Legendre’sProof.—LetABCbe a triangle, of which the sideACis the greatest. BisectBCinD. JoinAD. ThenADis less thanAC[I.xix.Ex.5]. Now, construct a new triangleAB′C′,having the sideAC′= 2AD, andAB′=AC. Again, bisectB′C′inD′, and form another triangleAB′′C′′, havingAC′′= 2AD′, andAB′′=AC′, &c. (1) The sum of the angles of the triangleABC= the sum of the angles ofAB′C′[I.xvi.Cor.1] = the sum of the angles ofAB′′C′′= thesum of the angles ofAB′′′C′′′, &c. (2) The angleB′AC′is less than halfBAC; the angleB′′AC′′is less than halfB′AC′, and so on; hence the angleB(n)A(n)will ultimatelybecome infinitely small. (3) The sum of the base angles of any triangle of the series isequal to the angle of the preceding triangle (seeDem.I.xvi.). Hence, if the annexeddiagram represent the triangleAB(n+1)C(n+1), the sum of the base anglesAandC(n+1)is
equal to the angleB(n)C(n); and whennis indefinitely large, this angle is an infinitesimal; hence thepointB(n+1)will ultimately be in the lineAC, and the angleAB(n+1)C(n+1)will become astraight angle [I.Def.x.], that is, it is equal to two right angles; but the sum of the angles ofAB(n+1)C(n+1)is equal to the sum of the anglesABC.Hence the sum of the three angles ofABCis equal to two right angles.
Hamilton’sQuaternionProof.—LetABCbe the triangle. ProduceBAtoD, and makeADequal toAC. ProduceCBtoE, and makeBEequal toBD; finally, produceACtoF, and makeCFequal toCE. Denote the exterior angles thus formed byA′,B′,C′. Now let the legACof theangleA′be turned round the pointAthrough the angleA′; then the pointCwill coincidewithD. Again, let the legBDof the angleB′be turned round the pointBthrough theangleB′, untilBDcoincides withBE; then the pointDwill coincide withE. Lastly, letCEbe turned roundC, through the angleC′, untilCEcoincides withCF, and thepointEwithF. Now, it is evident that by these rotations the pointChas been broughtsuccessively into the positionsD,E,F; hence, by a motion of mere translation along thelineFC, the lineCAcan be brought into its former position. Therefore it follows, sincerotation is independent of translation, that the amount of the three rotations is equal toone complete revolution round the pointA; thereforeA′+B′+C′= four right angles;but
Observation.—The foregoing demonstration is the most elementary that was ever given of thiscelebrated Proposition. I have reduced it to its simplest form, and without making any use of thelanguage of Quaternions. The same method of proof will establish the more general Proposition,that the sum of the external angles of any convex rectilineal figure is equal to four rightangles.
Mr. Abbott,f.t.c.d., has informed me that this demonstration was first given by Playfairin1826, so that Hamilton was anticipated. It has been objected to on the ground that, appliedverbatim to a spherical triangle, it would lead to the conclusion that the sum of the angles is tworight angles, which being wrong, proves that the method is not valid. A slight consideration willshow that the cases are different. In the proof given in the text there are three motions of rotation,in each of which a point describes an arc of a circle, followed by a motion of translation, in which thesame point describesa right line, and returns to its original position. On the surface of a sphere weshould have, corresponding to these, three motions of rotation, in each of which the pointwould describe an arc of a circle, followed by a motion of rotation about the centre of thesphere, in which the point should describean arc of a great circleto return to its originalposition. Hence, the proof for a plane triangle cannot be applied to a spherical triangle.
________________NOTE C.toinscribearegularpolygonofseventeensidesinacircle.
________________
NOTE C.
toinscribearegularpolygonofseventeensidesinacircle.
Analysis.—LetAbe one of the angular points,AOthe diameter,A1,A2,…A8the vertices atone side ofAO. ProduceOA3toM, andOA2toP, makingA3M=OA5, andA2P=OA8. Again,cut offA6N=OA7, andA1Q=OA4. Lastly, cut offOR=ON, andOS=OQ. Then we have[IV.Ex.40],
and performing the multiplication and substituting, we get
4R(OM − ON − OP + OQ)= 4R2.
Hence, the rectangle and the difference of the linesMRandPSbeing given, each is given; henceMRis given; butMR=OM−ON; thereforeOM−ONis given; and we have proved that therectangleOM.ON=R2; thereforeOMandONare each given. In like manner,OPandOQareeach given.
Again,
ρ6.ρ7 = R(ρ1− ρ4)= R.OQ, and ρ6− ρ7 = ON.
Hence, sinceOQandONare each given,ρ6andρ7are each given; therefore we can draw thesechords, and we have the arcA6A7between their extremities given; that is, the seventeenth part ofthe circumference of a circle.Hence the problem is solved.
The foregoing analysis is due toAmpere:seeCatalan,Théorèmes et Problèmes de GéométrieElémentaire. We have abridged and simplifiedAmpere’s solution.
________________NOTE D.tofindtwomeanproportionalsbetweentwogivenlines.
________________
NOTE D.
tofindtwomeanproportionalsbetweentwogivenlines.
The problem to find two mean proportionals is one of the most celebrated in Geometry onaccount of the importance which the ancients attached to it. It cannot be solved by the line andcircle, but is very easy by Conic Sections. The following is a mechanical construction by the Rulerand Compass.
Sol.—Let the extremesAB,BCbe placed at right angles to each other; complete the rectangleABCD, and describe a circle about it. ProduceDA,DC, and let a graduated ruler be made torevolve round the pointB, and so adjusted thatBEshall be equal toGF; thenAF,CEare twomean proportionals betweenAB,BC.
Dem.—SinceBEis equal toGF, the rectangleBE.GE=BF.GF. ThereforeDE.CE=DF.AF; henceDE:DF::AF:CE; and by similar triangles,AB:AF::DE:DF, andCE:CB::DE:DF. HenceAB:AF::AF:CE; andAF:CE::CE:CB. ThereforeAB,AF,CE,CBare continual proportions. Hence [VI.Def.iv.]AF,CEare two mean proportionalsbetweenABandBC.
The foregoing elegant construction is due to the ancient GeometerPhiloofByzantium. If wejoinDGit will be perpendicular toEF. The lineEFis called Philo’s Line; it possesses theremarkable property of being the minimum line through the pointBbetween the fixed linesDE,DF.
Newton’sConstruction.—LetABandLbe the two given lines of whichABis the greater.BisectABinC. WithAas centre andACas radius, describe a circle, and in it place the chordCDequal to the second lineL. JoinBD, and draw by trial throughAa line meetingBD,CDproducedin the pointsE,F, so that the interceptEFwill be equal to the radius of the circle.DEandFAare the mean proportionalsrequired.
Dem.—JoinAD. Since the lineBFcuts the sides of the△ACE, we have
Again, since the△ACDis isosceles, we have
HenceED(ED+CD) =FA(AB+FA),orDE2( ) 1 + CD DE=FA.AB( ) 1+ AF- AB,thereforeDE2=FA.AB, and we haveAB.CD=DE.FA.HenceAB,DE,FA,CDare in continuedproportion.