equal squares, and having every two contiguous faces at right angles to one another. The regular Dodecahedron. Since three angles each equal to the angle of a regular pentagon may form a solid angle; let ABCDE be a regular pentagon, and with two others each equal to this, let a solid angle at A be formed; the inclinations of every two contiguous faces will be equal. At the points B, C, D, E successively, let solid angles be formed by pentagons equal to ABCDE. The solid angles at B, C, D, E, are each equal to the solid angle at A, and the inclination of every two contiguous faces is the same. Thus is formed a figure with six equal and regular pentagons, having the inclination of every two contiguous faces equal, and the angles at the linear boundary of the figure alternately consisting of an angle of a pentagon and of two angles of two pentagons equally inclined to each other. Next, let another figure equal to this be constructed with six pentagons, each equal to the pentagon ABCDE. If these two figures be so placed that the angular points of the plane angles in the linear boundary of one, may coincide with the points at which there are two angles in the other figure; at each of these points will be formed ten solid angles, each equal to the angle at A, and having the inclination of every two contiguous faces equal to one another. Hence a regular figure is formed having twelve equal faces, and the inclinations of every two contiguous faces equal to one another. I. 5. Construct a circle the area of which shall have a given ratio to that of a given circle. 6. Divide a circle into any number of equal parts by means of concentric circles. 7. To divide a circle into any number of equal parts, the perimeters of which shall be equal to the circumference of the circle. 8. Let AB and DC be two diameters of a given circle, at right angles to each other; AEB a circular arc described with radius DB or DA; prove that the area of the lune AEBC = area of triangle ADB. 9. Two circles touch each other internally, and the area of the lune cut out of the larger is equal to twice the area of the smaller circle. Required the ratio of the diameters of these circles. 10. The diameter of a circle is divided into two parts, upon each of which as diameters circles are described; when the remaining area of the great circle is equal to that of one of these two circles, find the ratio which the parts of the diameter bear to one another. 11. The diameter of a semicircle ADB is divided into two parts in C (so that the length of AC is twice that of BC), and upon them are described the semicircles AEC, CFB. Compare the areas of the circles which are described on each side of the common tangent CD so as to touch it and the two semicircles. 12. The centers of three circles A, B, and C are in the same right line, B and C touch A internally, and each other externally; P, Q, being the points where A is touched by B, C respectively; to find a point Ron A such that the portion of the lune PR intercepted between B and A may be equal to the portion of QR between C and A. 13. On the chord of a quadrant a semicircle is described; required the area of the crescent thus formed. 14. Semicircles are described upon the radii CA, CB of a quadrant, and intersect each other in a point D, shew that the area common to both semicircles is equal to the area without them, and that the remaining areas of the two semicircles are equal, each onefourth of the square on AC. 15. If on one of the radii of a quadrant a semicircle be described; and on the other, another semicircle so described as to touch the former and the quadrantal arc; compare the area of the quadrant with the area of the circle described in the figure bounded by the three curves. 16. Any right-angled triangle BAC is inscribed in a semicircle, A being the right angle, and AD a perpendicular on the base BC. If circles be described on the sides BA, AC as diameters, prove that the areas of these circles will always be to each other in the same ratio as the segments into which the base is divided by the line AD. 17. If on the two sides of a right-angled triangle, semicircles be described, and a circle be described touching them both, it will include the circle whose diameter is the hypotenuse; and the space between the two circles will be to the outer circle as twice the rectangle of the sides of the triangle to the square on the sum of the sides. II. 18. In different circles the radii which bound equal sectors contain angles reciprocally proportional to their circles. 19. Prove that the sectors of two different circles are equal, when their angles are inversely as the squares on the radii. 20. If the arc of a semicircle be trisected, and from the points of section lines be drawn to either extremity of the diameter, the difference of the two segments thus made will be equal to the sector which stands on either of the arcs. 21. If AB be a circular arc, center O, and AD be drawn perpendicular to BO, and the arc AC taken equal to AD, then the sector BOC equals the segment ACB. 22. If two points B, D, be taken at equal distances from the ends of the arc of a quadrant, and perpendiculars BG, DH be drawn to the extreme radius; the space BGHD shall be equal to the sector BOD. 23. If circles be inscribed in the triangles formed by drawing the altitude of a triangle right-angled at the vertex, the circles and the triangles are proportional. 24. If a semicircle be described on the hypotenuse AB of a rightangled triangle ABC, and from the center E, the radius ED be drawn at right angles to AB, shew that the difference of the segments on the two sides equals twice the sector CED. 25. If semicircles be described upon the sides of a right-angled triangle on the interior, the difference between the sum of the circular segments thus standing upon the exterior of the sides and segments of the base, equals the space intercepted by the circumferences described on the sides. 26. AB, BD are two radii of a circle at right angles to each other. Produce BD to C, and make BC equal to the arc AD. Join AC cutting the circumference in E. Then the area EDC is equal to the area of the segment AE. 27. ABC is an isosceles right-angled triangle. On BC is described a semicircle BDEC, and BFC is a circle whose radius is AB and center A. The segment BCF is equal to the segments BAD, ACE 28. The circle inscribed in a square is equal to four equal circles touching one another and the sides of that square internally. 29. If the diagonals of a quadrilateral inscribed in a circle cut each other at right angles, and circles be described on the sides; prove that the sum of two opposite circles will be equal to the sum of the other two. 30. If two chords of a circle intersect each other either within or without the circle at right angles; and if on these segments as diameters, circles be described, the areas of these four circles are together equal to that of the original circle. 31. Shew that the semicircles described on the diagonals of a right-angled parallelogram together equal the sum of the semicircles described on the sides. 32. A quadrilateral inscribed in a circle has a diameter passing through the center; or has its two diameters at right angles to one another; on the sides of the quadrilateral semicircles are described; the four crescents outside are together equal to the quadrilateral. III. 33. Equal straight lines whose extremities are in the surface of a sphere, are equally distant from the center of the sphere. 34. The angle between the planes of two great circles of a sphere, is measured by the arc of a great circle which joins their poles. 35. Every section of a sphere made by a plane is a circle: and if two parallel planes cut a sphere so that the sections are equal, they are equidistant from the center. 36. A straight line or a plane can only touch a sphere at one point; and at that point the radius of the sphere will be perpendicular to the line or plane. 37. Shew that all lines drawn from an external point to touch a sphere are equal to one another; and thence prove that if a tetrahedron can have a sphere inscribed in it touching its six edges, the sum of every two opposite edges is the same. 38. If two equal circles cut one another in the diameter, and a plane cut them perpendicularly to the same diameter, the points of section of this plane with the circumferences, are in a circle. 39. If three straight lines intersect each other within a sphere at right angles, each at right angles to the plane of the other two; the sum of the squares on the six segments is equal to the square on the diameter of the sphere, together with twice the rectangle of the segments of the diameter made at the point of intersection. 40. Having given an irregular fragment, which contains a portion of spherical surface; shew how the radius of the sphere, to which the fragment belongs, may be practically determined. IV. 41. All the sections of a tetrahedron made by planes parallel to the base are similar to the base. 42. Find the inclination of two contiguous faces of a tetrahedron to each other. 43. If on the base of a regular tetrahedron lines be drawn from any two angles to bisect the opposite sides; the line joining their point of section with the vertex of the solid is at right angles with the base. 44. ABCD is a regular tetrahedron; from the vertex A, a perpendicular is drawn to the plane BCD meeting it in O. Shew that three times the square on 40 is equal to twice the square on AB. 45. If the shortest distances between opposite edges of a tetrahedron be mutually at right angles, they will bisect the edges. 46. Prove that the shortest distance between two opposite edges of a regular tetrahedron is equal to half the diagonal of the square described on an edge. 47. If in a tetrahedron the shortest distances between the opposite edges are mutually at right angles, prove that these distances meet in a point, that they bisect each other, and that the opposite edges of the tetrahedron are equal. 48. If the line joining the bisections of two edges of a tetrahedron which do not meet be bisected, the point so found is distant from the base one-fourth of the perpendicular altitude of the tetrahedron. 49. If the angles of a regular tetrahedron be joined to the centers of the circles inscribed in its faces, the joining lines will form the edges of a new tetrahedron parallel to those of the old. 50. The perpendicular drawn from any angle of a regular tetrahedron upon the opposite face, will meet that face in the centre of the circle which circumscribes that face. V. 51. The angles of inclination of the faces of a regular tetrahedron and of a regular octahedron are supplementary to each other. 52. Given the side of a regular octahedron, find the radius of the inscribed and circumscribed spheres. 53. Draw three diameters of a sphere each at right angles to the other two; then the six points where the extremities of the diameters meet the surface of the sphere, will be the angles of a regular octahedron, and the lines joining the adjacent points will be the edges, also the three diameters of the sphere its diagonals. HINTS, &c. 8. This is a particular case of Euc. 1. 22. The triangle however may be described by means of Euc. 1. 1. Let AB be the given base, produce AB both ways to meet the circles in D, E (fig. Euc. 1. L.); with center A, and radius AE, describe a circle, and with center B and radius BD, describe another circle cutting the former in G. Join GA, GB. 9. Apply Euc. 1. 6, 8. 10. This is proved by Euc. 1. 32, 13, 5. 11. Let fall also a perpendicular from the vertex on the base. 12. Apply Euc. 1. 4. 13. Let CAB be the triangle (fig. Euc. 1. 10.) CD the line bisecting the angle ACD and the base AB. Produce CD, and make DE equal to CD, and join AE. Then CB may be proved equal to AE, also AE to AC. 14. Let AB be the given line, and C, D the given points. From C draw CE perpendicular to AB, and produce it making EF equal to CE, join FD, and produce it to meet the given line in G, which will be the point required. 15. Make the construction as the enunciation directs, then by Euc. 1. 4, BH is proved equal to CK: and by Euc. 1. 13, 6, OB is shewn to be equal to OC. 16. Let C, D be the two given points one on each side of the given line AB, such that lines CE, DE drawn to any point E in the line are always equal. Join CD cutting AB in F, then FC is also equal to FD. Then by Euc. 1. 8. 17. The angle BCD may be shewn to be equal to the sum of the angles ABC, ADC. 18. The angles ADE, AED may be each proved to be equal to the complements of the angles at the base of the triangle. 19. The angles CAB, CBA, being equal, the angles CAD, CBE are equal, Euc. 1. 13. Then, by Euc. 1. 4, CD is proved to be equal to CE. And by Euc. 1. 5, 32, the angle at the vertex is shewn to be four times either of the angles at the base. 20. Let AB, CD be two straight lines intersecting each other in E, and let P be the given point, within the angle AED. Draw EF bisecting the angle AED, and through P draw PGH parallel to EF, and cutting ED, EB in G, H. Then EG is equal to EH. And by bisecting the angle DEB and drawing through P a line parallel to this line, another solution is obtained. It will be found that the two lines are at right angles to each other. 21. Let the two given straight lines meet in A, and let P be the given point. Let PQR be the line required, meeting the lines AQ, AR in Q and R, so that PQ is equal to QR. Through P draw PS parallel to AR and join RS. Then APSR is a parallelogram and AS, PR the diagonals. Hence the construction. 22. Let the two straight lines AB, AC meet in A. In AB take any point D, and from AC cut off AE equal to AD, and join DE. On DĚ, or DE produced, take DF equal to the given line, and through F draw FG parallel to AB meeting AC in G, and through G draw GH parallel to DE meeting AB in H. Then GH is the line required. |