A square whose side is twice the length of another, and a circle whose diameter is twice that of another will each have an area four times that of the original. And in the case of solids: A ball of twice the diameter will weigh eight times as much as the original, and a ball of three times the diameter will weigh twenty-seven times as much as the original. In attempting to calculate the side of a cube which shall have twice the volume of a given cube, we meet the old difficulty of incommensurability, and the solution cannot be effected geometrically, as it requires the construction of two mean proportionals between two given lines. III THE TRISECTION OF AN ANGLE This problem is not so generally known as that of squaring the circle, and consequently it has not received so much attention from amateur mathematicians, though even within little more than a year a small book, in which an attempted solution is given, has been published. When it is first presented to an uneducated reader, whose mind has a mathematical turn, and especially to a skilful mechanic, who has not studied theoretical geometry, it is apt to create a smile, because at first sight most persons are impressed with an idea of its simplicity, and the ease with which it may be solved. And this is true, even of many persons who have had a fair general education. Those who have studied only what is known as "practical geometry" think at once of the ease and accuracy with which a right angle, for example, may be divided into three equal parts. Thus taking the right angle ACB, Fig. 4, which may be set off more easily and accurately than any other angle except, perhaps, that of 60 deg., and knowing that it contains 90 deg., describe an arc ADEB, with C for the center and any convenient radius. Now every school-boy who has played with a pair of compasses knows that the radius of a circle will "step" round the circumference exactly six times; it will therefore divide the 360 deg. into six equal parts of 60 deg. each. This being the case, with the radius CB, and B for a center, describe a short arc crossing the arc ADEB in D, and join CD. The angle DCB will be 60 deg., and as the angle ACB is 90 deg., the angle ACD must be 30 deg., or one-third part of the whole. In the same way lay off the angle ACE of 60 deg., and ECB must be 30 deg., and the remainder DCE must also be 30 deg.. The angle ACB is therefore easily divided into three equal parts, or in other words, it is trisected. And with a slight modification of the method, the same may be done with an angle of 45 deg., and with some others. These however are only special cases, and the very essence of a geometrical solution of any problem is that it shall be applicable to _all_ cases so that we require a method by which _any_ angle may be divided into three equal parts by a pure Euclidean construction. The ablest mathematicians declare that the problem cannot be solved by such means, and De Morgan gives the