then the grade is an A.Contrapositive: If the score is less than 90%, then the grade is not an A.(b)Converse: If the grade is a passing grade, then the score is at least 50%.Contrapositive: If the grades is not a passing grade, then the score is less than50%.(c)Converse: If you score less than 50%, then you fail.Contrapositive: If you score at least 50%, then you pass.(d)Converse: If you succeed, then you tried hard.Contrapositive: If you do not succeed, then you did not try hard.2.5.6(a)The grade is an A but the score is less than 90%.

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Solutions to Exercises in Chapter 29(b)The grade is at least 50% but the grade is not a passing grade.(c)You fail and your score is at least 50%.(d)You try hard but do not succeed.2.5.8(a)H : I pass geometryC : I can take topology(b)H : it rainsC : I get wet(c)H : the numberxis divisible by 4C :xis even2.5.9(a)5trianglesT, the angle sum ofTis 180◦.(b)EtriangleTsuch that the angle sum ofTis less than 180◦.(c)EtriangleTsuch that the angle sum ofTis not equal to 180◦.(d)5great circlesαandβ,α∩βZ∅.(e)5pointPand5line`Elinemsuch thatPlies on`andm'`.2.5.10(a)5model for incidence geometry, the Euclidean Parallel Postulate does nothold in that model.(b)Ea model for incidence geometry in which there are not exactly 7 points (thenumber of points is either≤6 or≥8).(c)Ea triangle whose angle sum is not 180◦.(d)Ea triangle whose angle sum is at least 180◦.(e)It is not hot or it is not humid outside.(f)My favorite color is not red and it is not green.(g)The sun shines and (but?) we do not go hiking. (See explanation in last fullparagraph on page 36.)(h)Ea geometry student who does not know how to write proofs.2.5.11(a)Negation of Euclidean Parallel Postulate. There exist a line`and a pointPnoton`such that either there is no linemsuch thatPlies onmandmis parallelto`or there are (at least) two linesmandnsuch thatPlies on bothmandn,m‖`, andn‖`.(b)Negation of Elliptic Parallel Postulate. There exist a line`and a pointPthatdoes not lie on`such that there is at least one linemsuch thatPlies onmandm‖`.(c)Negation of Hyperbolic Parallel Postulate. There exist a line`and a pointPthat does not lie on`such that either there is no linemsuch thatPlies onmandm‖`or there is exactly one linemwith these properties.Note.You could emphasize the separate existence of`andPby starting each of thestatements above with, ‘‘There exist a line`and there exists a pointPnot on`suchthat ....’’2.5.12not(SandT)K(notS)or(notT).STSandTnot(SandT)notSnotT(notC) or (notH)TrueTrueTrueFalseFalseFalseFalseTrueFalseFalseTrueFalseTrueTrueFalseTrueFalseTrueTrueFalseTrueFalseFalseFalseTrueTrueTrueTrue

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10Solutions to Exercises in Chapter 2not(SorT)K(notS)and(notT).STSorTnot(SorT)notSnotT(notC) and (notH)TrueTrueTrueFalseFalseFalseFalseTrueFalseTrueFalseFalseTrueFalseFalseTrueTrueFalseTrueFalseFalseFalseFalseFalseTrueTrueTrueTrue2.5.13H*CK(notH)orC.HCH*CnotH(notH) orCTrueTrueTrueFalseTrueTrueFalseFalseFalseFalseFalseTrueTrueTrueTrueFalseFalseTrueTrueTrueBy De Morgan,not(Hand(notC)) is logically equivalent to (notH)orC.2.5.14not(H*C)K(Hand(notC)).HCH*Cnot(H*C)notCHand(notC)TrueTrueTrueFalseFalseFalseTrueFalseFalseTrueTrueTrueFalseTrueTrueFalseFalseFalseFalseFalseTrueFalseTrueFalse2.5.15If^ABCis right triangle with right angle at vertexC, anda,b, andcare the lengthsof the sides opposite verticesA,B, andCrespectively, thena2Cb2Dc2.2.5.16(a)Euclidean. If`is a line andPis a point that does not lie on`, then there existsexactly one linemsuch thatPlies onmandm‖`.(b)Elliptic. If`is a line andPis a point that does not lie on`, then there does notexist a linemsuch thatPlies onmandm‖`.(c)Hyperbolic. If`is a line andPis a point that does not lie on`, then there existat least two distinct linesmandnsuch thatPlies on bothmandnand`isparallel to bothmandn.2.6.1Converse to Theorem 2.6.2.Proof.Let`andmbe two lines (notation). Assume there exists exactly one pointthat lies on both`andm(hypothesis). We must show that`Zmand`∦m.Suppose`Dm(RAA hypothesis). There exists two distinct pointsQandRthatlie on`(Incidence Axiom 2).SincemD`,QandRlie on both`andm.Thiscontradicts the hypothesis that`andmintersect in exactly one point, so we canreject the RAA hypothesis and conclude that`Zm.Since`andmhave a point in common,`∦m(definition of parallel).Note.In the next few proofs it is convenient to introduce the notation('ABfor theline determined byAandB(Incidence Axiom 1). This notation is not defined in thetextbook until Chapter 3, but it fits with Incidence Axiom 1 and allows the proofsbelow to be written more succinctly.

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Solutions to Exercises in Chapter 2112.6.2Theorem 2.6.3.Proof.Let`be a line (hypothesis). We must prove that there exists a pointPsuchthatPdoes not lie on`. There exist three noncollinear pointsA,B, andC(Axiom 3).The three pointsA,BandCcannot all lie on`because if they did then they wouldbe collinear (definition of collinear). Hence at least one of them does not lie on`and the proof is complete.Note.Many (or most) students will get the last proof wrong. The reason is thatthey want to assume some relationship between the line given in the hypothesis ofthe theorem and the points given by Axiom 3.In particular, many students willstart with the first three sentences of the proof above, but will then either assert orassume that`D('AB.This is an important opportunity to point out that the axioms mean exactly what theysay and we must be careful not to impose our own assumptions on them. The factthat we are imposing an unstated assumption on the axioms is often covered up bythe fact that the same letters are used for the three points given by Axiom 3 and thetwo points given by Axiom 2. So this exercise is also an opportunity to discuss nota-tion and how we assign names to mathematical objects. (The fact that we happen touse the same letter to name two different points does not make them the same point.)Similar comments apply to the next three proofs.2.6.3Theorem 2.6.4.Proof.LetPbe a point (hypothesis). We must show that there are two distinctlines`andmsuch thatPlies on both`andm. There exist three noncollinear pointsA,B, andC(Axiom 3). There are two cases to consider: eitherPis equal to one ofthe three pointsA,B, andCor it is not.Suppose, first, thatPDA. Define`D('ABandmD('AC(Axiom 1). ObviouslyPDAlies on both these lines. It cannot be that`Dmbecause in that caseA,B,andCwould be collinear. So the proof of this case is complete. The proofs of thecases in whichPDBandPDCare similar.Now suppose thatPis distinct from all three of the pointsA,B, andC.In thatcase we can define three lines`D('PA,mD('PB, andnD('PC(Axiom 1). Thesethree lines cannot all be the same because if they were thenA,B, andCwould becollinear. Therefore at least two of them are distinct and the proof is complete.Note.Many students will assert that the three lines in the last paragraph are distinct.But that is not necessarily the case, as Fig. S2.5 shows.Outline of an alternative proof:Find one line`such thatPlies on`. Then use theprevious theorem to find a pointRthat does not lie on`. The line throughPandRis the second line. This proof is simpler and better than the one given above, butmost students do not think of it.You might want to lead them in that directionby suggesting that they prove the existence of one line first rather than proving theexistence of both at the same time.

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12Solutions to Exercises in Chapter 2CBPAFIGURE S2.5:Two of the lines could be the same2.6.4Theorem 2.6.5.Proof.Let`be a line (hypothesis). We must show that there exist two linesmandnsuch that`,mandnare distinct and bothmandnintersect`.There exist two distinct pointsPandQsuch thatPandQlie on`(Axiom 2). Thereexists a pointRsuch thatRdoes not lie on`(Theorem 2.6.3). Letmbe the linedetermined byPandRand letnbe the line determined byQandR(Axiom 1).SincePlies on both`amdm,mintersects`.SinceQlies on both`amdn,nintersects`. SinceRdoes not lie on`, neithermnornis equal to`. To complete theproof we must show thatmZn.SupposemDn(RAA hypothesis). Then bothPandQlie onmas well as`, somD`(the uniqueness part of Axiom 1). But this contradicts an earlier statementin the proof, so we must reject the RAA hypothesis and conclude thatmZn.2.6.5Theorem 2.6.6.Proof.LetPbe a point (hypothesis). We must prove that there is at least one linesuch thatPdoes not lie on that line.LetA; BandCbe three non-collinear points (Axiom 3).Define`D('AB,mD('ACandnD('BC(Axiom 1). There are two cases to consider: either P is equalto one of the three pointsA; BorCor it is not.Suppose, first, thatPDA. ThenPDAdoes not lie onnbecause if it did,A; BandCwould be collinear. So the proof for the first case is complete. There aresimilar proofs forPDBandPDC.Now assumePis distinct fromA; BandC. We will show thatPcannot lie onall three of the lines`,m, andn. SupposePlies on all three of`; m, andn(RAAhypothesis). The fact thatPandAlie on bothmand`implies thatmD`(Axiom 1).The fact thatPandBlie on bothnand`implies thatnD`(Axiom 1).Thus`DmDn. But this cannot be becauseA; BandCare non-collinear. Therefore wemust reject the RAA hypothesis and the proof is complete.2.6.6Theorem 2.6.7.Proof.LetA; BandCbe three non-collinear points (Axiom 3). Define`D('AB,mD('ACandnD('BC(Axiom 1). In order to complete the proof we must showthat`,m, andnare distinct lines and that there does not exist a pointPsuch thatPlies on all three of the lines.

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Solutions to Exercises in Chapter 213No two of the lines can be equal becauseA,B, andCare noncollinear.Theargument that there is no pointPthat lies on all three of the lines is the same as thatin the last two paragraphs of the preceding proof.2.6.7Theorem 2.6.8.Proof.LetPbe a point (hypothesis). We must prove that there exist pointsQandRsuch thatP,Q, andRare noncollinear.There exists a pointQsuch thatQZP(by Axiom 3 there are at least threepoints, soPis not the only point). There exists a pointRsuch thatRdoes not lie on('PQ(Theorem 2.6.3). There is no line on which all three ofP,Q, andRlie becausethe only line on which the first two lie is('PQandRdoes not lie on that line.2.6.8Theorem 2.6.9.Proof.LetPandQbe two points such thatPZQ(hypothesis). We must provethat there exists a pointRsuch thatP,Q, andRare noncollinear.There exists a unique line`such that bothPandQlie on`(Incidence Axiom 1).There is a pointRsuch thatRdoes not lie on`(Theorem 2.6.3). We will completethe proof by showing thatP,Q, andRare noncollinear. Suppose there exists a linemsuch that all three of the pointsP,Q, andRlie onm(RAA hypothesis). ThenmD`by the uniqueness part of Incidence Axiom 1. But this is impossible sinceRdoes not lie on`but does lie onm. Hence we must reject the RAA hypothesis andconclude thatP,Q, andRare noncollinear.

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