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25. Let POQ be any quadrant, O being the centre of the circle, and let BG; DH be drawn perpendicular to the radius PO, and OB, OD be joined. The triangle GBO is equal to DHO.
26. The segments on BC, BA, AC may be shewn to be similar. And similar segments of circles may be proved to be proportional to the squares of their radii, Euc. XII. 2, and to the squares of the chords on which they stand, Euc. vi. 6.
If Euc. vi. 31 be extended to any similar figures, the equality follows directly.
27. The triangles CEA, CEB are equal, and the difference of the two segments may be shewn to be equal to the difference of the parts of the semicircle made by CE. The difference of the same parts may also be shewn to be equal to double the sector DEC. 28.
Let AB be the hypothenuse of the right-angled triangle ABC, and let the semicircles described upon the sides AC, BC, intersect the hypothenuse in D. Join AD. AD is perpendicular to AB. The segments on AC, AD, and on one side of CD are similar; and the segments on AC may be shewn to be equal to the segments on AD, CD. Also the segment on BC may be shewn to be equal to the segments on BD and the other side of CD.
If however Euc. vi. 31 be true for all similar figures, the conclusions above stated follow at once from the right-angled triangles.
29. (a) Join BD, CD, DA, Euc. III. 31 ; 1. 14. (6) Produce CD to meet. the arc of the quadrant in E. Then the sector ACE is half of the quadrant: also the semicircle CDA may be shewn to be equal to half the quadrant. (c) The segments on CD and DA are similar and equal, if the figure bounded by DA, AC, and the arc CD be added to each, the remaining part of the semicircle on AC is equal to the triangle ACD which is a right-angled isosceles triangle.
30. This theorem is analogous to Euc. 111. 14.
31. Let D be the given point, and from D let DA be drawn through the centre E, and meeting the surface in C, A. Let DB be a line from D touching the sphere at B. Join BE. Then the triangle DBE (fig. Euc. 111. 36) is in a plane passing through D, and E the centre of the sphere, and the distances DE, EB are always the same. Hence it follows that BD is always of the same length. Euc. 1. 47.
The sphere which touches the six edges of any tetrahedron, has four circular sections touching the sides of the four triangles which form the surface of the tetrahedron.
32. Let the circle ADB cut the circle AEB in the diameter AB at any angle, C being their common centre. Next let the plane perpendicular to AB cut the circumference of the circle ADB in D, F, and the circumference of AEB in E, G. Then E, D, G, F may be proved to be in the circumference of a circle.
33. See the Géométrie par M. Vincent, p. 450.
34. Let ABCD be a regular tetrahedron. From A in the plane ABC draw AE perpendicular to BC, and join DE in the plane BCD, also from A draw AG perpendicular to the line DE. Then the angle AEG is the inclination of the two faces ABC, DBC of the tetrahedron, and the base EG is one-third of the hypothenuse AE in the right-angled triangle AGE.
Let abcdef be a regular octahedron whose faces are equal to those of the tetrahedron. Join a, f two opposite vertices. Draw ag in the plane abc perpendicular to bc, and ge perpendicular to af. Draw fg in the plane foc, and from f draw fh perpendicular to ag produced.
Then agf is the inclination of two faces of the octahedron. Also in the rightangled triangle fhg, gh may be proved to be one-third of fg, and fg is equal to AE. Hence the triangles fgh, AEF are equal in all respects. Therefore the angle fgh is equal to the angle AEB. Hence the angle AEF is the supplement of the angle agf, or the inclination of two contiguous faces of a tetrahedron, is the supplement of the inclination of two contiguous faces of an octahedron.
PROBLEM 16, page 333.
Inscribe in a circle a triangle whose sides or sides produced, shall pass through three given points in the same plane.
Lemma I. Let there be given two points A, B, and a circle DCE whose centre is S: from the point A, draw ADC to cut the circle; through B, C, D describe a circle cutting the line AB in K: then K is a fixed point, however the line ADC may be drawn; and if KD be drawn to meet the circle in H, the line HE, drawn to the intersection E of BC with the circle, will be parallel to AB.
For, first, BA. AK=CD. DA= a given magnitude, namely, the square of the tangent from A to the circle. Also BA is given, and hence AK is also given, and K is a fixed point, however ADC be drawn from A to cut the circle.
And, secondly, since CDHE is a quadrilateral inscribed in the circle CDE, the exterior angle made by producing EH, is equal to the interior opposite angle DCE. In the same way the angle DKA is equal to DCB. Whence these angles are equal, and HE is parallel to AB.
Lemma II. The same conditions as before being given, draw the diameter TV through K and S; and find the point F such that SF.SK =ST : then joining FD, the angle EDF will be equal to the difference between BKS and a right angle, however the point C be taken in the circle DCE.
For, draw HX parallel to KS, and from X draw the diameter XSG, and join HG, GD; also draw KP perpendicular to AB.
Then since XHG is an angle in a semicircle, it is a right angle; and since HX is parallel to KS, HG is perpendicular to KS; and the angle EHG is equal to PKS.
Moreover, since HG is perpendicular to KS, the line DG always passes through F, and hence the line FG makes with DE the constant angle PKS, however C may be taken in the circle DCE.
Having premised these two Lemmas, we may proceed to the construction of the Problem, as follows:
Let A, B, Q be the three given points. Find the points K, F as in the Lemmas, together with the angle PKS. Join QF, and on it describe a segment to contain the angle PKS; and let it cut the circle DCE in D: Then D is one of the angular points of the triangle. Join DA meeting the circle in C, and CB meeting the circle in E, and draw ED. It will pass through Q by the reasoning of the Lemmas.
The same principle may be applied to the solution of Problem 58, p. 324.
Any two points, A, B, being given within a circle CDE, it is required to find a point D, so that the difference of the angles BDE, ADC may equal a given angle.
The points A, B may be taken any where either within or without the circle ; and the construction will be the same.
Analysis. Let the point D be supposed to be found, such that the angle EDB - DCA = the given angle. Make ECL equal to that angle, and join LC. Then, obviously, if BD meet the circle in H, and DA meet it in G, the chord GH will be parallel to LC. Find F and K, by the Lemmas, then the angle FHG is given. Wherefore through F draw FQ parallel to LC, and make the angle HFG equal to the given angle. Draw BH meeting the circle in D, and join DA: then these are the lines required.
Note. The second Lemma is only a variation of the last Porism of Euclid's third Book on that subject.
The 57th proposition in Dr Simson's Restoration of the Porisms, leads directly to the construction in the manner here given.
The former Problem, though not mentioned directly by Pappus, nor found in any ancient author, was without doubt considered by the Greek geometers.
It has been regarded by modern geometers as an extension of the 117th Proposition of the Seventh Book of the Collections of Pappus, namely:-when the three points are not in the same straight line. The Problem itself, as well as Proposition 117
of Pappus, has engaged the attention of several distinguished modern geometers. Bonnycastle, in p. 348 of his Geometry, has given a concise account of the several solutions by mathematicians on the continent; as also Dr Traill in p. 95 of his life Simson. In p. 97, he has given Simson's solution of the Problem, which, from a note attached, appears to have been completed in 1731. Simson's “Opera Reliqua” was published by the munificence of Earl Stanhope in 1776.
In 1742, M. Cramer proposed the problem to M. de Castillon, but it was not till 1776 that Castillon published a geometrical solution in the Berlin Memoirs of that year. In the same volume is a solution by La Grange, by means of trigonometrical formulæ. Carnot, in p. 383 of his Géométrie de Position, has given a modified form of La Grange's Solution. In the Petersburgh Acts for 1780, are solutions of the same Problem by Euler, Lexell and Fuss. In the Memoirs of the Italian Society (Tom. iv. 1788) are two papers respecting this problem; one by Ottajano, in which is given a geometrical solution of the problem, and an extension to the case of a polygon of any number of sides, which he inscribes in a given circle, so that the sides respectively shall pass through the same number of points. Ottajano also gives a sketch of the history of the problem. The other paper is by Malfatti, and contains a solution of the general problem of the polygon last mentioned.
In the Berlin Memoirs for 1796 is a paper by Lhuilier, containing an algebraical solution of the most general case of the polygon. He also extends the problem to the conic sections, and adds a similar one respecting the sphere. An extension of this Problem to the Conic Sections has also been effected by Poncelet in his Traité des Propriétés Projectives. M. Brianchon has considered the Problem in the case where a conic section is substituted for the circle, and where the three points are in one line. His solution will be found in the Journal de l'Ecole Polytechnique. (Tom. Iv.)
Dr Wallace and Mr Lowry applied the 57th Porism to this problem in Vol. 17. of the Old Series of Leybourn's Repository. In Vol. 1. of the New Series of that work, a new and very elegant analysis of the Porism was given by Mr Noble, which has been the main guide in demonstrating the two Lemmas here used. The same method, slightly modified, applies to any inscribed polygon, The most elegant system of investigation, however, that has ever been published, is that of Mr Swale, in the second number of his Apollonius. Mr Lowry has also given the solution of the problem in the case where the ellipse is substituted for thowcircle, and where the polygon has any number of sides. See Leybourn's Repository, Vol. 11., New Series, p. 189.