Chapter 19

PROP.IX.—Problem.About a given square(ABCD)to describe a circle.

Sol.—Draw the diagonalsAC,BDintersecting inO(seediagram to Propositionvi.).Ois the centre of the required circle.

Dem.—SinceABCis an isosceles triangle, and the angleBis right, each of the other angles is half a right angle; thereforeBAOis half a right angle. In like mannerABOis half a right angle; hence the angleBAOequalABO; therefore [I.vi.]AOis equal toOB. In like mannerOBis equal toOC, andOCtoOD.Hence the circle described, withOas centre andOAasradius, will pass through the pointsB,C,D, and be described about thesquare.

PROP.X.—Problem.To construct an isosceles triangle having each base angle double the verticalangle.

Sol.—Take any lineAB. Divide it inC, so that the rectangleAB.BCshall be equal toAC2[II.xi.]. WithAas centre, andABas radius, describe the circleBDE, and in it place the chordBDequal toAC[i.]. JoinAD.ADBis a trianglefulfilling the required conditions.

Dem.—JoinCD. About the triangleACDdescribe the circleCDE[v.]. Then, because the rectangleAB.BCis equal toAC2(const.), and thatACis equal toBD(const.); therefore the rectangleAB.BCis equal toBD2. Hence [III.xxxii.]BDtouches the circleACD. Hence the angleBDCis equal to the angleAin the alternate segment [III.xxxii.]. To each addCDA, and we have the angleBDAequal to the sum of the anglesCDAandA; but the exterior angleBCDof the triangleACDis equal to the sum of the anglesCDAandA. Hence the angleBDAis equal toBCD; but sinceABis equal toAD, the angleBDAis equal toABD; therefore the angleCBDis equal toBCD. Hence [I.vi.]BDis equal toCD; butBDis equal toAC(const.); thereforeACis equal toCD, and therefore [I.v.] the angleCDAis equal toA; butBDAhas been proved to be equal to the sum ofCDAandA. HenceBDAis double ofA.Hence each of the base angles of the triangleABDis double of the verticalangle.

Exercises.

1.Prove thatACDis an isosceles triangle whose vertical angle is equal to three times each ofthe base angles.

2.Prove thatBDis the side of a regular decagon inscribed in the circleBDE.

3.IfDB,DE,EFbe consecutive sides of a regular decagon inscribed in a circle, proveBF−BD= radius of circle.

4.IfEbe the second point of intersection of the circleACDwithBDE,DEis equal toDB;and ifAE,BE,CE,DEbe joined, each of the trianglesACE,ADEis congruent withABD.

5.ACis the side of a regular pentagon inscribed in the circleACD, andEBthe side of aregular pentagon inscribed in the circleBDE.

6.SinceACEis an isosceles triangle,EB2−EA2=AB.BC—that is =BD2; thereforeEB2−BD2=EA2—that is,the square of the side of a pentagon inscribed in a circle exceedsthe square of the side of the decagon inscribed in the same circle by the square of theradius.

PROP.XI.—Problem.To inscribe a regular pentagon in a given circle(ABCDE).

Sol.—Construct an isosceles triangle [x.], having each base angle double the vertical angle, and inscribe in the given circle a triangleABDequiangular to it. Bisect the anglesDAB,ABDby the linesAC,BE. JoinEA,ED,DC,CB;thenthe figureABCDEis a regular pentagon.

Dem.—Because each of the base anglesBAD,ABDis double of the angleADB, and the linesAC,BEbisect them, the five anglesBAC,CAD,ADB,DBE,EBAare all equal; therefore the arcs on which they stand are equal; and therefore the five chords,AB,BC,CD,DE,EAare equal. Hence the figureABCDEis equilateral.

Again, because the arcsAB,DEare equal, adding the arcBCDto both, the arcABCDis equal to the arcBCDE, and therefore [III.xxvii.] the anglesAED,BAE, which stand on them, are equal. In the same manner it can be proved that all the angles are equal; therefore the figureABCDEis equiangular.Hence it is aregular pentagon.

Exercises.

1.The figure formed by the five diagonals of a regular pentagon is another regularpentagon.

2.If the alternate sides of a regular pentagon be produced to meet, the five points of meetingform another regular pentagon.

3.Every two consecutive diagonals of a regular pentagon divide each other in extreme and meanratio.

4.Being given a side of a regular pentagon, construct it.

5.Divide a right angle into five equal parts.

PROP.XII.—Problem.To describe a regular pentagon about a given circle(ABCDE).

Sol.—Let the five pointsA,B,C,D,Eon the circle be the vertices of any inscribed regular pentagon: at these points draw tangentsFG,GH,HI,IJ,JF:thefigureFGHIJis a circumscribed regular pentagon.

Dem.—LetObe the centre of the circle. JoinOE,OA,OB. Now, because the anglesA,Eof the quadrilateralAOEFare right angles [III.xviii.], the sum of the two remaining anglesAOE,AFEis two right angles. In like manner the sum of the anglesAOB,AGBis two right angles; therefore the sum ofAOE,AFEis equal to the sum ofAOB,AGB; but the anglesAOE,AOBare equal, because they stand on equal arcsAE,AB[III.xxvii.]. Hence the angleAFEis equal toAGB. In like manner the remaining angles of the figureFGHIJare equal.Therefore it isequiangular.

Again, joinOF,OG. Now the trianglesEOF,AOFhave the sidesAF,FEequal [III.xvii., Ex. 1], andFOcommon, and the baseAOequal to the baseEO. Hence the angleAFOis equal toEFO[I.viii.]. Therefore the angleAFOis half the angleAFE. In like mannerAGOis half the angleAGB; butAFEhas been proved equal toAGB; henceAFOis equal toAGO, andFAOis equal toGAO, each being right, andAOcommon to the two trianglesFAO,GAO; hence [I.xxvi.] the sideAFis equal toAG; thereforeGFis doubleAF. In like mannerJFis doubleEF; butAFis equal toEF; henceGFis equal toJF. In like manner the remaining sides are equal; therefore the figureFGHIJis equilateral, and it has been proved equiangular.Hence it is a regularpentagon.

This Proposition is a particular case of the following general theorem, of which the proof is thesame as the foregoing:—

“If tangents be drawn to a circle, at the angular points of an inscribed polygon of any number ofsides, they will form a regular polygon of the same number of sides circumscribed to thecircle.”

PROP.XIII.—Problem.To inscribe a circle in a regular pentagon(ABCDE).

Sol.—Bisect two adjacent anglesA,Bby the linesAO,BO;thenO, the point ofintersection of the bisectors, is the centre of the required circle.

Dem.—JoinCO, and let fall perpendiculars fromOon the five sides of the pentagon. Now the trianglesABO,CBOhave the sideABequal toBC(hyp.), andBOcommon, and the angleABOequal toCBO(const.). Hence the angleBAOis equal toBCO[I.iv.]; butBAOis halfBAE(const.). ThereforeBCOis halfBCD, and thereforeCObisects the angleBCD. In like manner it may be proved thatDObisects the angleD, andEOthe angleE.

Again, the trianglesBOF,BOGhave the angleFequal toG, each being right; andOBFequal toOBG, becauseOBbisects the angleABC(const.), andOBcommon; hence [I.xxvi.]OFis equal toOG. In like manner all the perpendiculars fromOon the sides of the pentagon are equal; hence the circle whose centre isO, and radiusOF, will touch all the sides of the pentagon,and will therefore beinscribed in it.

In the same manner a circle may be inscribed in any regular polygon.

PROP.XIV.—Problem.To describe a circle about a regular pentagon(ABCDE).

Sol.—Bisect two adjacent anglesA,Bby the linesAO,BO.ThenO, the point ofintersection of the bisectors, is the centre of the required circle.

Dem.—JoinOC,OD,OE. Then the trianglesABO,CBOhave the sideABequal toBC(hyp.),BOcommon, and the angleABOequal toCBO(const.). Hence the angleBAOis equal toBCO[I.iv.]; but the angleBAEis equal toBCD(hyp.); and sinceBAOis halfBAE(const.),BCOis halfBCD. HenceCObisects the angleBCD. In like manner it may be proved thatDObisectsCDE, andEOthe angleDEA. Again, because the angleEABis equal toABC, their halves are equal. HenceOABis equal toOBA; therefore [I.vi.]OAis equal toOB. In like manner the linesOC,OD,OEare equal to one another and toOA. Therefore the circle described withOas centre, andOAas radius, will pass through the pointsB,C,D,E,and be described about thepentagon.

In the same manner a circle may be described about any regular polygon.

Propositionsxiii.,xiv.are particular cases of the following theorem:—

“A regular polygon of any number of sides has one circle inscribed in it, and another describedabout it, and both circles are concentric.”

PROP.XV.—Problem.In a given circle(ABCDEF)to inscribe a regular hexagon.

Sol.—Take any pointAin the circumference, and join it toO, the centre of the given circle; then withAas centre, andAOas radius, describe the circleOBF, intersecting the given circle in the pointsB,F. JoinOB,OF, and produceAO,BO,FOto meet the given circle again in the pointsD,E,C. JoinAB,BC,CD,DE,EF,FA;ABCDEFis the required hexagon.

Dem.—Each of the trianglesAOB,AOFis equilateral (seeDem.,I.i.). Hence the anglesAOB,AOFare each one-third of two right angles; thereforeEOFis one-third of two right angles. Again, the anglesBOC,COD,DOEare [I.xv.] respectively equal to the anglesEOF,FOA,AOB. Therefore the six angles at the centre are equal, because each is one-third of two right angles. Therefore the six chords are equal [III.xxix.].Hence the hexagon isequilateral.

Again, since the arcAFis equal toED, to each add the arcABCD; then the whole arcFABCDis equal toABCDE; therefore the anglesDEF,EFAwhich stand on these arcs are equal [III.xxvii.]. In the same manner it may be shown that the other angles of the hexagon are equal. Hence it is equiangular,and is therefore aregular hexagon inscribed in the circle.

Cor. 1.—The side of a regular hexagon inscribed in a circle is equal to the radius.

Cor. 2.—If three alternate angles of a hexagon be joined, they form an inscribed equilateral triangle.

Exercises.

1.The area of a regular hexagon inscribed in a circle is equal to twice the area of an equilateraltriangle inscribed in the circle; and the square of the side of the triangle is three times the square ofthe side of the hexagon.

2.If the diameter of a circle be produced toCuntil the produced part is equal to the radius,the two tangents fromCand their chord of contact form an equilateral triangle.

3.The area of a regular hexagon inscribed in a circle is half the area of an equilateral triangle,and three-fourths of the area of a regular hexagon circumscribed to the circle.

PROP.XVI.—Problem.To inscribe a regular polygon of fifteen sides in a given circle.

Sol.—Inscribe a regular pentagonABCDEin the circle [xi.], and also an equilateral triangleAGH[ii.]. JoinCG.CGis a side of the requiredpolygon.

Dem.—SinceABCDEis a regular pentagon, the arcABCis2 5ths of the circumference; and sinceAGHis an equilateral triangle, the arcABGis1 3rd of the circumference. Hence the arcGC, which is the difference between these two arcs, is equal to2 5ths−1 3rd, or-1 15th of the entire circumference;and therefore, if chords equaltoGC[i.]be placed round the circle, we shall have a regular polygon of fifteen sides,or quindecagon, inscribed in it.

Scholium.—Until the year 1801 no regular polygon could be described by constructionsemploying the line and circle only, except those discussed in this Book, and those obtained fromthem by the continued bisection of the arcs of which their sides are the chords; but in that year thecelebrated Gauss proved that if 2n+ 1 be a prime number, regular polygons of 2n+ 1 sides areinscriptable by elementary geometry. For the casen= 4, which is the only figure of this classexcept the pentagon for which a construction has been given, see Note at the end of thiswork.

Questions for Examination on Book IV.

1.What is the subject-matter of Book IV.?

2.When is one rectilineal figure said to be inscribed in another?

3.When circumscribed?

4.When is a circle said to be inscribed in a rectilineal figure?

5.When circumscribed about it?

6.What is meant by reciprocal propositions?Ans. In reciprocal propositions, to every line inone there corresponds a point in the other; and, conversely, to every point in one there corresponds aline in the other.

7.Give instances of reciprocal propositions in Book IV.

8.What is a regular polygon?

9.What figures can be inscribed in, and circumscribed about, a circle by means ofBookIV.?

10.What regular polygons has Gauss proved to be inscriptable by the line and circle?

11.What is meant by escribed circles?

12.How many circles can be described to touch three lines forming a triangle?

13.What is the centroid of a triangle?

14.What is the orthocentre?

15.What is the circumcentre?

16.What is the polar circle?

17.When is the polar circle imaginary?

18.What is the “nine-points circle”?

19.Why is it so called?

20.Name the special nine points through which it passes.

21.What three regular figures can be used in filling up the space round a point?Ans.Equilateral triangles, squares, and hexagons.

22.If the sides of a triangle be 13, 14, 15, what are the values of the radii of its inscribed andescribed circles?

23.What is the radius of the circumscribed circle?

24.What is the radius of its nine-points circle?

25.What is the distance between the centres of its inscribed and circumscribed circles?

26.Ifrbe the radius of a circle, what is the area of its inscribed equilateral triangle?—of itsinscribed square?—its inscribed pentagon?—its inscribed hexagon?—its inscribed octagon?—itsinscribed decagon?

27.With the same hypothesis, find the sides of the same regular figures.

Exercises on Book IV.

1.If a circumscribed polygon be regular, the corresponding inscribed polygon is also regular,and conversely.

2.If a circumscribed triangle be isosceles, the corresponding inscribed triangle is isosceles, andconversely.

3.If the two isosceles triangles in Ex.2 have equal vertical angles, they are bothequilateral.

4.Divide an angle of an equilateral triangle into five equal parts.

5.Inscribe a circle in a sector of a given circle.

6.The lineDEis parallel to the baseBCof the triangleABC: prove that the circles describedabout the trianglesABC,ADEtouch atA.

7.The diagonals of a cyclic quadrilateral intersect inE: prove that the tangent atEto thecircle about the triangleABEis parallel toCD.

8.Inscribe a regular octagon in a given square.

9.A line of given length slides between two given lines: find the locus of the intersection ofperpendiculars from its extremities to the given lines.

10.If the perpendicular to any side of a triangle at its middle point meet the internal andexternal bisectors of the opposite angle in the pointsDandE; prove thatD,Eare points on thecircumscribed circle.

11.Through a given pointPdraw a chord of a circle so that the interceptEFmay subtend agiven angleX.

12.In a given circle inscribe a triangle having two sides passing through two given points, andthe third parallel to a given line.

13.Given four points, no three of which are collinear; describe a circle which shall beequidistant from them.

14.In a given circle inscribe a triangle whose three sides shall pass through three givenpoints.

15.Construct a triangle, being given—

16.IfFbe the middle point of the base of a triangle,DEthe diameter of the circumscribedcircle which passes throughF, andLthe point where a parallel to the base through the vertexmeetsDE: proveDL.FEis equal to the square of half the sum, andDF.LEequal to the square ofhalf the difference of the two remaining sides.

17.If from any point within a regular polygon ofnsides perpendiculars be let fall on the sides,their sum is equal tontimes the radius of the inscribed circle.

18.The sum of the perpendiculars let fall from the angular points of a regular polygon ofnsides on any line is equal tontimes the perpendicular from the centre of the polygon on the sameline.

19.IfRdenotes the radius of the circle circumscribed about a triangleABC,r,r′,r′′,r′′′theradii of its inscribed and escribed circles,δ,δ′,δ′′the perpendiculars from its circumcentre on thesides;μ,μ′,μ′′the segments of these perpendiculars between the sides and circumference of thecircumscribed circle, we have the relations—

The relation (3) supposes that the circumcentre is inside the triangle.

20.Through a pointD, taken on the sideBCof a triangleABC, is drawn a transversalEDF,and circles described about the trianglesDBF,ECD. The locus of their second point of intersectionis a circle.

21.In every quadrilateral circumscribed about a circle, the middle points of its diagonals andthe centre of the circle are collinear.

22.Find on a given line a pointP, the sum or difference of whose distances from two givenpoints may be given.

23.Find a point such that, if perpendiculars be let fall from it on four given lines, their feet maybe collinear.

24.The line joining the orthocentre of a triangle to any pointP, in the circumference of itscircumscribed circle, is bisected by the line of collinearity of perpendiculars fromPon the sides ofthe triangle.

25.The orthocentres of the four triangles formed by any four lines are collinear.

26.If a semicircle and its diameter be touched by any circle, either internally or externally,twice the rectangle contained by the radius of the semicircle, and the radius of the tangential circle,is equal to the rectangle contained by the segments of any secant to the semicircle, through the pointof contact of the diameter and touching circle.

27.Ifρ,ρ′be the radii of two circles, touching each other at the centre of the inscribed circle ofa triangle, and each touching the circumscribed circle, prove

1 1- 2 ρ + ρ′ = r,

and state and prove corresponding theorems for the escribed circles.

28.If from any point in the circumference of the circle, circumscribed about a regular polygonofnsides, lines be drawn to its angular points, the sum of their squares is equal to 2ntimes thesquare of the radius.

29.In the same case, if the lines be drawn from any point in the circumference of the inscribedcircle, prove that the sum of their squares is equal tontimes the sum of the squares of the radii ofthe inscribed and the circumscribed circles.

30.State the corresponding theorem for the sum of the squares of the lines drawn from anypoint in the circumference of any concentric circle.

31.If from any point in the circumference of any concentric circle perpendiculars be let fall onall the sides of any regular polygon, the sum of their squares is constant.

32.For the inscribed circle, the constant is equal to3n 2times the square of the radius.

33.For the circumscribed circle, the constant is equal tontimes the square of the radius of theinscribed circle, together with1 2ntimes the square of the radius of the circumscribedcircle.

34.If the circumference of a circle whose radius isRbe divided into seventeen equalparts, andAObe the diameter drawn from one of the points of division (A), and ifρ1,ρ2……ρ8denote the chords fromOto the points of division,A1,A2……A8on one side ofAO,then

ρ1ρ2ρ4ρ8 = R4; and ρ3ρ5ρ6ρ7 = R4.—Catalan.

Dem.—Let the supplemental chords corresponding toρ1,ρ2, &c., be denoted byr1,r2, &c.;then [III.xxxv.Ex.2], we have

And it may be proved in the same manner that

35.If from the middle point of the line joining any two of four concyclic points a perpendicularbe let fall on the line joining the remaining two, the six perpendiculars thus obtained areconcurrent.

36.The greater the number of sides of a regular polygon circumscribed about a given circle, theless will be its perimeter.

37.The area of any regular polygon of more than four sides circumscribed about a circle is lessthan the square of the diameter.

38.Four concyclic points taken three by three determine four triangles, the centres of whosenine-points circles are concyclic.

39.If two sides of a triangle be given in position, and if their included angle be equal to anangle of an equilateral triangle, the locus of the centre of its nine-points circle is a rightline.

40.If, in the hypothesis and notation of Ex.34,α,βdenote any two suffixes whose sum is lessthan 8, and of whichαis the greater,

ραρβ = R(ρα−β + ρα+β).

For instance,ρ1ρ4=R(ρ3+ρ5) [III.xxxv., Ex.7].

In the same case, if the suffixes be greater than 8,

ρα.ρβ = R(ρα−β − ρ17−α−β).

For instance,ρ8ρ2=R(ρ6−ρ7) [III.xxxv., Ex.6].

41.Two lines are given in position: draw a transversal through a given point, forming with thegiven lines a triangle of given perimeter.

42.Given the vertical angle and perimeter of a triangle, construct it with either of the followingdata: 1.The bisector of the vertical angle; 2.the perpendicular from the vertical angle on the base;3.the radius of the inscribed circle.

43.In a given circle inscribe a triangle so that two sides may pass through two given points, andthat the third side may be a maximum or a minimum.

44.Ifsbe the semiperimeter of a triangle,r′,r′′,r′′′, the radii of its escribed circles,

′′′ ′′′′′ ′′′′ 2 r r +r r +r r = s.

45.The feet of the perpendiculars from the extremities of the base on either bisector of thevertical angle, the middle point of the base, and the foot of the perpendicular from the vertical angleon the base, are concyclic.

46.Given the base of a triangle and the vertical angle; find the locus of the centre of the circlepassing through the centres of the escribed circles.

47.The perpendiculars from the centres of the escribed circles of a triangle on thecorresponding sides are concurrent.

48.IfABbe the diameter of a circle, andPQany chord cuttingABinO, and if the linesAP,AQintersect the perpendicular toABatO, inDandErespectively, the pointsA,B,D,Eareconcyclic.

49.If the sides of a triangle be in arithmetical progression, and ifR,rbe the radii of thecircumscribed and inscribed circles; then 6Rris equal to the rectangle contained by the greatest andleast sides.

50.Inscribe in a given circle a triangle having its three sides parallel to three givenlines.

51.If the sidesAB,BC, &c., of a regular pentagon be bisected in the pointsA′,B′,C′,D′,E′,and if the two pairs of alternate sides,BC,AE;AB,DE, meet in the pointsA′′,E′′, respectively,prove

′′ ′′ ′ ′ ′ ′ ′ ′′ △A AE − △A AE = pentagon A B CD E .

52.In a circle, prove that an equilateral inscribed polygon is regular, and also an equilateralcircumscribed polygon, if the number of sides be odd.

53.Prove also that an equiangular circumscribed polygon is regular, and an equiangularinscribed polygon, if the number of sides be odd.

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