34. To divide a circle into any number of concentric equal annuli. 35. To inscribe a square in a given semicircle.

36. Prove that if, on one side of an equilateral triangle, as a diameter, a semicircle be described, and from the opposite angle two straight lines be drawn to trisect that side, these lines produced will trisect the semi-circumference.

37. Draw straight lines across the angles of a given square, so as to form an equilateral and equiangular octagon.

38. Prove that the square of the side of an equilateral triangle, inscribed in a circle, is equal to three times the square of the radius.

39. To draw straight lines from the extremities of a chord to a point in the circumference of the circle, so that their sum shall be equal to a given line. N.B. The given line must evidently be limited.

40. In a given triangle to inscribe a rectangle of a given area. 41. Given the perimeter of a right-angled triangle, and the perpendicular from the right angle upon the hypothenuse, to construct the triangle.

42. In an isosceles triangle to inscribe three circles touching each other, and each touching two of the three sides of the triangle.

43. To construct a trapezoid when four sides are given.

44. The same when three sides and the sum of the angles at the base are given.

45. When the two sides not paralel, the altitude and an angle are given.

46. When the difference of the parallel sides, the diagonal, a third side, and an angle.

47. Prove that the line drawn to the middle of the hypothenuse from the vertex of the right angle in a right-angled triangle is equal to half the hypothenuse.

48. Prove that if the four angles of a parallelogram be bisected, and the points in which two of the bisecting lines adjacent one side meet be joined with that in which the two bisecting lines adjacent the opposite side meet; 10, that the joining line will be parallel to the other two sides; 2°, that it will be equal to the difference of two adjacent sides.

49. That in any quadrilateral the lines joining the middle points of the opposite sides, and the line joining the middle points of the diagonals, meet in the same point, and all three bisect one another.

50. That the rectangle of the two sides of any triangle is equal to the rectangle of the perpendicular upon the third side from the vertex opposite, and the diameter of the circle circumscribing the triangle.

51. Prove that the rectangle of the diagonals of an inscribed quadrilateral is equal to the sum of the rectangles of the opposite sides.

52. Prove that the square of the line bisecting the vertical angle of a triangle, together with the rectangle of the two segments of the base, is equal to the rectangle of the other two sides of the triangle. 53. To find two lines that shall have the same ratio as two given rectangles.

54. Draw a transverse line to two circles such that the parts comprehended within the circumferences shall be equal to a given line. 55. To inscribe in a circle (radius not given) a triangle of given base, vertical angle, and altitude.

56. In a given circle to place six others, so that each shall touch two others, and the given.

57. To construct a figure similar to two given similar figures, and equal to their sum or difference.

APPENDIX I.

ISOPERIMETRY.

Def. 1. A maximum is the greatest quantity among those of the same kind; a minimum the least.*

Def. 2. Isoperimetrical figures are those which have equal perim

eters.

THEOREM I.

Among all triangles of the same perimeter, that is, a MAXIMUM in which the undetermined sides are equal.

Let ABC, ABD be the two triangles. With C as center, and radius CA, describe a circle cutting AC produced in F; ABF, inscribed in a semicircle, will be a right angle; produce FB, making DE=DB, and draw the perpendiculars DH, CG. It will be easy to show of A the oblique lines that AE < AF ... BE < BF ... BH <BG; these last two being the altitudes of the given triangles which have a common base.

F

B

D

H

E

THEOREM II.

Of all isoperimetric polygons the maximum has its sides equal.

By drawing a diagonal in the polygon so as to cut off two sides forming a triangle; if these two sides be not equal, they may be replaced by two others which are equal, and which, by the last theorem, will inclose a greater triangle. This process may be repeated with all the sides of the polygon.

THEOREM III.

Of all triangles formed with two given sides, that is, a maximum in which the two given sides make a right angle.

For with the same base it will have the greater altitude.

*These definitions are suited to our present purpose.

THEOREM IV.

Of all polygons formed with sides all given except one side, that is, a maximum, of which the given sides are inscribed in a semicircle, of which the side not given is the diameter.

Let ABCDEF be the maximum polygon formed with sides all given except AB. Join AD, BD; then ADB must be a right F angle: otherwise, preserving the parts BCD and ADEF the same, A

E

D

C

B

the triangle ADB might be increased (th. 3); the point D must, therefore, be in the semicircumference described on AB as a diameter. In the same manner it may be proved that the points E, F, C, &c., must be in the semicircumference. Q. E. D.

Schol. There is but one way of forming the polygon; for if, after having found a circle which satisfies the requisition, a larger circle be supposed, the chords which are the sides of the polygon correspond to smaller angles at the center, and the sum of these will be less than two right angles.

THEOREM V.

Of all polygons formed with given sides, that is, a maximum which can be inscribed in a circle.

E

F

G

Let ABCEFG be an inscribed polygon, abcefg one which is not capable of being inscribed, and of equal sides with the former; draw the diameter EM; join AM, MB; upon ab make the triangle abm ABM, and join em. By th. 4, EFGAM> efgam and ECBM>ecbm by addition, EFGAMBC minus AMB > efgambc minus amb. Q.

...

E. D.

=

Corollary from the two last propositions. The regular polygon is the maximum among all isoperimetric polygons of the same number of sides.