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[PDF page 53] CH. II] GEOMETRICAL F ALLACIES. 37 the triangle OAE. Hence, by Euc. i. 8, the triangles are equal; and therefore the angle ODC is equal to the angleOAE. Again, since HO bisects DA and is perpendicular to it, we have the angle ODA equal to the angle OAD. Hence the angleADC (which is the difference ofODC and ODA) is equal to the angleDAE (which is the difference ofOAE and OAD). But ADC is a right angle, andDAE is necessarily greater than a right angle. Thus the result is impossible. Second Fallacy*. To prove that a part of a line is equal to the whole line.Let ABC be a triangle; and, to fix our ideas, let us suppose that the triangle is scalene, that the angleB is acute, and that the B C A ED angle A is greater than the angle C. From A draw AD making the angle BAD equal to the angleC, and cuttingBC in D. From A draw AE perpendicular to BC. The triangles ABC, ABD are equiangular; hence, by Euc.vi. 19, △ABC : △ABD = AC 2 : AD2 . Also the triangles ABC, ABD are of equal altitude: hence, by Euc. vi. 1, △ABC : △ABD = BC : BD , ∴ AC 2 : AD2 = BC : BD . ∴ AC 2 BC = AD2 BD . * See a note by M. Coccoz inL’Illustration, Paris, Jan. 12, 1895.

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[PDF page 54] 38 SOME GEOMETRICAL QUESTIONS. [CH. II Hence, by Euc. ii. 13, AB2 + BC 2 − 2BC · BE BC = AB2 + BD 2 − 2BD · BE BD . ∴ AB2 BC + BC − 2BE = AB2 BD + BD − 2BE . ∴ AB2 BC − BD = AB2 BD − BC . ∴ AB2 − BC · BD BC = AB2 − BC · BD BD . ∴ BC = BD , a result which is impossible. Third Fallacy. To prove that every triangle is isosceles.Let ABC be any triangle. BisectBC in D, and throughD draw DO perpendic- ular to BC. Bisect the angle BAC by AO. First. If DO and AO do not meet, then they are parallel. Therefore AO is at right angles toBC. Therefore AB = AC. Second. If DO and AO meet, let them meet in O. Draw OE perpendicular to AC. Draw OF perpendicular to AB. Join OB, OC. A B CD O E F Let us begin by taking the case where O is inside the triangle, in which case E falls on AC and F on1 AB. The triangles AOF and AOE are equal, since the sideAO is com- mon, angle OAF = angle OAE, and angle OF A = angle OEA. Hence AF = AE. Also, the trian- gles BOF and COE are equal. For since OD bisects BC at right angles, we haveOB = OC; also, since the triangles AOF and AOE are equal, we haveOF = OE; lastly, the angles at F and E are right angles. Therefore, by Euc.i. 47 andi. 8, the triangles BOF and COE are equal. Hence F B = EC. Therefore AF + F B = AE + EC, that is, AB = AC. The same demonstration will cover the case whereDO and AO meet at D, as also the case where they meet outsideBC but so near it that E and F fall on AC and AB and not onAC and AB produced. 1. corrected BC to AB

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[PDF page 55] CH. II] GEOMETRICAL F ALLACIES. 39 Next take the case whereDO and AO meet outside the triangle, and E and F fall onAC and AB produced. Draw OE perpendicular to AC produced. Draw OF perpendicularto AB produced. Join OB, OC. A B CD O EF Following the same argument as before, from the equality of the triangles AOF and AOE, we obtainAF = AE; and, from the equality of the triangles BOF and COE, we obtain F B = EC. Therefore AF − F B = AE − EC, that is, AB = AC. Thus in all cases, whether or notDO and AO meet, and whether they meet inside or outside the triangle, we have AB = AC: and therefore every triangle is isosceles, a result which is impossible. Fourth Fallacy. I am indebted to Captain Turton for the following ingenious fallacy; it appeared for the first time in the third edition of this work. On the hypothenuse, BC, of an isosceles right-angled triangle, DBC, describe an equilateral triangle ABC, the vertex A being on the same side of the base as D is. On CA take a point H so that CH = CD. Bisect BD in K. Join HK and let it cutCB (produced) in L. Join DL. Bisect DL at M, and throughM draw M Operpendic- ular to DL. Bisect HL at N, and throughN draw N O perpendicular to HL. Since DL and HL intersect, therefore M O and N O will also intersect; moreover, since BDC is a right angle, M O and N O both slope away from DC and therefore they will meet on the side ofDL remote from A. Join OC, OD, OH, OL.

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[PDF page 56] 40 SOME GEOMETRICAL QUESTIONS. [CH. II The triangles OM D and OM L are equal, henceOD = OL. Simi- larly the trianglesON Land ON H are equal, henceOL = OH. There- fore OD = OH. Now in the triangles OCD and OCH, we have OD = OH, CD = CH (by construction), andOC common, hence (by Euc. i. 8) the angleOCD is equal to the angleOCH, which is absurd. Fifth Fallacy*. To prove that, if two opposite sides of a quadri- lateral are equal, the other two sides must be parallel.Let ABCD be a quadrilateral such thatAB is equal to DC. Bisect AD in M, and through M draw M O at right angles to AD. Bisect BC in N, and draw N O at right angles to BC. If M O and N O are parallel, thenAD and BC (which are at right angles to them) are also parallel. If M O and N O are not parallel, let them meet inO; then O must be either inside the quadrilateral as in the left-hand diagram or outside A B C D N M O A B C D N M O the quadrilateral as in the right-hand diagram. JoinOA, OB, OC, OD. Since OM bisects AD and is perpendicular to it, we haveOA = OD, and the angle OAM equal to the angle ODM. Similarly OB = OC, and the angle OBN equal to the angle OCN. Also by hypothesis AB = DC, hence, by Euc.i. 8, the trianglesOAB and ODC are equal in all respects, and therefore the angleAOB is equal to the angle DOC. Hence in the left-hand diagram the sum of the anglesAOM, AOB is equal to the sum of the anglesDOM, DOC; and in the right-hand * Mathesis, October, 1893, series 2, vol.iii, p. 224.

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[PDF page 57] CH. II] GEOMETRICAL F ALLACIES. 41 diagram the difference of the anglesAOM, AOB is equal to the dif- ference of the angles DOM, DOC; and therefore in both cases the angle M OB is equal to the angleM OC, i.e. OM (or OM produced) bisects the angle BOC. But the angle N OB is equal to the angle N OC, i.e. ON bisects the angle BOC; hence OM and ON coincide in direction. Therefore AD and BC, which are perpendicular to this direction, must be parallel. This result is not universally true, and the above demonstration contains a flaw. Sixth Fallacy. The following argument is taken from a text-book on electricity, published in 1889 by two distinguished mathematicians, in which it was presented as valid. A given vectorOP of length l can be resolved in an infinite number of ways into two vectorsOM, M P, of lengths l′, l′′, and we can makel′/l′′ have any value we please from nothing to infinity. Suppose that the system is referred to rectangular axes Ox, Oy; and that OP, OM, M P make respectively anglesθ, θ′, θ′′ with Ox. Hence, by projection onOy and on Ox, we have l sin θ = l′ sin θ′ + l′′ sin θ′′ , l cos θ = l′ cos θ′ + l′′ cos θ′′ . Therefore tan θ = n sin θ′ + sin θ′′ n cos θ′ + cos θ′′ , where n = l′/l′′. This result is true whatever be the value of n. But n may have any value (ex. gr. n = ∞, or n = 0 ), hence tan θ = tan θ′ = tan θ′′, which obviously is impossible. Seventh Fallacy. Here is a fallacious investigation, to which Mr Chartres first called my attention, of the value ofπ: it is founded on well-known quadratures. The area of the semi-ellipse bounded by the minor axis is (in the usual notation) equal to1 2 πab. If the centre is moved off to an indefinitely great distance along the major axis, the ellipse degenerates into a parabola, and therefore in this particular limiting position the area is equal to two-thirds of the circumscribing rectangle. But the first result is true whatever be the dimensions of the curve. ∴ 1 2 πab = 2 3 a × 2b, ∴ π = 8/3, a result which is obviously untrue.

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[PDF page 58] 42 SOME GEOMETRICAL QUESTIONS. [CH. II Geometrical Paradoxes. To the above examples I may add the following questions, which, though not exactly fallacious, lead to results which at a hasty glance appear impossible. First Paradox. The first is a problem, sent to me by Mr Renton, to rotate a plane lamina (say, for instance, a sheet of paper) through four right angles so that the effect is equivalent to turning it through only one right angle. If it is desired that the effect shall be equivalent to turning it through a right angle about a pointO, the solution is as follows. De- scribe on the lamina a squareOABC. Rotate the lamina successively through two right angles about the diagonalOB as axis and through two right angles about the side OA as axis, and the required result will be attained. Second Paradox. As in arithmetic, so in geometry, the theory of probability lends itself to numerous paradoxes. Here is a very simple illustration. A stick is broken at random into three pieces. It is possible to put them together into the shape of a triangle provided the length of the longest piece is less than the sum of the other two pieces (cf. Euc. i. 20), that is, provided the length of the longest piece is less than half the length of the stick. But the probability that a fragment of a stick shall be half the original length of the stick is1 2. Hence the probability that a triangle can be constructed out of the three pieces into which the stick is broken would appear to be1 2. This is not true, for actually the probability is 1 4. Third Paradox. The following example illustrates how easily the eye may be deceived in demonstrations obtained by actually dissecting the figures and re-arranging the parts. In fact proofs by superposi- tion should be regarded with considerable distrust unless they are sup- plemented by mathematical reasoning. The well-known proofs of the propositions Euclidi. 32 and Euclidi. 47 can be so supplemented and are valid. On the other hand, as an illustration of how deceptive a non- mathematical proof may be, I here mention the familiar paradox that a square of paper, subdivided like a chessboard into64 small squares, can be cut into four pieces which being put together form a figure con- taining 65 such small squares*. This is effected by cutting the original * I do not know who discovered this paradox. It is given in various modern books, but I cannot find an earlier reference to it than one by Prof. G.H. Darwin, Messenger of Mathematics, 1877, vol.vi, p. 87.

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[PDF page 59] CH. II] GEOMETRICAL PARADOXES. 43 A B square into four pieces in the manner indicated by the thick lines in the first figure. If these four pieces are put together in the shape of a rectangle in the way shown in the second figure it will appear as if this rectangle contains 65 of the small squares. This phenomenon, which in my experience non-mathematicians find perplexing, is due to the fact that the edges of the four pieces of paper, which in the second figure lie along the diagonalAB, do not coincide exactly in direction. In reality they include a small lozenge or diamond-shaped figure, whose area is equal to that of one of the 64 small squares in the original square, but whose lengthAB is much greater than its breadth. The diagrams show that the angle between the two sides of this lozenge which meet atA is tan−1 2 5 − tan−1 3 8, that is, is tan−1 1 46, which is less than1 1 4 ◦. To enable the eye to distinguish so small an angle as this the dividing lines in the first figure would have to be cut with extreme accuracy and the pieces placed together with great care. The paradox depends upon the relation 5 × 13 − 82 = 1. Sim- ilar results can be obtained from the formulae 13 × 34 − 212 = 1 , 34×89−552 = 1,...; orfromtheformulae 52−3×8 = 1, 132−8×21 = 1, 342 − 21 × 55 = 1,.... These numbers are obtained by finding conver- gents to the continued fraction 1 + 1 1 + 1 1 + 1 1 + · · · . A similar paradox for a square of17 cells, by which it was shown that 289 was equal to288, was alluded to by Ozanam* who gave also the diagram for dividing a rectangle of 11 by 3 into two rectangles whose dimensions appear to be5 by 4 and 7 by 2. * Ozanam, 1803 edition, vol. i, p. 299.

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[PDF page 60] 44 SOME GEOMETRICAL QUESTIONS. [CH. II Turton’s Seventy-Seven Puzzle.A far better dissection puzzle was invented by Captain Turton. In this a piece of cardboard,11 inches by 7 inches, subdivided into77 small equal squares, each1 inch by1 inch, can be cut up and re-arranged so as to give78 such equal squares, each 1 inch by 1 inch, of which77 are arranged in a rectangle of the same dimensions as the original rectangle from one side of which projects a small additional square. The construction is ingenious, but cannot be described without the use of a model. The trick consists in utilizing the fact that cardboard has a sensible thickness. Hence the edges of the cuts can be bevelled, but in the model the bevelling is so slight as to be imperceptible save on a very close scrutiny. The play thus given in fitting the pieces together permits the apparent production of an additional square. Colouring Maps. I proceed next to mention the geometrical proposition that not more than four colours are necessary in order to colour a map of a country (divided into districts) in such a way that no two contiguous districts shall be of the same colour. By contiguous districts are meant districts having a common line as part of their boundaries: districts which touch only at points are not contiguous in this sense. The problem was mentioned by A.F. Möbius* in his Lectures in 1840, but it was not until Francis Guthrie†communicated it to De Mor- gan about 1850 that attention was generally called to it: it is said that the fact had been familiar to practical map-makers for a long time pre- viously. Through De Morgan the proposition became generally known; and in 1878 Cayley‡recalled attention to it by stating that he did not know of any rigorous proof of it. Probably the following argument, though not a formal demonstra- tion, will satisfy the reader that the result is true. Let A, B, C be three contiguous districts, and letX be any other district contiguous with all of them. Then X must lie either wholly outside the external boundary of the areaABC or wholly inside the internal boundary, that is, it must occupy a position either likeX or * Leipzig Transactions(Math.-phys. Classe), 1885, vol.xxxvii, pp. 1–6. † Proceedings of the Royal Society of Edinburgh, July 19, 1880, vol.x, p. 728. ‡ Proceedings of the London Mathematical Society, 1878, vol.ix, p. 148, andPro- ceedings of the Royal Geographical Society, 1879, N.S., vol.i, p. 259.

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[PDF page 61] CH. II] THE FOUR-COLOUR THEOREM.. 45 like X ′. In either case every remaining occupied area in the figure is enclosed by the boundaries of not more than three districts: hence there is no possible way of drawing another areaY which shall be contiguous with A, B, C, and X. In other words, it is possible to draw on a plane four areas which are contiguous, but it is not possible to draw five such areas. If A, B, C are not contiguous, each with the other, or if X is not contiguous withA, B, and C, it is not necessary to colour them all differently, and thus the most unfavourable case is that already treated. Moreover any of the above areas may diminish to a point and finally disappear without affecting the argument. That we may require at least four colours is obvious from the dia- gram on this page, since in that case the areasA, B, C, and X would have to be coloured differently. A proof of the proposition involves difficulties of a high order, which as yet have baffled all attempts to surmount them. The argument by which the truth of the proposition was formerly supposed to be demonstrated was given by A.B. Kempe* in 1879, but * He sent his first demonstration across the Atlantic to theAmerican Journal of Mathematics, 1879, vol. ii, pp. 193–200; but subsequently he communicated it

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[PDF page 62] 46 SOME GEOMETRICAL QUESTIONS. [CH. II there is a flaw* in it. In 1880, Tait published a solution†depending on the theorem that if a closed network of lines joining an even number of points is such that three and only three lines meet at each point then three colours are sufficient to colour the lines in such a way that no two lines meeting at a point are of the same colour; a closed network being supposed to exclude the case where the lines can be divided into two groups between which there is but one connecting line. His deduction therefrom that four colours will suffice for a map was given in the last edition of this work. The demonstration appeared so straightforward that at first it was generally accepted, but it would seem that it too involves a fallacy‡. The proof however leads to the interesting corollary that four colours may not suffice for a map drawn on a multiply-connected surface such as an anchor ring. Although a proof of the theorem is still wanting, no one has suc- ceeded in constructing a plane map which requires more than four tints to colour it, and there is no reason to doubt the correctness of the state- ment that it is not necessary to have more than four colours for any plane map. The number of ways which such a map can be coloured with four tints has been also considered§, but the results are not sufficiently interesting to require mention here. Physical Configuration of a Country. As I have been alluding to maps, I may here mention that the theory of the repre- sentation of the physical configuration of a country by means of lines drawn on a map was discussed, by Cayley and Clerk Maxwell∥. They in simplified forms to the London Mathematical Society,Transactions, 1879, vol. x, pp. 229–231, and toNature, Feb. 26, 1880, vol.xxi, pp. 399–400. * See articles by P.J. Heawood in theQuarterly Journal of Mathematics, London, 1890, vol. xxiv, pp. 332–338; and 1897, vol.xxxi, pp. 270–285. † Proceedings of the Royal Society of Edinburgh, July 19, 1880, vol.x, p. 729; and Philosophical Magazine, January, 1884, series 5, vol.xvii, p. 41. ‡ See J. Peterson of Copenhagen, L’Intermédiaire des mathématiciens, vol. v, 1898, pp. 225–227; and vol.vi, 1899, pp. 36–38. § See A.C. Dixon, Messenger of Mathematics, Cambridge, 1902–3, vol.xxxii, pp. 81–83. ∥ Cayley on ‘Contour and Slope Lines,’Philosophical Magazine, London, October, 1859, series 4, vol. xviii, pp. 264–268;Collected Works, vol. iv, pp. 108–111. J. Clerk Maxwell on ‘Hills and Dales,’Philosophical Magazine, December, 1870, series 4, vol.xl, pp. 421–427;Collected Works, vol. ii, pp. 233–240.

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[PDF page 63] CH. II] HILLS AND DALES.. 47 showed that a certain relation exists between the number of hills, dales, passes, &c. which can co-exist on the earth or on an island. I proceed to give a summary of their nomenclature and conclusions. All places whose heights above the mean sea level are equal are on the same level. The locus of such points on a map is indicated by a contour-line. Roughly speaking, an island is bounded by a contour-line. It is usual to draw the successive contour-lines on a map so that the difference between the heights of any two successive lines is the same, and thus the closer the contour-lines the steeper is the slope, but the heights are measured dynamically by the amount of work to be done to go from one level to the other and not by linear distances. A contour-line in general will be a closed curve. This curve may enclose a region of elevation: if two such regions meet at a point, that point will be a crunode (i.e. a real double point) on the contour-line through it, and such a point is called apass. The contour-line may enclose a region of depression: if two such regions meet at a point, that point will be a crunode on the contour-line through it, and such a point is called afork or bar. As the heights of the corresponding level surfaces become greater, the areas of the regions of elevation become smaller, and at last become reduced to points: these points are thesummits of the corresponding mountains. Similarly as the level surface sinks the regions of depression contract, and at last are reduced to points: these points are thebottoms (or immits) of the corresponding valleys. Lines drawn so as to be everywhere at right angles to the contour- lines are calledlines of slope. If we go up a line of slope generally we shall reach a summit, and if we go down such a line generally we shall reach a bottom: we may come however in particular cases either to a pass or to a fork. Districts whose lines of slope run to the same summit are hills. Those whose lines of slope run to the same bottom aredales. A watershed is the line of slope from a summit to a pass or a fork, and it separates two dales. Awatercourse is the line of slope from a pass or a fork to a bottom, and it separates two hills. If n + 1 regions of elevation or of depression meet at a point, the point is a multiple point on the contour-line drawn through it; such a point is called a pass or a fork of thenth order, and must be counted as n separate passes (or forks). If one region of depression meets another in several places at once, one of these must be taken as a fork and the rest as passes.

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[PDF page 64] 48 SOME GEOMETRICAL QUESTIONS. [CH. II Having now a definite geographical terminology we can apply geo- metrical propositions to the subject. Leth be the number of hills on the earth (or an island), then there will be alsoh summits; let d be the number of dales, then there will be alsod bottoms; let p be the whole numberofpasses, p1 thatofsinglepasses, p2 ofdoublepasses, andsoon; let f be the whole number of forks,f1 that of single forks,f2 of double forks, and so on; letw be the number of watercourses, then there will be also w watersheds. Hence, by the theorems of Cauchy and Euler, h = 1 + p1 + 2p2 + · · · , d = 1 + f1 + 2f2 + · · · , and w = 2(p1 + f1) + 3(p2 + f2)) + · · · . The above results can be extended to the case of a multiply- connected closed surface. Games. Leaving now the question of formal geometrical propo- sitions, I proceed to enumerate a few games or puzzles which depend mainly on the relative position of things, but I postpone to chapteriv the discussion of such amusements of this kind as necessitate any con- siderable use of arithmetic or algebra. Some writers regard draughts, solitaire, chess and such like games as subjects for geometrical treat- ment in the same way as they treat dominoes, backgammon, and games with dice in connection with arithmetic: but these discussions require too many artificial assumptions to correspond with the games as actu- ally played or to be interesting. The amusements to which I refer are of a more trivial description, and it is possible that a mathematician may like to omit the remainder of the chapter. In some cases it is difficult to say whether they should be classified as mainly arithmetical or geometrical, but the point is of no importance. Statical Games of Position. Of the innumerable statical games involving geometry of position I shall mention only three or four. Three-in-a-row. First, I may mention the game of three-in-a-row, of which noughts and crosses, one form of merrilees, and go-bang are well-known examples. These games are played on a board—generally in the form of a square containingn2 small squares or cells. The common practice is for one player to place a white counter or piece or to make

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[PDF page 65] CH. II] THREE-IN-A-ROW.. 49 a cross on each small square or cell which he occupies: his opponent similarly uses black counters or pieces or makes a nought on each square which he occupies. Whoever first gets three (or any other assigned number) of his pieces in three adjacent cells and in a straight line wins. The mathematical theory for a board of9 cells has been worked out completely, and there is no difficulty in extending it to one of16 cells: but the analysis is lengthy and not particularly interesting. Most of these games were known to the ancients*, and it is for that reason I mention them here. Three-in-a-row. Extension. I may, however, add an elegant but difficult extension which has not previously found its way, so far as I am aware, into any book of mathematical recreations. The problem is to place n counters on a plane so as to form as many rows as possible, each of which shall contain three and only three counters†. It is easy to arrange the counters in a number of rows equal to the integral part of 1 8(n − 1)2. This can be effected by the following construction. Let P be any point on a cubic. Let the tangent atP cut the curve again inQ. Let the tangent atQ cut the curve inA. Let P A cut the curve inB, QB cut it inC, P C cut it inD, QD cut it inE, and so on. Then the counters must be placed at the pointsP, Q, A, B, . . .. Thus 9 counters can be placed in 8 such rows; 10 counters in 10 rows; 15 counters in 24 rows; 81 counters in 800 rows; and so on. If however the pointP is a pluperfect point of thenth order on the cubic, then Sylvester proved that the above construction gives a number of rows equal to the integral part of1 6(n − 1)(n − 2). Thus 9 counters can be arranged in 9 rows, which is a well-known and easy puzzle; 10 counters in 12 rows; 15 counters in 30 rows; and so on. Even this however is an inferior limit and may be exceeded—for instance, Sylvester stated that 9 counters can be placed in 10 rows, each containing three counters; I do not know how he placed them, but one way of so arranging them is by putting them at points whose coor- dinates are (2, 0), (2, 2), (2, 4), (4, 0), (4, 2), (4, 4), (0, 0), (3, 2), (6, 4); another way is by putting them at the points(0, 0), (0, 2), (0, 4), (2, 1), (2, 2), (2, 3), (4, 0), (4, 2), (4, 4); more generally, the angular points of * Becq de Fouquières, Les jeux des anciens , second edition, Paris, 1873, chap. xviii. † Educational Times Reprints, 1868, vol.viii, p. 106;Ibid.1886, vol.xl v, pp. 127– 128.
